August 8, 2024

# Factoring Polynomials Formula and Factoring Polynomials Worksheet with Answers

Factoring Polynomials Formula and Factoring Polynomials Worksheet with Answers

# What is Factorisation?

Representation of an algebraic expression  (  polynomials  )as the product of two or more expressions is called factorisation. Each such expression is called a factor of the given algebraic expression.  These factors may be numbers, algebraic variables or algebraic expressions.

When  the factors  of the polynomial are multiplied together, you will get  the original polynomial.

## Methods of Factoring Polynomials

There are a certain number of methods by which we can factorise polynomials. Let us discuss these methods.

(i) Method of common factors

The first method for factoring polynomials will be factoring out the greatest common factor. It means look at all the terms and determine if there is a factor that is in common to all the terms. If there is, we will factor it out of the polynomial.  For example ,Suppose we have to factorise  $3x+&space;9$ . We shall write each term as a product of irreducible factors;
$3x=&space;3&space;\times&space;x$ ,  9 = 3 x 3 .  Hence $3x+9=&space;(3&space;\times&space;x)&space;+(3&space;\times&space;3)$

Now the  distributive law states that  a(b+c) = ab + bc . By using this law

$3(x+3)=&space;(3\times&space;x)&space;+(3\times&space;3)$

Therefore, we can write  $3x+9&space;=&space;3(x+3)$ . Thus  factors of $3x+9$ are  3 and $x+3$ .

Note that we can always check our factoring by multiplying the terms back out to make sure we get the original polynomial.

Let’s take a look at some more examples.

Example 1:  Factorise    $\mathbf{8x^{4}-4x^{3}+10x^{2}}$ .

Solution :  $8x^{4}=2&space;\times&space;2\times&space;2\times&space;x\times&space;x\times&space;x\times&space;x$

$4x^{3}=&space;2\times&space;2\times&space;x\times&space;x\times&space;x$

$10x^{2}=2&space;\times&space;5\times&space;x\times&space;x$

Thus the common factors are $2,&space;x$ and  $x$   i.e.  $2x^{2}$ .

Therefore $8x^{4}-4x^{3}+10x^{2}=&space;2x^{2}(4x^{2}-2x+5)$ .

Answer: Thus the factors of $8x^{4}-4x^{3}+10x^{2}$  are $2x^{2}$  and $(4x^{2}-2x+5)$ .

Example 2 : Factorise $\mathbf{9xy+18y}$ .

Solution:  $9xy=&space;3\times&space;3&space;\times&space;x\times&space;y$

$18y=&space;2\times&space;3\times&space;3\times&space;y$

Thus the  common factors are 3 , 3 and $y$ i.e. 9y.

Therefore  $9xy+18y=&space;9y(x+2)$.

Answer: The factors of $9xy+18y$ are $9y$ and $(x+2)$ .

Example 3:  Factorise $\mathbf{12a^{2}bc+15ab^{2}c+9abc^{2}}$ .

Solution:  $12a^{2}bc=&space;2\times&space;2\times&space;3\times&space;a\times&space;a\times&space;b\times&space;c$

$15ab^{2}c=3\times&space;5\times&space;a\times&space;b\times&space;b\times&space;c$

$9abc^{2}=3\times&space;3\times&space;a\times&space;b\times&space;c\times&space;c$

Thus the common factors are 3, a, b and c i.e. 3abc.

Therefore $12a^{2}bc+15ab^{2}c+9abc^{2}=3abc(4a+5b+3c)$

Answer: Thus the factors of $12a^{2}bc+15ab^{2}c+9abc^{2}$  are $3abc$  and $(4a+5b+3c)$ .

Example 4: Factorise  $\mathbf{&space;9x^{2}(2x+7)-12x(2x+7)}$.

Solution : $9x^{2}(2x+7)=3&space;\times&space;3\times&space;x\times&space;x\times&space;(2x+7)$

$12x(2x+7)=&space;3&space;\times&space;4\times&space;x\times&space;(2x+7)$

Thus the common factors are $3&space;,&space;x$ and $(2x+7)$ i.e. $3x(2x+7)$.

