Here we are giving the details on how to calculate the greatest common factor of two polynomials using the long division method. Students must have a look at the complete procedure to find the H.C.F of Polynomials by Long Division Method and few solved examples on it. We are using this long division method especially when there is no scope to find the highest common factor of polynomials by factorization.

How to find H.C.F of Polynomials by Long Division?

The step by step procedure to evaluate the G.C.F of two polynomials by long division method manually is provided. For the sake of your comfort and convenience, the steps are listed below make use of them.

• Arrange the given two polynomials in the descending order of powers of any of its variables.
• If there is any common factor in the polynomials, then separate it. Multiply the common factors to get the H.C.F of them.
• Just like the determination of G.C.F by the method of division in arithmetic, here also division is not complete. In every step, the division of that step is divided by the obtained remainder. At any step, if any common factor is available in the remainder then it should be taken out, then perform division.
• In every step, the term in the quotient should be found by comparing the first term of the dividend with the first term of the divisor. Sometimes, if necessary, the dividend may be multiplied by a multiplier of a factor.

Highest Common Factor of Polynomials by Long Division Method

Example 1.

Find the H.C.F of 6x³ – 17x² – 5x + 6, 6x³ – 5x² – 3x + 2 and 3x³ – 7x² + 4 by using the long division method.

Solution:

Given three polynomials are 6x³ – 17x² – 5x + 6, 6x³ – 5x² – 3x + 2 and 3x³ – 7x² + 4

All the polynomials are arranged in the descending order of the powers of the variable ‘x’. And the polynomials have no common factors between them. So, by the long division method

The H.C.F. of 6x³ – 17x² – 5x + 6, 6x³ – 5x² – 3x + 2 is 6x² + x – 2.

Now, it is to be seen whether the third expression 3x³ – 7x² + 4 is divisible by 6x² + x – 2 or not. If it is not, then the H.C.F. of them is to be determined by the division method.

Therefore, the H.C.F. of 6x³ – 17x² – 5x + 6, 6x³ – 5x² – 3x + 2 and 3x³ – 7x² + 4 is 3x + 2.

Example 2.

Calculate the Highest Common Factor of polynomials 7a³ – 28a² + 49a – 42, 5a³ – 25a² + 50a – 40 by using the long division method?

Solution:

Given two polynomials are 7a³ – 28a² + 49a – 42, 5a³ – 25a² + 50a – 40

Consider f(a) = 7a³ – 28a² + 49a – 42, g(a) = 5a³ – 25a² + 50a – 40

Take out the common factors,

f(a) = 7(a³ – 4a² + 7a – 6)

g(a) = 5(a³ – 5a² + 10a – 8)

At the time of writing the final result the H.C.F. of 7 and 5 i.e. 35 is to be multiplied with the divisor of the last step.

Therefore, the H.C.F of 7a³ – 28a² + 49a – 42, 5a³ – 25a² + 50a – 40 is 35 * 2(a – 2) = 70(a – 2).

Example 3.

Find the G.C.F of x⁶ + 2x⁵ + 5x³ + 4x² + 6 and x³ + 2 by using the long division method?

Solution:

Given two polynomials are x⁶ + 2x⁵ + 5x³ + 4x² + 6 and x³ + 2.

By arranging the two polynomials in the descending order of powers of x we get,

x⁶ + 2x⁵ + 0x⁴ + 5x³ + 4x² + 0x + 6 and x³ + 0x² + 0x + 2

There are no common factors in those polynomials. So, dive them by using the long division method

Therefore, the H.C.F. of x⁶ + 2x⁵ + 5x³ + 4x² + 6 and x³ + 2 is x³ + 2x² + 3.

The post H.C.F. of Polynomials by Long Division Method | G.C.F of Two Polynomials by Long Division appeared first on Learn CBSE.

Learn about the relation between the highest common factor and least common factor of two polynomials on this page. First of all, know the definitions of H.C.F, L.C.M, and the relationship between them. Check the solved example questions in the below sections. We have listed the step-by-step solutions for all the example problems provided making it easy for you to understand the concepts.

H.C.F and L.C.M Definition

The highest common factor (H.C.F) is defined as the largest term that divides evenly into all the terms from a group.

Example: H.C.F of 12 and 15 is 3. Because multiples of 12 are 2, 2, 3, and multiples of 15 are 5, 3.

The Lowest Common Multiple (L.C.M) is defined as the smallest term that is a multiple of all numbers from a group.

Example: L.C.M of 12, 15 is 60. Because multiples of 12 are 2, 2, 3, multiples of 15 are 3, 5. And the L.C.M is the multiplication of the smallest common number and remaining numbers from the group.

Relation Between H.C.F. and L.C.M. of Two Polynomials

The relation between L.C.M and H.C.F of polynomials is the product of polynomials is equal to the product of its H.C.F and L.C.M. This relationship can be expressed as follows.

p(x) * q(x) = {L.C.M of p(x) and q(x)} * {H.C.F of p(x) and q(x)]}.

So, find the highest common factor, least common factor of polynomials by using the factorization method. And multiply them to find the product of two polynomial expressions. Using this formula, you can easily find one polynomial when L.C.M, H.C.F of both polynomials, and other polynomial is given.

Solved Examples on Relationship Between H.C.F. and L.C.M. of Two Polynomials

Example 1.

Find H.C.F and L.C.M of two polynomials 2x² – x – 1 and 4x² + 8x + 3 and prove that the product of polynomials is the product of their L.C.M and H.C.F?

Solution:

Given two polynomials are 2x² – x – 1 and 4x² + 8x + 3.

By factoring 2x² – x – 1, we get

= 2x² – 2x + x -1

= 2x(x – 1) + 1(x – 1)

= (x – 1) ( 2x + 1)

By factoring 4x² + 8x + 3, we get

= 4x² + 6x + 2x + 3

= 2x (2x + 3) + 1(2x + 3)

= (2x + 3) (2x + 1)

The common factor in both polynomials is (2x + 1).

Therefore, H.C.F = common factor = (2x + 1)

L.C.M = common factor * remaining factors

= (2x + 1) * (2x + 3) (x – 1) = (2x + 1) (2x + 3) (x – 1)

The relation between H.C.F. and L.C.M. of Two Polynomials is Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

(2x² – x – 1) (4x² + 8x + 3) = (2x + 1) (2x + 3) (x – 1) (2x + 1)

(x – 1) ( 2x + 1) (2x + 3) (2x + 1) = (2x + 1)²(2x + 3) (x – 1)

Hence proved.

Example 2.

Find the G.C.F of polynomials x³ + y³ and x⁴ + x²y² + y⁴ whose L.C.M is (x³ + y³) (x² + xy + y²)?

Solution:

Given two polynomials are x³ + y³ and x⁴ + x²y² + y⁴

L.C.M of polynomials is (x³ + y³) (x² + xy + y²)

By factoring x⁴ + x²y² + y⁴, we get

= (x² + y²)² – (xy)²

= (x² + xy + y²)² (x² – xy + y²)²

Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

(x³ + y³) (x⁴ + x²y² + y⁴) = (x³ + y³) (x² + xy + y²) * H.C.F

Cancel the common term (x³ + y³)

(x⁴ + x²y² + y⁴) = (x² + xy + y²) * H.C.F

(x² + xy + y²)² (x² – xy + y²)² = (x² + xy + y²) * H.C.F

Therefore, H.C.F = (x² – xy + y²)²

Example 3.

L.C.M and G.C.F of two polynomials are x³ – 10x² + 11x + 70 and x – 7. And one of the polynomial is x² – 12x + 35, find the other polynomial?

Solution:

Given that,

L.C.M, H.C.F of two polynomials is x³ – 10x² + 11x + 70, x – 7

First polynomial = x² – 12x + 35

Product of Two Polynomials = Product of Polynomials L.C.M and H.C.F

x² – 12x + 35 * Second polynomial = (x³ – 10x² + 11x + 70) * (x – 7)

Second polynomial = (x³ – 10x² + 11x + 70) * (x – 7) / (x² – 12x + 35)

So, second polynomial = (x + 2) (x – 7)

= x² – 7x + 2x -14 = x² -5x – 14.

The post Relation Between HCF and LCM of Two Polynomials appeared first on Learn CBSE.

Algebraic fractions are nothing but fractions where its numerator and denominator are algebraic polynomial expressions. These fractions are subject to the laws of arithmetic expressions. Reducing algebraic fractions to their lowest term means the fraction has no common factor other than 1. Get the definition, step by step process to reduce any type of algebraic fractions to its lowest term easily, few example questions and answers in the following sections.

Reduce Algebraic Fractions to its Lowest Term

Reducing algebraic fractions to its lowest term and simplifaction of algebraic fractions are one and the same. Here, we are changing the numerator, denominator of the given fraction so that it should not have a common factor between the numerator and denominator. Actually, the reduced form of the fraction is always equal to the original fraction. While we reduce an algebraic fraction to its lowest term we need to remember one important thing i.e if the numerator and denominator of the fractions are multiplied or divided by the same quantity, then the fraction value remains unchanged.

How to Reduce Algebraic Fractions to Lowest Terms?

Go through the below-mentioned steps to get help with reducing a fraction to its lowest term manually. You will find the procedure quite easy with the detailed explanation provided. Remember that the simplified algebraic fraction numerator, denominator H.C.F is always 1.

• Let us take any polynomial algebraic fraction having numerator and denominator.
• Find the factors of numerator and denominator separately.
• Cancel the common factors in both numerator and denominator.
• Reduce the algebraic fraction to the lowest term.

Solved Examples on Simplifying Algebraic Fractions

Example 1.

Simplify (x² + 5x) / (x² – 5).

Solution:

Given algebraic fraction is (x² + 5x) / (x² – 5).

We see that the numerator and denominator of the given algebraic fraction is polynomial, which can be factorized.

= [x (x + 5)] / [(x – 5) (x + 5)]

Cancel the common factor (x + 5) in the numerator, denominator.

= x / (x – 5)

Example 2.

Reduce the algebraic fraction (x² + 15x + 56) / (x² + 5x – 24) to its lowest term.

Solution:

Given algebraic fraction is (x² + 15x + 56) / (x² + 5x – 24)

Both numerator and denominator of the fraction are polynomials, which can be factorized.

