# Factor Polynomials by Common Factor

Questions With Solutions

How to factor a polynomial using a common factor? Questions are presented along with detailed solutions and explanations are included.

## What is factorization by common factor?It is a factorization method based on the law of distributivity which is a(b + c) = a · b + a · c and used in reverse as follows a · b + a · c = a(b + c)a is a common factor to a b and a c is therefore factored out.
Factoring a polynomial is to write it as the product of simpler polynomials. Example: 2 x + 4 = 2(x + 2)
3 x^{ 2} - x = x(3x - 1)
NOTE: it is very easy to check if your factorization is correct by expanding the factored form to check if you get the original polynomial Example: check that 3 x^{ 2} - x = x(3x - 1)
Expand x(3x - 1) using the law of distributivity.
x(3x - 1) = (x)(3x) +(x)(-1) = 3x^{2} - x , which is correct.
## More ExamplesFind a common factor and use the method of distributivity in reverse to factor the polynomials completely.a) 9 x - 6b) x^{ 2} - xc) 3 x + 12 x yd) 16 x^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}e) 2 x^{ 4}(x + 5) + x^{ 2}(x + 5)## Solution to the above examplesa) Find any common factors in the two terms of 9 x - 6 by expressing both terms 9x and 6 in the given binomial as prime factorization. Hence
9 x - 6 = 3 ·3 ·x - 2 ·3The greatest common factor is 3 and is factored out. Hence 9 x - 6 = 3 (3 x - 2)b) The prime factorization of x and ^{ 2}x is needed to find the greatest common factor in x.
^{ 2} - xx^{ 2} - x = x · x - x = x · x - 1 · x The greatest common factor is x and is therefore factored out. Hence
x^{ 2} - x = = x (x - 1)c) The prime factorizations of 3 x and 12 x y are needed to find the greatest common factor in 3 x + 12 x y.
3 x + 12 x y = 3 · x - 3 · 4 · x · y = 3 · x · 1 - 3 x · 4 · y The greatest common factor is 3 x. Hence
3 x + 12 x y = 3 x (1 + 4 y)d) The prime factorization of 16 x , ^{ 3} 8 x and ^{ 2} y 4 x y are needed to find the greatest common factor in ^{ 2}16 x.
^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}16 x
= ^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}2 · 2 · 2 · 2 · x · x · x + 2 · 2 · 2 · x · x · y + 2 · 2 · x · y · yThe greatest common factor is 2 · 2 · x = 4 x. Hence
16 x) = 4 x (4 x^{ 3} + 8 x^{ 2} y + 4 x y^{ 2} = 4 x ( 2 · 2 · x · x + 2 · x · y + y · y^{ 2} + 2 x y + y^{ 2})e) We note that x + 5 is a common factor which can be factored out as follows:
2 x^{ 4}(x + 5) + x^{ 2}(x + 5) = (x + 5)(2 x^{ 4} + x^{ 2})We now find the greatest common factor of the terms 2 x and ^{ 4}x and factor it out.
^{ 2}2 x^{ 4} + x^{ 2} = 2 · x · x · x · x + x · x = x^{ 2}(2 x^{ 2} + 1)The complete factoring of 2 x is written as follows:
^{ 4}(x + 5) + x^{ 2}(x + 5)2 x^{ 4}(x + 5) + x^{ 2}(x + 5) = x^{ 2}(x + 5)(2 x^{ 2} + 1)
## QuestionsUse common factors to factor completely the following polynomialsa) - 3 x + 9 b) 28 x + 2 x ^{ 2}c) 11 x y + 55 x ^{ 2} y d) 20 x y + 35 x ^{ 2} y - 15 x y ^{ 2}e) 5 y (x + 1) + 10 y ^{ 2}(x + 1) - 15 x y (x + 1)## Solutions to the Above Questionsa) Find any common factors in the two terms of- 3 x + 9 by expressing both terms 3 x and 9 in the given binomial as prime factorization.- 3 x + 9 = - 3 · x - 3 · 3The greatest common factor is 3 and is factored out. Hence - 3x + 9 = 3 (- x + 3) = - 3 (x - 3)b) Write the prime factorization of each of the terms in the given polynomial 28 x + 2 x .^{ 2}28 x + 2 x ^{ 2} = 2 · 2 · 7 · x + 2 · x · x The greatest common factor is 2 x and is factored out. Hence28 x + 2 x ^{ 2} = 2 x (14 + x)c) Write the prime factorization of each of the terms in the given polynomial 11 x y + 55 x .^{ 2} y11 x y + 55 x ^{ 2} y = 11 · x · y + 5 · 11 · x · x · yThe greatest common factor is 11 x y and is factored out. Hence11 x y + 55 x ^{ 2} y = 11 x y(1 + 5 x)d) Write the prime factorization of each of the terms in the given polynomial 20 x y + 35 x .^{ 2} y - 15 x y ^{ 2}20 x y + 35 x ^{ 2} y - 15 x y ^{ 2} = 2 · 2 · 5 · x · y + 5 · 7 · x · x · y - 3 · 5 · x · y · y The greatest common factor is 5 x y and is factored out. Hence 20 x y + 35 x ^{ 2} y - 15 x y ^{ 2} = 5 x y( 4 + 7 x - 3 y)e) We start by factoring out the common factor (x + 1) in the given polynomial.5 y (x + 1) + 10 y ^{ 2}(x + 1) - 15 x y (x + 1) = (x + 1)(5y + 10y^{2} - 15 x y)We now factor the polynomial 5y + 10y using the GCF to all three terms.^{2} - 15 x y5 y + 10y + 2 · 5 · y · y - 3 · 5 · y · x = 5 · y (1 + 2 y - 3 x)
^{2} - 15 x y = 5 · yThe given polynomial may be factored as follows. 5 y (x + 1) + 10 y ^{ 2}(x + 1) - 15 x y (x + 1) = 5 y(x + 1)(1 + 2y - 3 x) |

### More References and links

Introduction to PolynomialsFactor Polynomials

Factor Polynomials by Grouping

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Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers

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