Therefore  $9x^{2}(2x+7)-12x(2x+7)=3x(2x+7)(3x-4)$ .

Answer :  Factors of $9x^{2}(2x+7)-12x(2x+7)$ are $3x,$ $(2x+7)$ and $(3x-4)$ .

(ii)Factoring By Grouping

Let us try to understand grouping for factorizing with the help of the following examples.

Example 1:  Factorize the polynomial $6xy-4y+6-9x$ using the method of regrouping of factoring a polynomial.

Solution: For factoring polynomials we observe that we have no common factor among all the terms in the expression

$6xy-4y+6-9x$. Let’s try regrouping them as $(6xy&space;-&space;4y)$ and (6-9x).

$(6xy&space;-&space;4y)&space;+&space;(6&space;-&space;9x)&space;=&space;2y(3x&space;-&space;2)-&space;3(3x&space;-&space;2)$

$=&space;(3x&space;-&space;2)(2y&space;-&space;3)$

Answer: Therefore the factors of $6xy-4y+6-9x$ are $(3x-2)$ and $(2y-3)$ .

Example 2:  Factorise $\mathbf{2xy&space;+&space;3x&space;+&space;2y&space;+&space;3&space;.}$

Solution: There is no common factor among all the terms.  Notice that first two terms have a common factor $x$ and the next two

terms have common factor  1.

So, $2xy&space;+&space;3x&space;+&space;2y&space;+&space;3&space;=(2xy&space;+&space;3x)&space;+&space;(2y&space;+&space;3&space;)$

$=x(2y&space;+&space;3)&space;+&space;1(2y&space;+&space;3&space;)$

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

The factors of $2xy&space;+&space;3x&space;+&space;2y&space;+&space;3$ are $(x+1)$ and $(2y&space;+&space;3).$

Example 3 :  Factorise $x^{2}&space;+&space;xy&space;+&space;8x&space;+&space;8y$

Solution: There is no common factors in all the terms of   $x^{2}&space;+&space;xy&space;+&space;8x&space;+&space;8y$  , So Grouping the terms, we have

$x^{2}&space;+&space;xy&space;+&space;8x&space;+&space;8y$ = $x(x&space;+&space;y)&space;+&space;8(x&space;+&space;y)$

$=&space;(x&space;+&space;y)(x&space;+&space;8)$

Hence, the required factors  are  $(x&space;+&space;y)$ and $(x&space;+&space;8)$ .

Example 4: Factorise $\mathbf{15pq&space;+&space;15&space;+&space;9q&space;+&space;25p}$

Solution: Grouping the terms, we have

$15pq&space;+&space;15&space;+&space;9q&space;+&space;25p$ $=&space;(15pq&space;+&space;25p)&space;+&space;(9q&space;+&space;15)$

$=&space;5p(3q&space;+&space;5)&space;+&space;3(3q&space;+&space;5)$

$=&space;(3q&space;+&space;5)&space;(5p&space;+&space;3)$

Hence, the required factors = $(3q&space;+&space;5)$ and  $(5p&space;+&space;3).$

Example 5: Factorise $\mathbf{2x^{2}+8x+3x+12}$.

Solution: $2x^{2}+8x+3x+12$ =$(2x^{2}+8x)+(3x+12)$

=$2x(x+4)+3(x+4)$

= $(2x+3)(x+4)$

Thus the required factors are $(2x+3)$ and $(x+4)$.

(iii) Factorisation using  Algebraic Identities (Factoring Polynomials Formulas)

There are some nice special forms of some polynomials that can make factoring easier for us.  Some of them are given below.

Example 1: Factorise  $\mathbf{x^{2}+6x+9}$ .

Sol.  The polynomial  $x^{2}+6x+9&space;=&space;x^{2}+&space;2&space;\times&space;3\times&space;x&space;+3^{2}$   . So,  it is of the form $a^{2}+2ab+b^{2}$  where $a=&space;x$ and $b=3$.