= (x² + 8x + 7x + 56) / (x² + 8x – 3x – 24)

= (x(x + 8) + 7 (x + 8)) / (x(x + 8) – 3 (x + 8))

= ((x + 8) (x + 7)) / ((x + 8) (x – 3))

Cancel the common term (x + 8) in the numerator, denominator.

= (x + 7) / (x – 3)

(x² + 15x + 56) / (x² + 5x – 24) = (x + 7) / (x – 3).

Example 3.

Reduce the algebraic fraction (2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1) to the lowest term.

Solution:

Given algebraic fraction is (2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1)

Both numerator and denominator are polynomials, factorize them.

= (2x³ (x² – x – 2)) / ((x²)² – 1²)

= (2x³ (x² – 2x + x – 2)) / ((x² – 1) (x² + 1))

= (2x³ (x (x – 2) + 1 (x – 2)) / ((x² – 1²) (x² + 1))

= (2x³ (x + 1) (x – 2)) / ((x – 1) (x + 1) (x² + 1))

cancel the common term (x + 1) in the numerator, denominator.

= (2x³ (x – 2)) / ((x – 1) (x² + 1))

(2x⁵ – 2x⁴ – 4x³) / (x⁴ – 1) = (2x³ (x – 2)) / ((x – 1) (x² + 1))

The post Reduce Algebraic Fractions to its Lowest Term | Simplifying Algebraic Fractions appeared first on Learn CBSE.

Simple Interest is an easy method of finding Interest Over a Loan/ Principal Amount. It is necessary to recall the concept of Interest and Ways to Calculate it. The Concept of Simple Interest is Applicable in Many Sectors such as Automobile, Finance, Banking, etc. Whenever you borrow money from a bank or someone you need to repay along with an extra amount. Get to know the Simple Interest Definition, Formula, and How to find the Simple Interest in the later modules.

What is Simple Interest?

Simple Interest is the method of finding the Interest Over a Certain Amount. When you borrow from a bank you need to return back some extra money while paying. While returning back the amount it depends on the rate of interest as well as the time for which you borrow.

How to find the Simple Interest?

Formula to Calculate the Simple Interest will give you the Interest Amount if Principal, Rate of Interest, Time Duration are Given.

SI = (P*T*R)/100

Where, P is the Principal Amount

T is the Time Duration

R is the Rate of Interest

Amount(A) = Principal(P)+Interest(I)

The formula above is to calculate the Simple Interest on a yearly basis. To Calculate the Simple Interest when time duration is months the formula is as such

Simple Interest in Months = (P × n × R)/ (12 ×100)

Solved Examples of Calculating Simple Interest

1. Calculate the Simple Interest if the principal amount is Rs. 4000, the time period is 2 years and the rate is  5%. Also, calculate the total amount at the end of 2 years?

Solution:

Principal = Rs. 4000

Time Period= 2 Years

Rate of Interest = 5%

Simple Interest = (P*T*R)/100

= (4000*2*5)/100

= 400

Amount to be returned = Principal + Interest

= 4000+400

= 4400

2. Manoj Pays Rs. 8000 as an amount on the Sum of Rs. 5000 that he had borrowed for 2 years. Find the rate of interest?

Solution:

From the given data

Amount = Rs 8000

Principal = Rs 5000

SI = A – P = 8000 – 5000 = Rs 3000

T = 2 years

We need to find the Value of R

SI = (P × R ×T) / 100

R = (SI × 100) /(P× T)

R = (3000 × 100 /5000 × 2) =

Thus, R = 30%

Manoj Pays Rs. 8000 as an amount on the Sum of Rs. 5000 at a rate of 30% for 2 Years.

The post Calculate Simple Interest | How to Find Simple Interest? appeared first on Learn CBSE.

Are you looking eagerly everywhere to find a Practice Test on Simple Interest with Solutions? You have come the right way we have curated the Solved Examples explaining how to calculate Simple Interest. We have listed the Formula to Calculate Simple Interest in the coming modules. Check out the Step by Step Solution provided for the Problems finding the Related Terms like Interest, Time Duration, Principal, etc. Assess your preparation standard using the Simple Interest Practice Test available and improve on the areas you are lagging.

Simple Interest Formula Questions | Word Problems on Simple Interest

1. Find the Principal when

(a) S.I. = 100 Rate = 5% per annum Time = 2 years

(b) S.I = 1200 Rate = 2% per annum Time = 10 Months

Solution:

(a) Formula to Calculate the Simple Interest SI = PTR/100

Rearranging the basic formula we get P = SI*100/T*R

= 100*100/2*5

= 1000

Prinicipal = Rs. 1000

(b) S.I = 1200 Rate = 2% per annum Time = 10 Months

Simple Interest in Months = (P × n × R)/ (12 ×100)

Rearranging we have the P as under

1200 = (P*10*2)/(12*100)

(1200*1200) = 20P

P = 1440000/20

= Rs. 72, 000

2. Shubash deposited Rs. 10000 at simple interest for 2 years. Had the interest been 2%, how much amount he would have got?

Solution:

P = 10000

R = 2%

T = 2 Years

Simple Interest = PTR/100

= 10000*2*2/100

= 400

Amount = Principal + Interest

= 10000+400

= 10400

Shubash gains an amount of 10400 after 2 years.

3. A sum fetched a total simple interest of Rs. 4000 at the rate of 10 %.p.a. in 5 years. What is the sum?

Solution:

Formula for SI = PTR/100

Rearranging we have P = SI*100/(TR)

= 4000*100/(10*5)

= 40000/5

= 8000

Sum taken is Rs. 8000

4. At what rate percent annum will a sum of money double itself in 5 years?

Solution:

We know A = P(1+TR/100)

From given data Amount doubles in 5 years

2P = P(1+R*5/100)

2-1= 5R/100

100 = 5R

R = 20%

Sum of the amount doubles in 5 years at a rate of 20%.

5. What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 5 years and that for 10 years?

Solution:

Let the Principal in both the cases be P and Rate of Interest be R %

From given data

Ratio of Simple Interest = (P*5*R)/100 : (P*10*R)/100

= 5:10

= 1:2

6. The simple interest on a certain sum for 6 months at 8% per annum is Rs. 150 less than the simple interest on the same sum for 15 months at 5% per annum. The sum is?

Solution:

Consider the sum to be x

From the given info we have the equation as under

(x*5*15/(100*12)) – (x*8*6/(100*12)) = 150

(75x-48x)/1200 = 150

27x = 1200*150

x = (1200*150)/27

= Rs. 6666.667

Therefore, the Sum is Rs. 6666.667

7. A sum of money at simple interest amounts to Rs. 875 in 3 years and to Rs. 900 in 4 years. The sum is?

Solution:

From given data Simple Interest per 1 Year = 900 – 875

= Rs. 25

SI for 3 Years = (25*3) = 75

Principal = Amount – Interest

= 875 – 75

= Rs. 800

Therefore the Sum is Rs. 800

The post Practice Test on Simple Interest | Simple Interest Questions and Answers appeared first on Learn CBSE.

Compound Interest, in general, is the Interest Calculated on a Principal and the Interest Accumulated over the Previous Period. It is not similar to the Simple Interest where Interest is not added. Get to know the Compound Interest Formula, Procedure on How to find the Compound Interest on a Daily, Monthly, Quarterly, Yearly Basis. Check out the Solved Examples explained step by step for a better understanding of the concept.

Compound Interest Definition

In Simple Words, Compound Interest is nothing but the Interest that adds back to the Principal Sum so that Interest will be earned during the next compounding period.

Formula to Calculate Compound Interest

Compound Interest Formula is given by Compound Interest = Amount – Principal

Amount A = P(1+r/n)^nt

where,

A= Amount

P= Principal

R= Rate of Interest

n = number of times the interest is compounded per year.

Compound Interest When Interest is Compounded Annually

The Amount Formula mentioned above is the general formula for the number of times the principal is compounded in a year. If the Amount is Compounded Yearly or Annually then Amount Formula is given by

A= P(1+R/100)^t

Compound Interest (CI) when Interest is Compounded Half-Yearly

In this case, if  R is the Rate of Interest Per Annum then it is clearly R/2 per half-year.
A = P(1 + (R/2)/100)^2*1

R/2 = R^‘

CI = A – P

= P(1 + (R/2)/100)^2*1 – P

In the Cases, When the Rate is compounded Half-Yearly, we divide the rate by 2 and multiply the time by 2 before using the general formula for the amount in case of the compound interest.

Compound Interest when Interest is Compounded Quarterly

Let us consider the Compound Interest on a Principal P kept for 1 Year and Interest Rate is R %. Since the Interest is Compounded Quarterly Principal Amount will be changed after 3 months. Interest for the Next Three Months will be calculated on the Amount after 3 first months.

In the same way, Interest for Third Quarter will be calculated on the amount left after the first 6 months. Last Quarter will be calculated on the amount remaining after the first 9 months.

A = P(1+ (R/4)/100)^4T

CI = A – P

= P(1+ (R/4)/100)^4T – P

Solved Example Questions on Compound Interest

1. A town had 15,000 residents in 2000. Its population declines at a rate of 10% per annum. What will be its total population in 2004?

Solution:

The population of a town decreases by 10% every year. Thus, the population of a town next year is calculated on the current year. For Decrease, we have the formula

A = P(1-R/100)ⁿ

= 15000(1-10/100)⁴

= 15000(0.9)⁴
= 9841

The population of the town in 2004 is 9841.

2. The price of a radio is Rs 2000 and it depreciates by 5% per month. Find its value after 4 months?

Solution:

For depreciation, the Amount is A = P(1-R/100)ⁿ

= 2000(1-5/100)⁴

= 2000(1-0.05)⁴

=2000(0.95)⁴
= 1629

Price of radio is Rs. 1629 after 4 months.

3. Calculate the compound interest (CI) on Rs. 10000 for 2 years at 5% per annum compounded annually?

Solution:

We know the formula for Compound Interest Annually is

A= P(1+R/100)^t

= 10000(1+5/100)²

= 10000(105/100)²

= 10000(1.1025)

= 11025

CI = A – P

= 11025 – 10000

= Rs. 1025

4. Calculate the compound interest to be paid on a loan of Rs. 5000 for 3/2 years at 10% per annum compounded half-yearly?

