# Factorization of Polynomials – For class 8, 9

Factorization of Polynomials

## Factorization

The process of finding two or more factors of a given expression is called factorization.

For example, find the factor of (6p2y + 3py)

= 6p2y + 3py

= 2 x 3 x p x p x y + 3 x p x y

= 3py (2p+1)

Hence, 3py and (2p+1) are the factors of (6p2y + 3py).

### Method of Factorization of polynomials

There are 6 methods of Factorization:

• Common method
• Grouping method
• Identity method
• Middle splitting term method
• Completing square method
• Trial and error method

1. #### Common Method of Factorization of polynomials

Find common constant and variable of each term of polynomial.

Example: Factorize

1. 4a3b2c – 2ab2c

= 2x2xaxaxaxbxbxc – 2xaxbxbxc

= 2ab2c (2a2-1)

1. 18x2y – 24xyz

= 6xy (3x-4z)

1. 4a(x+y) – 3b(x+y)

= (x+y) (4a-3b)

1. 8(3a-2b)2 – 10(3a-2b)

= 2x2x2x (3a-2b)x(3a-2b) – 2x5x(3a-2b)

= 2(3a-2b) {4(3a-2b) – 5}

= 2(3a-2b) (12a-8b-5)

1. 8(a+b) – 6(a+b)2

= 2x2x2x(a+b) – 2x3x(a+b)x(a+b)

= 2(a+b) {4 – 3(a+b)}

= 2(a+b) (4-3a-3b)

#### 2. Grouping Method of Factorization

Minimum four term of polynomial is required for grouping method of polynomial.

Example: Factorize

1. X3+2X2+5X+10

= (X3+2X2) + (5X+10)

= X2(X+2) + 5(X+2)

= (X+2) (X2+5)

1. Px-5q+pq-5x

= (px+pq) – (5q+5x)

= p(x+q) – 5(q+x)

= p(x+q) – 5(x+q)

= (x+q) (p-5)

1. (3a-1)2 – 6a+2

= (3a-1)x(3a-1) – 2(3a-1)

= (3a-1) {(3a-1) – 2}

= (3a-1) (3a-3)

= 3(3a-1) (a-1)

1. 2a2+bc-2ab-ac

= (2a2-2ab) + (bc-ac)

= 2a(a-b) + c(b-a)

= 2a(a-b) – c(a-b)

= (a-b) (2a-c)

1. X2 + 1/X2 -2 -3X +3/X

= (X2+1/X2-2) – 3(X-1/X)

= (X-1/X)2 – 3(X-1/X)

= (X-1/X) (X-1/X -3)

#### 3. Identity method of Factorization

Some important identities:

1. (a+b)2      = a2+b2+2ab
2. (a-b)2       = a2+b2-2ab
3. a2-b2         = (a+b)(a-b)
4. (a+b+c)2   = a2+b2+c2+2ab+2bc+2ca
5. (a+b)3       = a3+b3+3ab(a+b)
6. (a-b)3        = a3-b3-3ab(a-b)
7. a3+b3        = (a+b)(a2-ab+b2)
8. a3-b3         = (a-b) (a2+ab+b2)
9. (a3+b3+c3-3abc)  = (a+b+c)(a2+b2+c2-ab-bc-ca)
10. If (a+b+c)=0 then (a3+b3+c3) = 3abc

Factorize:

1. (2a+b)2

We know that (a+b)2 = a2+b2+2ab

= (2a)2+b2+2x2axb

= 4a2+b2+4ab

1. (3x-2y)2

We know that (a-b)2=a2+b2-2ab

= (3x)2+(2y)2-2X(3x)X(2y)

= 9x2+4y2-12xy

1. 16x2-25y2

We know that a2-b2=(a+b)(a-b)

= (4x)2-(5y)2

= (4x+5y) (4x-5y)

1. 150-6x2

= 6(25-x2)

= 6 [(5)2-(x)2]

= 6 (5+x)(5-x)

1. x4-625

= (x2)2 – (25)2

= (x2+25) (x2-25)

= (x2+25) (x+5) (x-5)

1. (a+2b+5c)2

We know that (a+b+c)2=a2+b2+c2+2ab+2bc+2ca

= a2+4b2+25c2+4ab+20bc+10ca

1. (2a-5b-7c)2

= (2a)2+(-5b)2+(-7c)2+(2x2ax-5b)+(2x-5bx-7c)+(2x-7cx2a)

= 4a2+25b2+49c2-20ab+70bc-28ca

1. (3x+2)3

We know that (a+b)3=a3+b3+3a2b+3ab2

OR

(a+b)3=a3+b3+3ab(a+b)

= (3x)3+23+3(3x)2x2+3x(3x)x22

= 27x3+8+54x2+36x

1. (5a-3b)3

= (5a)3-(3b)3+3x(5a)2x(-3b)+3x(5a)x(-3b)2

=125a3-27b3-225a2b+135ab2

1. 64a3+343

We know that a3+b3= (a+b)(a2+b2-ab)

= (4a)3+73

= (4a+7)(16a2+49-28a)

1. 32a4-500a

= 4a(8a3-125)

= 4a [(2a)3-53]

= 4a (2a-5)(4a2+25+10a)

1. 125a3+b3+64c3-60abc

We know that a3+b3+c3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ca)

= (5a)3+b3+(4c)3-3x(5a)xbx(4c)

= (5a+b+4c)(25a2+b2+16c2-5ab-4bc-20ca)

1. (3a-2b)3+(2b-5c)3+(5c-3a)3

We know that, If a+b+c=0 then a3+b3+c3=3abc

= (3a-2b)+(2b-5c)+(5c-3a)

= 3a-2b+2b-5c+5c-3a

= 0

Then,

(3a-2b)3+(2b-5c)3+(5c-3a)3 = 3(3a-2b)(2b-5c)(5c-3a)

#### 4. Middle splitting term method of factorization

Example: Factorize

1. x2+11x+30

= x2+5x+6x+30

= (x2+5x) + (6x+30)

= x(x+5) + 6(x+5)

= (x+5)(x+6)

1. x2-4x+3

= x2-3x-x+3

=(x2-3x) – (x-3)

= x(x-3) -1(x-3)

= (x-3)(x-1)

#### 5. Completing Square method of Factorisation

Example: Factorize

1. y2+6y+8

= y2+2xyx3+32-32+8

= (y2+6y+32) -9+8

= (y+3)2 – 12

= (y+3+1)(y+3-1)

= (y+4)(y+2)

1. 3m2+24m+36

= 3(m2+8m+12)

= 3{m2+2xmx4+42-42+12}

= 3{(m2+8m+42) -16+12}

= 3{(m+4)2 -22}

= 3{(m+4+2)(m+4-2)}

=3(m+6)(m-2)

I hope you like this article “Factorization of Polynomials”. Comment below for other chapters.

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