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 $3x+ 9$ . We shall write each term as a product of irreducible factors;
$3x= 3 \times x$ , 9 = 3 x 3 . Hence $3x+9= (3 \times x) +(3 \times 3)$

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

$3(x+3)= (3\times x) +(3\times 3)$

Therefore, we can write $3x+9 = 3(x+3)$ . 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 : $8x^{4}=2 \times 2\times 2\times x\times x\times x\times x$

$4x^{3}= 2\times 2\times x\times x\times x$

$10x^{2}=2 \times 5\times x\times x$

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

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

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: $9xy= 3\times 3 \times x\times y$

$18y= 2\times 3\times 3\times y$

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

Therefore $9xy+18y= 9y(x+2)$ .

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

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

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

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

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

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 $\mathbf{ 9x^{2}(2x+7)-12x(2x+7)}$ .

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

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

Thus the common factors are $3 , x$ 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 $(6xy - 4y)$ and (6-9x).

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

$= (3x - 2)(2y - 3)$

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

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

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, $2xy + 3x + 2y + 3 =(2xy + 3x) + (2y + 3 )$

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

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

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

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

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

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

$= (x + y)(x + 8)$

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

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

Solution: Grouping the terms, we have

$15pq + 15 + 9q + 25p$ $= (15pq + 25p) + (9q + 15)$

$= 5p(3q + 5) + 3(3q + 5)$

$= (3q + 5) (5p + 3)$

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

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.

factoring polynomials formula

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

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

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

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

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

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