Solution:

Rate of Interest when compounded half-yearly we need to divide R by 2 and multiply Time with 2.

A = P(1+R/100)ⁿ

= P(1+R/2*100)^2*n

= 5000(1+10/2*100)^2*3/2

= 5000(1+5/100)³

= 5000(105/100)³

= 5000(1.157)

= 5788

CI = A – P

= 5788 – 5000

= 788

The Compounded Interest to be paid on a loan is Rs.788

The post Compound Interest – Definition, Formula, Solved Examples appeared first on Learn CBSE.

Let us learn in detail how to calculate compound interest with the growing principal in this article. If the Interest is due at the end of a certain period be it after a year, half-year, quarterly duration, etc. to a moneylender then Interest is added to the sum borrowed. Thus, Amount becomes the Principal for the next period of borrowing. The Procedure continues further till the specific time is found.

Refer to the Solved Examples on finding Compound Interest with Growing Principal and understand the concept better. We have provided step by step solutions for all the Compound Interest with Increasing Principal Problems.

Solved Examples on Compound Interest with Growing Principal

1. A man takes a loan of $ 20,000 at a compound interest rate of 5% per annum.

(i) Find the amount after 1 year?

(ii) Find the compound interest for 2 years?

(iii) Find the amount required to clear the debt at the end of 2 years?

Solution:

(i) Given P = $20, 000, R = 5%, T = 1 year

We know the Formula for Amount A = P(1+R/100)ⁿ

Substitute the given input data in the formula to obtain the amount

A = 20000(1+5/100)¹

= 20000(1+0.05)

= 20000(1.05)

= 21000

Amount to be paid after 1 year is $21, 000.

SI = A – P

= 21, 000 – 20, 000

= $1, 000

(ii) For Second Year New Principal = $21,000

Thus, Interest in Second Year = 5% of 21, 000

= 5/100*21, 000

= $1050

Compound Interest for 2 Years = Interest of 1st Year + Interest of 2nd Year

= $1000+$1050

= $2050

(iii) Amount required to clear the debt at the end of 2 years = Principal + Compound Interest for 2 Years

= $20, 000 + $2050

= $22, 050

2. At 8% per annum, find the compound interest for 2 years on a certain sum of money?

Solution:

Let the sum be x

Interest for 1st year = 8% of x

= 8x/100

Amount after 1 year = Principal + Interest after 1 year

= x+8x/100

= 108x/100

For second-year New Principal is 108x/100

Interest for 2nd Year = 8% of 108x/100

= 8/100*108x/100

=864x /10000

Amount after 2 years = Principal + Interest after 2 Years

= x +864x/100

= 964x/100

CI = Interest for 1st year + Interest for 2nd Year

=8x/100 + 864x /10000

= 8x/100+54x/625

= (200x+216x)/2500

= 416x/2500

= 104x/625

3. Calculate the compound interest to be paid on a loan of Rs. 5000 for 5/2 years at 10% per annum compounded half-yearly?

Solution:

From the given data Principal = Rs. 5000

T = 5/2

R = 10%

A = P(1+R/100)ⁿ

When Interest is Compounded Half Yearly we need to multiply T with 2 and divide R with 2 thus equation becomes as such

A = 5000(1+10/2*100)^2*5/2

= 5000(1+5/100)⁵

= 5000(105/100)⁵

= 5000(1.2762)

= 6381

CI = A – P

= 6381 – 5000

= Rs. 1381

Therefore, Compound Interest on a sum Rs. 5000 for 5/2 years at 10%, when compounded half-yearly, is Rs. 1381.

The post Compound Interest with Growing Principal appeared first on Learn CBSE.

No matter how many polynomials are given, you can easily find the great common factor of those polynomials by finding the factors. Find the factors of numerical and literal coefficients. Multiply the common factors out of all polynomials to get the highest common factor. Check out some example questions and answers on HCF of Polynomials by Factorization.

Solved Examples on G.C.F of Polynomials

Example 1.

Find the Greatest Common Factor of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6?

Solution:

Factorizing x² – 2x + 2 by using (a – b)².

= (x – 1)²

= (x – 1) ( x – 1)

Factorizing x⁴ – 1 by using a² – b².

= (x²)² – 1²

= (x² – 1) (x² + 1)

= ((x)² – 1²) (x² + 1)

= (x + 1) (x – 1) (x² + 1)

Factorizing x³ – 2x² – 5x + 6 by splitting the middle terms

= x³ – x² – x² – 6x + x + 6

= x²(x – 1) – x (x – 1) – 6 (x – 1)

= (x – 1) (x² – x – 6)

= (x – 1) (x² – 3x + 2x – 6)

= (x – 1) (x(x – 3) + 2 (x – 3))

= (x – 1) (x – 3) (x + 2)

The common factor of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6 is (x – 1)

Therefore, H.C.F of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6 is (x – 1).

Example 2.

Calculate the Highest Common Factor of x²y² – x² and xy² – 2xy – 3x by factorization.

Solution:

Factorizing x²y² – x² by taking the x² common.

= x² (y²- 1)

= x² (y² – 1²)

= x² ( y + 1) ( y – 1)

= x * x ( y + 1) ( y – 1)

Factorizing xy² – 2xy – 3x by taking the variable x common.

= x (y² – 2y – 3)

= x (y² – 3y + y – 3)

= x (y(y – 3) + 1( y – 3))

= x (y – 3) ( y + 1)

The common factors of x²y² – x² and xy² – 2xy – 3x are x , (y – 1).

Therefore, the H.C.F of x²y² – x² and xy² – 2xy – 3x is x * (y – 1).

Example 3.

Find the H.C.F of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x.

Solution:

Factorizing x⁴ – y⁴ by using a² – b² formula.

= (x²)² – (y²)²

= (x² + y²) (x² – y²)

= (x² + y²) (x + y) ( x – y)

Factorizing x²(x – y) + y²(x – y) + y – x by taking (x – y) common

= (x – y) (x² + y² – 1)

The common factor of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x are (x – y)

Therefore, the H.C.F of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x is (x – y).

The post HCF of Polynomials by Factorization | HCF of Polynomials appeared first on Learn CBSE.

Do you need any help to find the least common multiple of polynomials? Then read this article to find the L.C.M of polynomials by factorization method. All you need to do is get the factors of each polynomial, multiply the common and remaining terms to get the L.C.M. You can also check the example questions in the below sections.

LCM of Polynomials Solved Examples

Example 1.

Find the lowest common multiple of (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³)?

Solution:

Given polynomials are (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³).

First Polynomial = (2x² – 4x)

= 2x(x – 2), by taking 2x common

Second Polynomial = (3x⁴ – 12x²)

= 3x² (x² – 4), by taking 2x² common.

= 3x²(x² – 2²), by using the formula a² – b² = (a + b) (a – b)

= 3x²(x + 2) (x – 2)

Third Polynomial = (2x⁵ – 2x⁴ – 4x³)

= 2x³(x² – x -2), by taking 2x³ common

= 2x³(x² – 2x + x – 2), by splitting the middle term -x = -2x + x

= 2x³(x(x – 2) + 1(x – 2))

= 2x³(x + 1) ( x – 2)

The common terms of (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³) = x(x – 2)

Extra common terms are 2, 3x(x + 2), 2x²(x + 1)

Therefore, the required L.C.M. = x(x – 2) * 2 * 3x(x + 2) * 2x²(x + 1)

= 12x⁴ (x – 2) (x + 2) (x + 1).

Example 2.

Find the L.C.M of 3y³ – 18y²x + 27yx², 4y⁴ + 24y³x + 36y²x² and 6y⁴- 54y²x² by factorization.

Solution:

Given polynomials are 3y³ – 18y²x + 27yx², 4y⁴ + 24y³x + 36y²x² and 6y⁴- 54y²x²

First polynomial = 3y³ – 18y²x + 27yx²

= 3y(y² – 6yx +9x²) by taking 3y common

= 3y(y² – 3yx -3yx +9x²) by splitting the middle term -6yx = -3yx – 3yx

= 3y(y(y – 3x) -3x(y – 3x))

= 3y(y – 3x) (y – 3x)

Second Polynomial = 4y⁴ + 24y³x + 36y²x²

= 4y²(y² + 6yx + 9x²), by taking 4y² common

= 4y²(y² + 3yx + 3yx + 9x²) by splitting the middle term 6yx = 3yx + 3yx

= 4y²(y(y + 3x) + 3y(y + 3x))

= 4y²(y + 3x) (y + 3x)

Third Polynomial = 6y⁴- 54y²x²

= 6y²(y² – 9x²), by taking 6y² common

= 6y²(y² – (3x)²) by using a² – b² formula

= 6y²(y + 3x) (y – 3x)

The common factors of the above three expressions is ‘y’ and other common factors of first and third expressions are ‘3’ and ‘(y – 3x)’. The common factors of second and third expressions are ‘2’, ‘y’ and ‘(y + 3x)’. Other than these, the extra common factors in the first expression is ‘(y – 3x)’ and in the second expression are ‘2’ and ‘(y + 3x)’

Therefore, the required L.C.M. = y × 3 × (y – 3x) × 2 × y × (y + 3x) × (y – 3x) × 2 × (y + 3x)

= 12y²(y + 3x)² (y – 3x)²

Example 3.

Claculate the L.C.M of 2x² -x -1 and 4x² + 8x +3?

Solution:

Given polynomials are 2x² -x -1 and 4x² + 8x +3.

First Polynomial = 2x² -x -1

= 2x² – 2x + x -1, by splitting the middle term -x = -2x + x

= 2x(x – 1) + 1 (x – 1)

= (x – 1) (2x + 1)

Second Polynomial = 4x² + 8x +3

= 4x² + 6x + 2x + 3, by splitting the middle term 8x = 6x + 2x

= 2x(2x + 3) + 1(2x + 3)

= (2x + 1)(2x + 3)

The common factors of the above two expressions is (2x + 1). The extra common factor are (2x + 3), (x – 1).

Therefore, the required L.C.M. = (2x + 1) * (x – 1) * (2x + 3)

= (2x + 1) (x – 1) (2x + 3)

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Monomial is an algebra expression that contains one term. It includes numbers, whole numbers, variables, and all are multiplied together. The highest common factor (HCF) or the greatest common factor (GCF) of two monomials are the product of numerical coefficients and literal coefficients.