Now using the identity $(a+b)^{2}=a^{2}+2ab+b^{2}$  we get,

$x^{2}+6x+9&space;=&space;x^{2}+&space;2&space;\times&space;3\times&space;x&space;+3^{2}=(x+3)^{2}$.

Example 2: Factorise $\mathbf{49p^{2}&space;-&space;36}$ .

Solution:  $49p^{2}&space;-49$  can be written as $49p^{2}&space;-36&space;=&space;(7p)^{2}&space;-&space;(6)^{2}$ . The expression is of the form $a^{2}&space;-b^{2}$  where $a=7p$ and $b=6$ ,

Now using the identity  $a^{2}-b^{2}&space;=(a+b)(a-b)$ we get,

$49p^{2}&space;-36&space;=&space;(7p)^{2}&space;-&space;(6)^{2}$ = $(7p+6)(7p-6)$

Example 3: Factorise $\mathbf{x^{4}-81}$ .

Solution:  $x^{4}-81=(x^{2})^{2}-9^{2}$ .

The expression is of the form $a^{2}&space;-b^{2}$  where $a=x^{2}$ and $b=9$ ,

Now using the identity  $a^{2}-b^{2}&space;=(a+b)(a-b)$ we get,

$x^{4}-81=(x^{2})^{2}-9^{2}$ = $(x^{2}+9)(x^{2}-9)$

Now, $(x^{2}+9)$ cannot be factorised further, but ($x^{2}-9$) is factorisable.

Again using the same identity , it follows that

$x^{4}-81=(x^{2}+9)(x^{2}-9)=(x^{2}+9)(x^{2}-3^{2})&space;=(x^{2}+9)(x+3)(x-3)$.

Answer : Therefore  the factors of $x^{4}-81$ are $(x^{2}+9)$ , $(x+3)$ and $(x-3)$ .

Example 4: Factorise $\mathbf{x^{2}-12x+36-y^{2}}$ .

Solution: $x^{2}-12x+36-y^{2}=(x^{2}-2\times&space;6\times&space;x&space;+6^{2})-y^{2}$

=$(x-6)^{2}&space;-y^{2}$            [ Applying Identity $(a-b)^{2}&space;=a^{2}-2ab&space;+b^{2}$ with $a=x&space;\,&space;and&space;\,&space;b=6$]

=$(x-6+y)(x-6-y)$      [Applying Identity $a^{2}-b^{2}&space;=(a+b)(a-b)$ with $a=(x-6)$ and $b=y$]

Answer  : Thus the factors of $x^{2}-12x+36-y^{2}$  are $(x-6+y)$ and $(x-6-y)$.

Example 5: Factorise  $\mathbf{x^{3}-64}$ .

Solution: $x^{3}-64=&space;x^{3}-4^{3}$ .

The expression is of the form $a^{3}-b^{3}$ with $a=x$ and $b=4$ .

Now applying the identity $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, we get

$x^{3}-64=&space;x^{3}-4^{3}=(x-4)(x^{2}+4x+16)$ .

Answer : Therefore the factors of $x^{3}-64$ are $(x-4)$ and $(x^{2}+4x+16)$ .

Example 6: Factorise $\mathbf{x^{2}&space;+&space;5x&space;+&space;6}$ .

Solution : $x^{2}+5x+6=x^{2}+(3+2)x+3\times&space;2$ . Now using the identity $(x+a)(x+b)=x^{2}+(a+b)x+ab$,

we get  $x^{2}+5x+6=x^{2}+(3+2)x+3\times&space;2=(x+3)(x+2)$.

Answer : Factors of $x^{2}+5x+6$  are $(x+3)$ and $(x+2)$ .

### Factoring polynomial worksheet PDF

You might also be interested in:

#### Bina singh

View all posts by Bina singh →
Some tips for mathematics students SSC CHSL 2024 Exam Date ssc chsl 2023 tier 1 cut off NIRF Rankings 2023 : Top 10 Engineering colleges in India CBSE Compartment Exam 2023 Application Form