Steps to Find HCF of Monomials

The following are the simple steps to calculate the greatest common factor of two monomials easily.

• Get the factors of both monomial numerical coefficients.
• From that, find the HCF of the numerical coefficient.
• Coming to the literal coefficients, check the lowest power of each variable.
• And calculate the HCF of literal coefficients.
• Multiply the HCF of numerical coefficients and literal coefficients.

Example Questions on Highest Common Factor of Monomials

Example 1.

Find the H.C.F. of 25x³y² and 5xy³z.

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 25 and 5.

Since, 25 = 5 × 5 = 5² and 5 = 1 × 5 = 5¹

Therefore, the H.C.F. of 25 and 5 is 5

The H.C.F. of literal coefficients = The H.C.F. of x³y² and xy³z = xy²

Since, in x³y² and xy³z, x and y are common.

The lowest power of x is x.

The lowest power of y is y².

Therefore, the H.C.F. of x³y² and xy³z is xy².

Thus, the H.C.F. of 25x³y² and 5xy³z.

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 5 × (xy²)

= 5xy².

Example 2.

Find the G.C.F of 32x²yz² and 72xy²z

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 32 and 72.

Since, 32 = 2 x 2 x 2 x 2 x 2 = 2⁵ and 72 = 2 x 2 x 2 x 9 = 2³ x 9

Therefore, the H.C.F. of 32 and 72 is 2³ = 8

The H.C.F. of literal coefficients = The H.C.F. of x²yz² and xy²z = xyz

Since, in x²yz² and xy²z, x, y and z are common.

The lowest power of x is x.

The lowest power of y is y.

The lowest power of z is z.

Therefore, the H.C.F. of x³y² and xy³z is xyz.

Thus, the H.C.F. of 32x²yz² and 72xy²z.

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 8 × (xyz)

= 8xyz.

Example 3.

Find the highest common factor of 8p²q⁵r³ and 6p⁴q³r².

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 8 and 6.

Since, 8 = 2 x 2 x 2 = 2³ and 6 = 2 × 3 = 2¹ x 3¹

Therefore, the H.C.F. of 8 and 6 is 2

The H.C.F. of literal coefficients = The H.C.F. of p²q⁵r³ and p⁴q³r² = p²q³r²

Since, in p²q⁵r³ and p⁴q³r², p, q, and r are common.

The lowest power of p is p².

The lowest power of q is q³.

The lowest power of r is r²

Therefore, the H.C.F. of p²q⁵r³ and p⁴q³r² is p²q³r².

Thus, the H.C.F. of 8p²q⁵r³ and 6p⁴q³r².

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 2 × (p²q³r²)

= 2p²q³r².

The post Highest Common Factor of Monomials | GCF of Monomials appeared first on Learn CBSE.

We are using the factorization method to calculate the highest common factor of monomials. Factorization is the process of breaking an entity into a product of another entity which when multiplied together gives the original entity. Get the example questions on finding the greatest common factor or Highest Common Factor of Monomials by Factorization.

HCF of Monomials Solved Examples

Example 1.

Find the H.C.F. of the monomials 16x²y³and 22x²y.

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 16, 22

Since, 16 = 2 * 2 * 2 * 2, 22 = 2 * 11

Therefore, the H.C.F. of 16, 22 is 2.

Now, the variables x and y are present in all the quantities. Out of these the highest common power of x is 2 and the highest common power if y is 1.

Therefore, the required H.C.F. = 2x²y¹ = 2x²y

The method by which the Highest common factor of the monomials is determined can be formulated as follows:

(i) The G.C.F. of the numerical coefficients are to be determined at first.

(ii) Then the variables are to be written beside the coefficient with their greatest common power or highest common power.

Example 2.

Find the Greatest Common Factor of the monomials 90x⁴y²z³, 54x²y³z⁴, and 18x³y²z³.

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 90, 54, and 18

Since, 90 = 2 * 5 * 9, 54 = 2 * 3 * 9, 18 = 2 * 9

Therefore, the H.C.F. of 90, 54, and 18 is 2 * 9 = 18.

Now, the variables x, y, and z are available in all the quantities. The highest common power of x is 2, the highest common power of y is 2, and the greatest common power of z is 3.

Therefore, the required greatest common factor = 18x²y²z³

Example 3.

Find the G.C.F of 20abc, 22a²b²c², and 24a³b³c².

Solution:

20abc = 2 * 2 * 5 * a * b * c

22a²b²c² = 2 * 11 * a * a * b * b * c * c

24a³b³c² = 2 * 2 * 2 * 3 * a * a * a * b * b * b * c * c

From the resolved factors of the above three monomials, the common factors are 2, a, b, c

Therefore, the required H.C.F. = 2 * a * b * c = 2abc

The post Highest Common Factor of Monomials by Factorization | HCF of Monomials appeared first on Learn CBSE.

The lowest common multiple of monomials or L.C.M of monomials is the least occurred term in the monomials. Find the numerical coefficients LCM, literal coefficients LCM, and multiply them to get the result. You can check out the following sections to get the simple steps to calculate the least common multiple of two or monomials. Also, find the solved example questions for a better understanding of the concept.

How to Find L.C.M of Monomials?

Learn about how to get the lowest common multiple of monomials by using the steps mentioned below. Follow these easy to use steps to get the answer easily.

• Find the factors of given monomials numerical coefficients.
• And get the least common multiple from those factors.
• Get the highest power of each variable from the monomials.
• Calculate the LCM of literal coefficients.
• Multiply the LCM of numerical coefficients and literal coefficients.

Examples on Lowest Common Multiple of Monomials

Example 1:

Find the LCM of 12x²y³z and 18xy²z.

Solution:

The L.C.M of Numerical coefficients = The L.C.M. of 12, 18.

Since, 12 = 2 * 2 * 3 and 18 = 2 * 3 * 3

Therefore, the LCM of 12, 18 is 2 * 2 * 3 * 3= 36

The L.C.M. of literal coefficients = The L.C.M. of x²y³z, xy²z = x²y³z

Since in x²y³z and xy²z

The highest power of x is 2.

The highest power of y is 3.

The highest power of z is 1.

Therefore, the L.C.M. of x²y³z, xy²z = x²y³z

Thus, the L.C.M. of 12x²y³z and 18xy²z = The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 36 * x²y³z

= 36x²y³z

Example 2:

Find the L.C.M of 26p⁴q²r³ and 16p³q²r².

Solution:

The L.C.M. of numerical coefficients = The L.C.M. of 26, 16

Since, 26 = 2 * 13, 16 = 2 * 2 * 2 * 2

Therefore, the L.C.M. of 26, 16 is 2 * 13 * 8 = 206.

The L.C.M. of literal coefficients = The L.C.M. of p⁴q²r³, p³q²r²

Since in p⁴q²r³ and p³q²r²,

The highest power of p is 4.

The highest power of q is 2.

The highest power of r is 3.

Therefore, the L.C.M. of p⁴q²r³ and p³q²r² = p⁴q²r³

Thus, the L.C.M of 26p⁴q²r³ and 16p³q²r² = The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 206 * p⁴q²r³

= 206p⁴q²r³

Example 3:

Find the LCM of 24y³x, 40x³y.

Solution:

The L.C.M. of numerical coefficients = The L.C.M. of 24 and 40.

Since, 24 = 2 * 2 * 2 * 3 and 40 = 2 * 2 * 2 * 5

Therefore, the L.C.M. of 24 and 40 is = 2 * 2 * 2 * 3 * 5 = 120

The L.C.M. of literal coefficients = The L.C.M. of y³x and x³y = x³y³

Since, in y³x and x³y,

The highest power of x is 3.

The highest power of y is 3.

Therefore, the L.C.M. of y³x and x³y = x³y³.

Thus, the L.C.M. of 24y³x, 40x³y

= The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 120 × (x³y³)

= 120x³y³.

The post Lowest Common Multiple of Monomials | Solved Example Questions appeared first on Learn CBSE.

You can calculate the least common multiple of two or more monomials by using the factorization method. Get the factors of numerical coefficients and literal coefficients. From the factors, observe the least common multiples and multiply them. And find the product of the lowest common multiples of numerical and literal coefficients of monomials. Have a look at the sample example questions on Lowest Common Multiple of Monomials by Factorization in the below sections.

Example Questions on LCM of Monomials

Example 1:

What is the least common multiple of 5x³y² and 7x²y³?

Solution:

5x³y² = 5 * x * x * x * y * y

7x²y³ = 7 * x * x * y * y * y

From the resolved factors of the above two monomials, the common factors are x, x, y, y.

The extra factors in the first monomial are 5 and in the second monomial are 7, y.

Therefore, the required L.C.M. = Common factors among two monomials × Extra common factors among two monomials.

= (x * x * y * y) x (5 * 7 * y)

= 35x²y³

Hence, the lowest common multiple of the monomials 5x³y² and 7x²y³ = 35x²y³.

Example 2:

Find the L.C.M of 14xy, 21y², 28y.

Solution:

The L.C.M. of numerical coefficients = The L.C.M. of 14, 21, and 28.

Since, 14 = 2 * 7, 21 = 7 * 3, and 28 = 2 * 2 * 7

Therefore, the L.C.M. of 14, 21 and 28 is 2 * 7 * 2 * 3 = 84

The L.C.M. of literal coefficients = The L.C.M. of xy, y², and y = xy²

Since, in xy, y², and y

The highest power of x is 1.

The highest power of y is 2.

Therefore, the L.C.M. of xy, y², and y = xy².

Thus, the L.C.M. of 14xy, 21y², 28y

= The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 84 × (xy²)

= 84xy²

Example 3:

Find the L.C.M of 27u⁴, 18u², 27u².

Solution:

The L.C.M. of numerical coefficients = The L.C.M. of 27, 18, and 27.

Since, 27 = 3 * 3 * 3, 18 = 3 * 3 * 2, 27 = 3 * 3 * 3

Therefore, the L.C.M. of 27, 18, and 27 is 2 * 3 * 3 * 3 = 54

The L.C.M. of literal coefficients = The L.C.M. of u⁴, u², u² = u⁴

Since, in u⁴, u², u²

The highest power of u is 4.

Therefore, the L.C.M. of u⁴, u², u² = u⁴.

Thus, the L.C.M. of 27u⁴, 18u², 27u²

= The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 54 × (u⁴)

= 54u⁴

The post Lowest Common Multiple of Monomials by Factorization | Monomials LCM appeared first on Learn CBSE.

You can compute the highest common factor or the greatest common factor of any number of polynomials by reading this article. Here we are giving the step by step explanation to get the H.C.F of polynomials along with the solved examples. Go through the below sections to solve the questions easily.

Step By Step Process to get G.C.F of Polynomials

Students can check out the below sections to find the detailed procedure of calculating the greatest common factor of polynomials,

• First of all, find the factors of polynomials.
• Identify the expression which is occurring more times.
• Separate the common factors from given polynomials and multiply them.

Example Questions on H.C.F of Polynomials

Example 1:

Find the Highest Common Factor of the polynomials x² – 6x + 9 and x² – 9.

Solution:

Factorizing x² – 6x + 9 by using the identities (a – b)², we get

(x)² – 2(x)(3) + (3)²

= (x – 3)²

= (x – 3) (x – 3)

Also, factorizing x² – 9, we get

(x)² – (3)², by using the identities of a² – b².

= (x + 3) (x – 3)

Therefore, H.C.F. of x² – 6x + 9 and x² – 9 is (x – 3).

Example 2:

Find the H.C.F of (a + b)² and (a² – b²).

Solution:

Factors of (a + b)² = (a + b) (a + b)

Factors of (a² – b²) = (a + b) (a – b)

The common factor is (a + b)

Therefore, the highest common factor of (a + b)² and (a² – b²) is (a + b).

Example 3:

Find the highest common factor of polynomials x² + 15x + 56, x² + 5x – 24 and x² + 8x.

Solution:

Factorizing x² + 15x + 56 by splitting the middle term, we get

= x² + 8x + 7x + 56

= x(x + 8) + 7(x + 8)

= (x + 7) (x + 8)

Also, factorizing x² + 5x – 24, we get

= x² + 8x – 3x – 24

= x (x + 8) – 3 (x + 8)

= (x + 3) (x + 8)

Factoring x² + 8x by taking x common.

= x (x + 8)

In all three polynomials the common factor is (x + 8)

Therefore, H.C.F. of x² + 15x + 56, x² + 5x – 24 and x² + 8x is (x + 8).

The post Highest Common Factor of Polynomials | G.C.F of Polynomials appeared first on Learn CBSE.

To calculate the Lowest Common Multiple of polynomials, you must find the factors of each polynomial and multiply the least occurred term. Check the steps to solve the L.C.M of polynomials and example questions in the further sections of this page.

Steps to Calculate L.C.M of Polynomials

Follow the detailed procedure to get the lowest common factor of polynomials mentioned below. These guidelines make it easy for you during the calculations.

• Get the factors of each polynomial.
• Identify common factors.
• Multiply the common factors and extra factors.

Some Examples on the Lowest Common Multiple of Polynomials

Example 1:

Find the lowest common multiple of x² − x, x², (x − 1)².

Solution:

Factorizing x² − x by taking the common factor ‘x’ we get,

= x(x − 1)

Also, factorizing x² we get

= x * x

Also, factorizing (x − 1)² we get,

= (x – 1) * (x – 1)

Therefore, the L.C.M. of x² − x, x², (x − 1)² is x²(x – 1)².

Example 2:

Find the L.C.M of n² − 3n + 2, n² − 4.

Solution:

Factorizing n² − 3n + 2 by splitting the middle term.

= n² – 2n – n + 2

= n (n – 2) -1 (n – 2)

= (n – 1) (n – 2)

Factorizing n² − 4 by using a² – b² formula.

= n² – 2²

= (n – 2) (n + 2)

Therefore, the L.C.M. of n² − 3n + 2, n² − 4 is (n – 2) (n + 2) (n – 1).

Example 3:

Find the least common multiple of 8x − 4, 6x² + x − 2.

Solution:

Factorizing 8x – 4 by taking common ‘4’.

= 4(2x – 1)

Factorizing 6x² + x – 2 by splitting the middle term.

= 6x² + 4x – 3x – 2

= 2x(3x + 2) – 1(3x + 2)

= (3x + 2) (2x – 1)

Therefore, the L.C.M of 8x − 4, 6x² + x − 2 is 4(2x-1) (3x+2).

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Calculating Multiplication of Algebraic Expression is not that easy as you think. Students must follow some rules to find Multiplication of Algebraic Expression. If students to do any small mistakes, then it may lead to wrong answers. So, every operation is important while finding Multiplication of Algebraic Expression. Different types of problems and methods to solve problems are given in this article clearly. Students can easily understand the method of solving the Multiplication of Algebraic Expression after reading this article completely.

Rules to Find Multiplication of Algebraic Expression

1. Product of two same signs is positive, and also the product of different signs is negative.
2. If a is a variable and x, y are two positive integers, then (aᵐ × aⁿ) = a^(m+n)

Types of Algebraic Expression Multiplication

There are different types of multiplication occurs while finding Algebraic Expressions
1) Multiplication of Two Monomials
2) Multiplication of a Polynomial by a Monomial
3) Multiplication of Two Binomials
4) Multiplication by Polynomial

How to Find Multiplication of Two Monomials?

1. Write the two numbers along with the multiplication sign
2. Multiply the numbers.
3. If you find the bases are the same then add the exponents.

Product of two monomials = (Multiplication of their numerical coefficients) × (Multiplication of their variable parts)

Solved Examples

1. Find the product of 4xy and -6x²y³

Solution:
Given that 4xy and -6x²y³
4xy × -6x²y³
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
4 × -6 = -24
Multiply the variables. If the base of the variables is the same, then add the powers.
xy × x²y³
x^(1 + 2)y^(1 + 3)
x³y⁴
Multiply coefficient and variables.
-24x³y⁴

The required answer is -24x³y⁴

2. Find the product of 7ab², -4a²b, and -5abc?

Given that 4ab², -6a²b, and -7abc
4ab² × -6a²b × -7abc
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
4 × -6 × -7 = 168
Multiply the variables. If the base of the variables is the same, then add the powers.
ab² × a²b × abc
a^(1 + 2 + 1)b^(2 + 1 + 1)c
a⁴b⁴c
Multiply coefficient and variables.
168a⁴b⁴c

The required answer is 168a⁴b⁴c

3. Find the product of 5ab and 3a³b²

Given that 5ab and 3a³b²
5ab × 3a³b²
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
5 × 3 = 15
Multiply the variables. If the base of the variables is the same, then add the powers.
ab × a³b²
a^(1 + 3)b^(1 + 2)
a⁴b³
Multiply coefficient and variables.
15a⁴b³

The required answer is 15a⁴b³

4. Find the product of 6x²y, 9z²x, and -6xy²z?

Given that 6x²y, 9z²x, and -6xy²z
6x²y × 9z²x × -6xy²z
Multiply the coefficient. If the signs are the same, then the resultant coefficient is positive. Or else, if the signs are not the same, then the resultant coefficient is negative.
6 × 9 × -6 = -324
Multiply the variables. If the base of the variables is the same, then add the powers.
x²y × z²x × xy²z
x^(2 + 1 + 1)y^(1 + 3)z^(2 + 1)
x⁴y⁴z³
Multiply coefficient and variables.
-324x⁴y⁴z³

The required answer is -324x⁴y⁴z³

How to Find Multiplication of a Polynomial by a Monomial?

1. Use a distributive law and multiply a polynomial by a monomial.
2. Multiply each individual term in the parenthesis by a monomial.

Multiply each term of the polynomial by the monomial, using the distributive law a × (b + c) = a × b + a × c.

Solved Examples

1. 6a²b² × (2a² – 6ab + 8b²)

Solution:
Given that 6a²b² × (2a² – 6ab + 8b²)
Apply distributive law of multiplication for the given terms.
a × (b + c) = a × b + a × c.
6a²b² × (2a² – 6ab + 8b²) = (6a²b²) × (2a²) + (6a²b²) × (-6ab) + (6a²b²) × (8b²)
Find the final expression by multiplying each term.
12a⁴b² – 36a³b³ + 48a²b⁴.

The required expression is 12a⁴b² – 36a³b³ + 48a²b⁴.

2. (-9x²y) × (2x²y – 5xy² + x – 7y)

Solution:
Given that (-9x²y) × (2x²y – 5xy² + x – 7y)
Apply distributive law of multiplication for the given terms.
a × (b + c) = a × b + a × c.
(-9x²y) × (2x²y – 5xy² + x – 7y) = (-9x²y) × (2x²y) + (-9x²y) × (-5xy²) + (-9x²y) × (x) + (-9x²y) × (-7y)
Find the final expression by multiplying each term.
-18x⁴y² + 45x³y³ – 9x³y + 63x²y²

The required expression is -18x⁴y² + 45x³y³ – 9x³y + 63x²y².

3. 0(x^4 + 2x^3 + 3x^2 + 9x + 1)

Solution:
Given that 0(x^4 + 2x^3 + 3x^2 + 9x + 1)
Any polynomial multiply by zero is zero.
Therefore, multiplying (x^4 + 2x^3 + 3x^2 + 9x + 1) with 0 is 0.

The answer is 0.

4. 1 (5 x^4 – 8 )

Solution:
Given that 1 (5 x^4 – 8 )
Any polynomial multiply by 1 is the polynomial itself.
Therefore, multiplying (5 x^4 – 8 ) with 1 is (5 x^4 – 8).

The required expression is (5 x^4 – 8).

3. How to Find Multiplication of Two Binomials?

Use two methods to find the Multiplication of Two Binomials. Students can use a horizontal method or Column wise multiplication to find Multiplication of Two Binomials.

How to Find Multiplication of Two Binomials using Horizontal method?

1. Firstly, note down two binomials.
2. Apply the distributive law of multiplication over addition twice.
3. Find the final expression of multiplication.

How to Find Multiplication of Two Binomials using Column wise multiplication?

1. Write one binomial expression under another expression.
2. Multiply the first binomial expression with the first term of the second binomial expression.
3. Multiply the first binomial expression with the second term of the second binomial expression.
4. Note down the first resultant expression and write the second resultant expression below the first resultant expression with like terms comes at the same column.
5. Add the first and second expressions to get the final expression.

Solved Examples

1. Multiply (m + n) × (r + s)

Solution:

Horizontal Method:
Note down two binomials.
(m + n) × (r + s)
Apply the distributive law of multiplication over addition twice.
m × (r + s) + n × (r + s)
(m × r + m × s) + (n × r + n × s)
mr + ms + nr + ns

The required expression is mr + ms + nr + ns.

Column wise multiplication

Write one binomial expression under another expression
m + n
× (r + s)
—————————-
rm + rn —> Multiplication of r with (m + n)
+ ms + ns —> Multiplication of s with (m + n)
—————————-
mr + ms + nr + ns

The required expression is mr + ms + nr + ns.

2. Multiply (2x + 4y) and (4x – 6y)

Solution:

Horizontal Method:
Note down two binomials.
(2x + 4y) and (4x – 6y)
Apply the distributive law of multiplication over addition twice.
2x × (4x – 6y) + 4y × (4x – 6y)
(2x × 4x – 2x × 6y) + (4y × 4x – 4y × 6y)
(8x² – 12xy) + (16xy – 24y²)
8x² – 12xy + 16xy – 24y²
8x² + 4xy – 24y².

The required expression is 8x² + 4xy – 24y².

Column wise multiplication
Write one binomial expression under another expression
(2x + 4y)
× (4x – 6y)
—————————-
8x² + 16xy —> Multiplication of 4x with (4x – 6y)
– 12xy – 24y² —> Multiplication of 4y with (4x – 6y)
—————————-
8x² + 4xy – 24y².

The required expression is 8x² + 4xy – 24y².

3. Multiply (4x² + 2y²) by (3x² + 5y²)

Solution:

Horizontal Method:
Note down two binomials.
(4x² + 2y²) × (3x² + 5y²)
Apply the distributive law of multiplication over addition twice.
4x² (3x² + 5y²) + 2y² (3x² + 5y²)
(12x⁴ + 20x²y²) + (6x²y² + 10y⁴)
12x⁴ + 20x²y² + 6x²y² + 10y⁴
12x⁴ + 26x²y² + 10y⁴

The required expression is 12x⁴ + 26x²y² + 10y⁴

Column wise multiplication
Write one binomial expression under another expression
(4x² + 2y²)
× (3x² + 5y²)
—————————-
12x⁴ + 20x²y² —> Multiplication of 3x² with (4x² + 2y²)
6x²y² + 10y⁴ —> Multiplication of 5y² with (4x² + 2y²)
—————————-
12x⁴ + 26x²y² + 10y⁴

The required expression is 12x⁴ + 26x²y² + 10y⁴

4. How to Find Multiplication by Polynomial?

Apply horizontal method or column multiplication to find Multiplication by Polynomial.

Solved Examples:

1. Multiply (6x² – 4x + 9) by (3x – 7)

Solution:

Horizontal Method:
Given that (6x² – 4x + 9) by (3x – 7)
(3x – 7) × (6x² – 4x + 9)
3x (6x² – 4x + 9) – 7 (6x² – 4x + 9)
(18x³ – 12x² + 27x ) + (- 42x² + 28x – 63)
18x³ – 12x² + 27x – 42x² + 28x – 63
18x³ – 54x² + 55x – 63

The required expression is 18x³ – 54x² + 55x – 63

Column wise multiplication
Write one binomial expression under another expression
(6x² – 4x + 9)
× (3x – 7)
—————————-
18x³ – 12x² + 27x —> Multiplication of 3x with (6x² – 4x + 9)
– 42x² + 28x – 63 —> Multiplication of -7 with (6x² – 4x + 9)
—————————-
18x³ – 54x² + 55x – 63

The required expression is 18x³ – 54x² + 55x – 63

2. Multiply (3x² – 6x + 2) by (2x² + 9x – 5)

Solution:

Horizontal Method:
Given that (3x² – 6x + 2) by (2x² + 9x – 5)
2x² (3x² – 6x + 2) + 9x (3x² – 6x + 2) – 5 (3x² – 6x + 2)
(6x⁴ – 12x³ + 4x²) + ( + 27x³ – 54x² + 18x ) + (- 15x² + 30x – 10 )
6x⁴ – 12x³ + 4x² + 27x³ – 54x² + 18x – 15x² + 30x – 10
6x⁴ + 15x³ – 65x² + 48x – 10

The required expression is 6x⁴ + 15x³ – 65x² + 48x – 10

Column wise multiplication
Write one binomial expression under another expression
(3x² – 6x + 2)
× (2x² + 9x – 5)
—————————-
6x⁴ – 12x³ + 4x² —> Multiplication of 2x² with (6x² – 4x + 9)
+ 27x³ – 54x² + 18x —> Multiplication of 9x with (6x² – 4x + 9)
– 15x² + 30x – 10 —> Multiplication of – 5 with (6x² – 4x + 9)
—————————-
6x⁴ + 15x³ – 65x² + 48x – 10

The required expression is 6x⁴ + 15x³ – 65x² + 48x – 10

3. Multiply (3x³ – 4x² – 2x + 9) by (5 – 6x + 7x²)

Solution:

Horizontal Method:
Given that (3x³ – 4x² – 2x + 9) by (5 – 6x + 7x²)
5 (3x³ – 4x² – 2x + 9) – 6x (3x³ – 4x² – 2x + 9) + 7x² (3x³ – 4x² – 2x + 9)
(15x³ – 20x² – 10x + 45) + (-18x⁴ + 24x³ + 12x² – 54x ) + (21x⁵ – 28x⁴ – 14x³ + 63x²)
15x³ – 20x² – 10x + 45 -18x⁴ + 24x³ + 12x² – 54x + 21x⁵ – 28x⁴ – 14x³ + 63x²
21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

The required expression is 21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

Column wise multiplication
Write one binomial expression under another expression
(3x³ – 4x² – 2x + 9)
× (5 – 6x + 7x²)
—————————-
15x³ – 20x² – 10x + 45 —> Multiplication of 5 with (3x³ – 4x² – 2x + 9)
-18x⁴ + 24x³ + 12x² – 54x —> Multiplication of – 6x with (3x³ – 4x² – 2x + 9)
21x⁵ – 28x⁴ – 14x³ + 63x² —> Multiplication of 7x² with (3x³ – 4x² – 2x + 9)
—————————-
21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

The required expression is 21x⁵ -46x⁴ + 25x³ + 55x² – 10x + 45

The post Multiplication of Algebraic Expression | Product of Algebraic Expression appeared first on Learn CBSE.

A fraction represents equal parts of a collection or whole. Algebraic fraction means a fraction whose numerator or denominator is a polynomial expression. Check the examples of algebraic fractions, its definition, and solved example questions. You can also see how to perform addition, multiplication, subtraction, and division between algebraic fractions.

Algebraic Fraction Definition

Fractions that have a polynomial expression in the numerator and denominator are called algebraic fractions. Denominators of the algebraic fractions can never be zero. You can write every polynomial as an algebraic fraction with a denominator.

Examples of Algebraic Fractions

• (x² + 2x + 3) / 3 is an algebraic fraction with integral denominator 3.
• (x + 3) / 5 is an algebraic fraction with integral denominator 5.
• 6 / (a + 5b + 3) is an algebraic fraction with integral numerator 6.
• 2 / (x + 3y) is an algebraic fraction with integral numerator 2.
• (x + y) / (x² + 10x + 7) is an algebraic fraction with numerator as linear polynomial and denominator as a quadratic polynomial.
• (y² – 11y + 32) / (y + 2) is an algebraic fraction with numerator as quadratic polynomial and denominator as a linear polynomial.

Operations with Algebraic Fractions

Below mentioned are the various operations that can be performed on algebraic fractions.

1. Reducing Fraction

To reduce the algebraic fraction, first, find the factors of numerator and denominator. Later cancel the common factors.

Example: (12x³y²) / (28xy)

(12x³y²) / (28xy) = (4 * 3 * x * x * x * y * y) / (7 * 4 * x * y)

= (3 * x * x * y) / 7 = (3x²y) / 7 = (3/7) x²y

2. Multiplying Algebraic Fractions

While multiplying 2 fractions, get the factors of the fraction and then reduce it to the lowest terms. Then multiply the numerators together, and denominators together to check the result.

Example: [(x + 1) / (5y + 10)] * [(y+2) / (x² + 2x + 1)]

= [(x + 1) / ((5(y + 2))] * [(y+2) / (x² + x + x + 1)]

= [(x + 1) / (5] * [1 / (x (x + 1) + 1 (x + 1)] = [(x + 1) / (5] * [1 / ((x + 1) (x + 1)]

= (1/5) * (1 / (x + 1)) = 1 / (5(x + 1))

3. Adding Fractions

You must have a common denominator to add two fractions. Simply, get the lcm or multiply those two denominators. Then, add the numerators.

Example: [(x – 4) / (x + 1)] + [3 / (x + 2)]

= [(x – 4) (x + 2) / (x + 1) (x + 2)] + [(3(x + 1)) / (x + 1) (x + 2)]

= [(x² – 4x + 2x – 8) / (x + 1) (x + 2)] + [(3x + 3) / (x + 1) (x + 2)]

= [(x² – 2x – 8) / (x + 1) (x + 2)] + [(3x + 3) / (x + 1) (x + 2)] = [(x² – 2x – 8 + 3x + 3) / (x + 1) (x + 2)]

= (x² + x – 5) / [(x + 1) (x + 2)]

4. Subtracting fractions

To subtract fractions, make a common denominator by finding the LCD and changing each fraction to an equivalent fraction. Then subtract those fractions.

Example: (2 / x) – (3 / y)

= (2y / xy) – (3x / xy) = (2y – 3x) / xy

5. Dividing Fractions

To divide algebraic fractions, invert the second fraction and multiply.

Example: (2 / x²) / (5 / x)

= (2 / x²) * (x / 5) = 2x / 5x² = 2 / 5x

Example Questions

Example 1. 

Solve [5 / (x + 6)] + [10 / (x + 1)]

Solution:

Given fraction is [5 / (x + 6)] + [10 / (x + 1)]

Both fractions denominators are different. So find the lcm

LCM of (x + 6) and (x + 1) is (x + 6) * (x + 1)

[5 / (x + 6)] + [10 / (x + 1)] = [(5(x + 1)) / (x + 6) (x + 1)] + [(10(x + 6)) / (x + 6) (x + 1)]

= [(5x + 5) / (x + 6) (x + 1)] + [(10x + 60) / (x + 6) (x + 1)]

= [(5x + 5 + 10x + 60) / (x + 6) (x + 1)] = [(15x + 65) / (x + 6) (x + 1)]

[5 / (x + 6)] + [10 / (x + 1)] = [5(3x + 13) / (x + 6) (x + 1)]

Example 2.

Solve [(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)]

Solution:

Given fraction is [(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)]

[(x² + 13x + 35) / (x + 4)] / [(x² – 3x – 40) / (x – 6)] = [(x² + 13x + 35) / (x + 4)] * [(x – 6) / (x² – 3x – 40)]

= [(x² + 7x + 5x + 35) / (x + 4)] * [(x – 6) / (x² – 8x + 5x – 40)] = [(x(x + 7) + 5 (x + 7) / (x + 4)] * [(x – 6) / (x(x – 8) + 5(x – 8)]

= [((x + 7) (x + 5) / (x + 4)] * [(x – 6) / ((x – 8) (x + 5)]

= [(x + 7) (x + 5) (x – 6)] / [(x + 4) (x – 8) (x + 5)] = [(x + 7) (x – 6)] / [(x + 4) (x – 8)]

Example 3.

Perform the indicated operation.

(i) [10 / y] – [15 / y² – 10y + 25]

(ii) [x / (x² + 5x + 6)] – [2 / (x² + 3x + 2)]

Solution:

(i) Given fraction is [10 / y] – [15 / y² – 10y + 25]

= [10 / y] – [15 / (y² – 5y – 5y + 25]

= [10 / y] – [15 / (y(y – 5) – 5(y – 5)] = [10 / y] – [15 / (y – 5)(y – 5)]

= [10 (y – 5)² / y(y – 5)²] – [15y / y(y – 5)²]

= (10 (y – 5)² – 15y) / y(y – 5)² = 10(y² – 10y + 25) / y(y – 5)²

= [10(y² + 25) – 100y – 15y] / [y(y – 5)²] = [10(y² + 25) – 115y] / [y(y – 5)²]

(ii) Given fraction is [x / (x² + 5x + 6)] – [2 / (x² + 3x + 2)]

= [x / (x² + 3x + 2x + 6)] – [2 / (x² + 2x + x + 2)]

= [x / (x(x + 3) + 2(x + 3)] – [2 / (x(x + 2) + 1(x + 2)]

= [x / ((x + 3)(x + 2)] – [2 / (x + 2)(x + 1)] = [x(x + 1) / (x + 2)(x + 1)(x + 3)] – [2(x + 3) / (x + 2)(x + 1)(x + 3)]

= [x² + x – 2x – 6] / [(x + 2)(x + 1)(x + 3)] = [x² – x – 6] / [(x + 2)(x + 1)(x + 3)]

= [(x – 3) (x + 2)] / [(x + 2)(x + 1)(x + 3)]

= [x – 3] / [(x + 1)(x + 3)]

The post Algebraic Fractions | Operations with Algebraic Fractions Examples appeared first on Learn CBSE.

Arithmetic Fraction is a fraction which is in the form of a / b. Algebraic fractions contain polynomials either in the numerator or denominator. Get the definition of Arithmetic Fraction and Algebraic Fraction, some solved example questions in the further sections of this page.

Arithmetic Fraction Definition

Arithmetic fractions are represented as p / q. Where p is the numerator and q is the denominator which is not equal to zero. A number or expression in the form of a numerator/denominator is called the arithmetic fraction. It can also be expressed as p ÷ q. Some examples are 1/2, 3/10, 5/4, etc

When the numerator, denominator of arithmetic fractions are multiplied or divided by the same quantity, then the value of the fraction remains the same. Arithmetic fractions are mostly monomial quantities or they can be reduced to monomials.

Some Examples of Arithmetic Fraction

• 12/16 is an arithmetic fraction having numerator 12 and denominator 16.
• 17/85 is also an arithmetic fraction having numerator 17 and denominator 85.
• 18/81 = 2/9
• 16/32 = 1/2

Algebraic Fraction Definition

An algebraic Fraction is also a fraction whose numerator, denominators are algebraic expressions. Both numerator and denominator are polynomials.

Examples of Algebraic Fraction

• If both denominator and numerator are monomials.
  • a/b, x/y, ax²/bc, my²/n, etc
• If the denominator is monomial and the numerator is polynomial or binomial.
  • x+y / z, a²+bc+b² / ab, ab+bc+ca / a, etc
• If the denominator is binomial/polynomial and the numerator is monomial.
  • a/b+c, x/y-z, a/ab+bc+ca, etc
• If numerator and denominator are binomial or polynomial.
  • x+y/x-y, a+b+c/a+c, a²+2bc+c/a+b, etc

The post Arithmetic Fraction and Algebraic Fraction Definition | Examples with Answers appeared first on Learn CBSE.

Simplification of algebraic fractions is nothing but finding the factors of both numerator and denominator and canceling the like terms. A fraction is a real number that represents the part of a single object from a group of objects. It is a combination of the numerator, denominator, and separator (/). The denominator of a fraction can not be zero. We can represent fractions as multiple fractions without changing the original one simply by altering its numerator, denominator values. Get the two simple methods to simply the fractions and solved examples in the below sections.

Algebraic Fractions

Fractions which has polynomial expressions in the numerator, denominator are called algebraic fractions. The algebraic fractions denominator can never be a zero. Every polynomial may be represented as an algebraic fraction with a denominator.

Adding Fractions:

To add two or more fractions, you must have a common denominator. To make a common denominator, just find the lcm of denominators or find the product of denominators. After making the denominator common, just add the numerators to get the addition of fractions.

Example: Solve 5 / 6 + 1 / 12?

LCM of 6, 12 is 12.

(5 * 2) / (6 * 2) + 1 / 12 = 10 / 12 + 1 / 12

= (10 + 1) / 12 = 11 / 12

Subtracting Fractions:

Addition and subtrcation of fractions is same. Here also, make a common denominator and subtract the numerator values.

Example: Solve [(x + 3) / (x – 3)] – (3 / (x – 3))

(x + 3 – 3) / (x – 3) = x / x – 3.

Multiplying Fractions:

Multiply numerators of the fractions and denominators of the fractions separately. Write the obtained numerator and denominator as a fraction to get the multiplication of fractions.

Example: Evaluate 1 / 2 * 23 / 25?

(1 * 23) / (2 * 25) = 23 / 50.

Dividing Fractions:

When you divide two fractions (a/b) / (c/d), you will get the answer as (ad / bc).

Example: Find 1/2 / 2/1?

(1 * 1) / (2 * 2) = 1 / 4.

Equivalent Fractions:

We can say that two or more fractions are equivalent when their numerator, denominators are equal. To make the fractions equal, you can divide or multiply the fractions by the same value.

Example: Check whether 12/42 and 2/7 are equivalent or not?

To make the numerator, the denominator of a second fraction equal to the first fraction, multiply the second fraction with 6.

(2 * 6) / (7 * 6) = 12 / 42

So, both are equivalent fractions.

What is Meant by Simplification of Algebraic Fractions?

Simplification of algebraic fractions means reducing a fraction to its lowest value. Both the numerator and denominator of a fraction are reduced so that it should have no common factor between them. The original value of the fraction will never change after the simplification. Therefore, it also called the equivalent fractions where numerator, denominator have no common factor except 1. An example is 25 / 15 = (5 * 5) / (5 * 3) = 5 / 3

Methods to Simplify Algebraic Fractions

Actually, we have two simple methods to simplify algebraic fractions. The one and only most important rule in the simplification of algebraic fractions is you must divide numerator, denominator with the same number at a time. Learn those steps in the below sections.

Method 1:

Simply, divide numerator and denominator by 2, 3, 4, 5… so on until you cannot find the common factor for those numbers.

Method 2:

• Find the highest common factor (HCF) for both the numerator, denominator.
• Divide both by the HCF number.
• The obtained result is a simplified fraction.

Whenever a polynomial expression is given in a fraction, then you can use this better and shorter way to simplify the fractions.

• Find the factors of both the numerator, denominator.
• Cancel the common factors in the numerator, denominator.
• Multiply the remaining values to get the result.

Example Questions on Simplification of Algebraic Fractions

Example 1.

Simplify the algebraic fraction: (10x³y³z²) / (2xy²z)

Solution:

Given fraction is (10x³y³z) / (2xy²z²)

The factors of the fraction are

(5 * 2 * x * x * x * y * y * y * z * z) / (2 * x * y * y * z)

We can see that ‘2’, ‘x’, ‘y * y’, ‘z’ are the common factors in the numerator and denominator. So, we cancel the common factors from the numerator and denominator.

= 5x²y / z

Example 2.

Simplify (2x – 10)⁶ / (x – 5)⁷.

Solution:

Given fraction is (2x – 10)⁶ / (x – 5)⁷

The common factors are (2 * (x – 5))⁶ / (x-5)⁷

= [2⁶ * (x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5)] / [(x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5) * (x-5)]

We can see that (x-5) is the common factor in the numerator, denominator. So, we cancel the common factors.

= 2⁶ / (x-5) = 64 / (x-5)

Example 3.

Reduce the algebraic fraction to its lowest term:

(x² + 2x – 35) / (x² – 25)

Solution:

Given fraction is (x² + 2x – 35) / (x² – 25)

x² + 2x – 35 = x² + 7x – 5x – 35 = x (x + 7) – 5(x + 7)

= (x – 5) (x + 7)

x² – 25 = (x – 5) (x + 5)

The common factors of the fraction is

[(x – 5) (x + 7)] / [(x – 5) (x + 5)]

Cancel the like terms

(x + 7) / (x + 5)

Example 4.

Simplify by adding and subtracting algebraic fractions:

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2]

Solution:

Given that,

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2]

The lcm of 3 , 6, 2 is 6.

Make the denominator as 6 for all parts of the expression.

[(2x – 1) / 3] – [(x – 5) / 6] + [(x – 4) / 2] = [2(2x – 1) / 6] – [(x – 5) / 6] + [3(x – 4) / 6]

= [(4x – 2) / 6] – [(x – 5) / 6] + [(3x – 12) / 6]

As, all denominators are equal perform arithmetic operations on numerator.

= [(4x – 2) – (x – 5) + (3x – 12)] / 6

= [4x – 2 – x + 5 + 3x – 12] / 6

= [6x – 9] / 6

= 3(x – 3) / 6

= 3(x – 3) / (3 * 2) = (x – 3) / 2

Example 5.

Simplify the following:

(i) [(x²y² + 3xy) / (4x² – 1)] / [(xy + 3) / (2x + 4)]

(ii) [(a² – 4b²) / (ab + 2b²)] * [(2b) / (a – 2b)]

Solution:

(i) Given that,

[(x²y² + 3xy) / (4x² – 16)] / [(xy + 3) / (2x + 4)]

(a/b) / (c/d) = (ad / bc)

[(x²y² + 3xy) / (4x² – 16)] / [(xy + 3) / (2x + 4)] = [(x²y² + 3xy) * (2x + 4)] / [(4x² – 16) * (xy + 3)]

The common factors are

= [(xy (xy + 3) * (2x + 4)] / [(2x -4) * (2x + 4) * (xy + 3)]

Cancel the common factors in both numerator, denominator.

= (xy) / (2x + 4)

= (xy) / [2(x + 2)]

(ii) Given that,

[(a² – 4b²) / (ab + 2b²)] * [(2b) / (a – 2b)]

Multiply numerator, denominator

= [(a² – 4b²) * (2b)] / [(ab + 2b²) * (a – 2b)]

The common factors are

= [(a + 2b) * (a – 2b) * 2b] / [(b(a + 2b) * (a – 2b)]

Cancel the like terms in both numerator, denominator

= (2b) / b = 2

FAQs on Simplification of Algebraic Fractions

1. How do you simplify an algebraic fraction?

In order to simplify an algebraic fraction, find the common factors for the numerator, denominator. Cancel the like terms in the numerator and denominator to get the simplified form.

2. How do you simplify algebraic fractions with quadratics?

Get the factors of polynomial expressions by using factorization. Find the common factors and cancel them. Perform the required operations to get the simplified fraction.

The post Simplification of Algebraic Fractions | How to Simplify Fractions? appeared first on Learn CBSE.

An algebraic expression is an expression that is the combination of constants and variables along with different algebraic operations such as addition, subtraction, etc. We included all types of algebraic expression problems imposed in the exams. So, students can prepare perfectly for the exam with our Algebraic Expression material. Also, find the quick links in this article where you can get the detailed concepts, questions, answers, along with explanations.

Example:

1. 2x + 3 = 6

• In 2x + 3 = 6, x is an unknown variable and
• The coefficient is assigned for the variable. 2x + 3 = 6, here 2 is the coefficient of x.
• The contant term is a definite value. 3 and 6 are the constant in 2x + 3 = 6.

Types of Algebraic Expressions

There are different types of algebraic expressions available. Let us have a look at different algebraic expressions with detailed examples.

Monomial Algebraic Expression

Monomial Algebraic Expression is an algebraic expression that contains only one term.

Example:
2x, -2xy, 3y² are some of the examples for Monomial Algebraic Expression.

Binomial Expression

Binomial Algebraic Expression is an algebraic expression that contains two terms.

Example:
6y + 8, 3y + 9, 8y³ + 2, etc. are some of the examples for Binomial Algebraic Expression.

Trinomial Expression

Trinomial Algebraic Expression is an algebraic expression that contains three terms.

Example:
2x – 3y + 6, 4x + 2y – 7z, 5a³ + 8b² + 9c⁴, etc. are some of the examples for Trinomial Algebraic Expression.

Multinomial Expression

Multinomial Algebraic Expression is an algebraic expression that contains two-term.

Example:
3x³ y² + 7x²y – 5xy + 4, 4a² + 9b² – 5c² – 2d², l + 9m + 7n – 5p, etc. are some of the examples for Multinomial Algebraic Expression.

Polynomial Expression

Polynomial Expression is an algebraic expression that contains the power of variables with a non-negative integer.

Example:
2x² + 4x + 6 is a polynomial.
x² + 4/x is not a polynomial. Because 4/x is negative.

Find the quick links of different algebraic expression concepts below. Simply, click the required link and prepare that particular concept with clear examples.

• Addition of Algebraic Expressions
• Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expression
• Division of Algebraic Expressions

Algebraic Expression Examples

1. Express the following algebraic expressions with the help of signs and symbols.

(i) The sum of a and b
(ii) The subtraction of x from y.
(iii) The product of c and d.
(iv) x divided by 6.
(v) 5 divided by m.
(vi) The sum of 4 and p.
(vii) The product of z and 12.
(viii) 4 less than 6 times x.
(ix) Half of the product of 5 and x.
(x) One-tenth of x.
(xi) 4 less than the sum of m and n.
(xii) The values of c and d are equal.
(xiii) The values of a is greater than of b.
(xiv) 7 is less than y.

Solution:

(i) The sum of a and b
a + b
(ii) The subtraction of x from y.
y – x
(iii) The product of c and d.
cd
(iv) x divided by 6.
x/6
(v) 5 divided by m.
5/m
(vi) The sum of 4 and p.
4 + p
(vii) The product of z and 12.
12 × z
(viii) 4 less than 6 times x.
6x – 4
(ix) Half of the product of 5 and x.
5x/2
(x) One-tenth of x.
x/10
(xi) 4 less than the sum of m and n.
(m + n) – 4
(xii) The values of c and d are equal.
c = d
(xiii) The values of a is greater than of b.
a > b
(xiv) 7 is less than y.
7 < y

2. Express the following algebraic expressions in words

(i) c + d
(ii) 4a
(iii) a/6
(iv) a + b + 4
(v) 2m + n.
(vi) x + 3n
(vii) b – 5d
(viii) 3l – m
(ix) (m + 4n)/3
(x) s/3 + 9
(xi) 7 > 3x
(xii) a + b < 10

Solution:
(i) c + d
The sum of c and d
(ii) 4a
4 times of a
(iii) a/6
1/6 the part of a.
(iv) a + b + 4
The sum of a, b and 4
(v) 2m + n.
The sum of n and two times of m
(vi) x + 3n
The sum of x and three times of n
(vii) b – 5d
Deduction of 5 times of d from b
(viii) 3l – m
Deduction of m from 3 times of l
(ix) (m + 4n)/3
1/3 of the sum of m and four times n
(x) s/3 + 9
Sum of 1/3 rd portion of s and 9
(xi) 7 > 3x
7 is greater than three times of x
(xii) a + b < 10
Sum of a + b is less than 10

3. Express the following algebraic expressions using symbol if it is necessary.

(i) Sam has $14, Arun has $a more. How many dollars does Arun possess?
(ii) You worked out m sums yesterday. Today you have worked out 8 sums less. How many sums have you worked out today?
(iii) A car driver had earned L dollar on a day and $5 less on the next day. How much money has he earned on the next day?
(iv) Kyle has 7 pens. His father bought b more pens for her. How many pens now Kyle has?
(v) Ben had 15 chocolates, he lost x chocolates. How many chocolates are now remaining with him?
(vi) Anil is S years older than Arun. The present age of Arun is R years. How old is Anil now? What will be their ages after 5 years?
(vii) A painter earns $m daily. How much will he earn in 4 days?
(viii) There are C rows of trees in Akhil’s garden. In each row, there are 5 trees. How many trees are there in the garden?
(ix) You have two pencils. Your father gave you some more pencils? How many pencils are there with you now?

Solution:

(i) Sam has $14, Arun has $a more. How many dollars does Arun possess?
14 + a
(ii) You worked out m sums yesterday. Today you have worked out 8 sums less. How many sums have you worked out today?
m – 8
(iii) A car driver had earned L dollar on a day and $5 less on the next day. How much money has he earned on the next day?
L – 5
(iv) Kyle has 7 pens. His father bought b more pens for her. How many pens now Kyle have?
7 + b
(v) Ben had 15 chocolates, he lost x chocolates. How many chocolates are now remaining with him?
15 – x
(vi) Anil is S years older than Arun. The present age of Arun is R years. How old is Anil now? What will be their ages after 5 years?
Anil = S + R
Arun = R + 5
Anil = S + R + 5
(vii) A painter earns $m daily. How much will he earn in 4 days?
4m
(viii) There are C rows of trees in Akhil’s garden. In each row, there are 5 trees. How many trees are there in the garden?
5C
(ix) You have two pencils. Your father gave you some more pencils? How many pencils are there with you now?
2 + x

4. Write the algebraic expressions using symbols for the given problems?

(i) Monal had 5 color pens. She has lost some of them. How many color pencils she has now?
(ii) Sam’s age is 16 years.
(i) What was her age a years before?
(ii) What will be her age b years hence?
(iii) Five less than one-fourth of x
(iv) One-fifth of x.
(v) William is 5 years older than his brother Sonu. If Sonu’s age is m years, what will be William’s age?
(vi) The price of a dozen bananas is $ n. What will be the price of 6 dozen bananas?
(vii) The difference between the two numbers is L, the greater number is 20. Find a smaller number?
(viii) The product of two numbers is 25. One of them is c. Find the other?
(ix) Your age is 16 years now. What was your age h year ago? What will be your age after h years?

Solution:

(i) Monal had 5 color pens. She has lost some of them. How many color pencils she has now?
5 – x
(ii) Sam’s age is 16 years.
(i) What was her age a years before?
(ii) What will be her age b years hence?
(i) (16 – a) years
(ii) (16 + a) years
(iii) Five less than one-fourth of x
x/4 – 5
(iv) One-fifth of x.
x/5
(v) William is 5 years older than his brother Sonu. If Sonu’s age is m years, what will be William’s age?
(m + 5) years
(vi) The price of a dozen bananas is $ n. What will be the price of 6 dozen bananas?
6n
(vii) The difference between the two numbers is L, the greater number is 20. Find a smaller number?
Smaller number = 20 – L
(viii) The product of two numbers is 25. One of them is c. Find the other?
Other number = 25/c
(ix) Your age is 16 years now. What was your age h year ago? What will be your age after h years?
Age before h years = (16 – y) years
Age after h years = (16 + y) years

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