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
Common Method of Factorization of polynomials
Find common constant and variable of each term of polynomial.
Example: Factorize
- 4a3b2c – 2ab2c
= 2x2xaxaxaxbxbxc – 2xaxbxbxc
= 2ab2c (2a2-1)
- 18x2y – 24xyz
= 6xy (3x-4z)
- 4a(x+y) – 3b(x+y)
= (x+y) (4a-3b)
- 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)
- 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
- X3+2X2+5X+10
= (X3+2X2) + (5X+10)
= X2(X+2) + 5(X+2)
= (X+2) (X2+5)
- Px-5q+pq-5x
= (px+pq) – (5q+5x)
= p(x+q) – 5(q+x)
= p(x+q) – 5(x+q)
= (x+q) (p-5)
- (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)
- 2a2+bc-2ab-ac
= (2a2-2ab) + (bc-ac)
= 2a(a-b) + c(b-a)
= 2a(a-b) – c(a-b)
= (a-b) (2a-c)
- 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:
- (a+b)2 = a2+b2+2ab
- (a-b)2 = a2+b2-2ab
- a2-b2 = (a+b)(a-b)
- (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
- (a+b)3 = a3+b3+3ab(a+b)
- (a-b)3 = a3-b3-3ab(a-b)
- a3+b3 = (a+b)(a2-ab+b2)
- a3-b3 = (a-b) (a2+ab+b2)
- (a3+b3+c3-3abc) = (a+b+c)(a2+b2+c2-ab-bc-ca)
- If (a+b+c)=0 then (a3+b3+c3) = 3abc
Factorize:
- (2a+b)2
We know that (a+b)2 = a2+b2+2ab
= (2a)2+b2+2x2axb
= 4a2+b2+4ab
- (3x-2y)2
We know that (a-b)2=a2+b2-2ab
= (3x)2+(2y)2-2X(3x)X(2y)
= 9x2+4y2-12xy
- 16x2-25y2
We know that a2-b2=(a+b)(a-b)
= (4x)2-(5y)2
= (4x+5y) (4x-5y)
- 150-6x2
= 6(25-x2)
= 6 [(5)2-(x)2]
= 6 (5+x)(5-x)
- x4-625
= (x2)2 – (25)2
= (x2+25) (x2-25)
= (x2+25) (x+5) (x-5)
- (a+2b+5c)2
We know that (a+b+c)2=a2+b2+c2+2ab+2bc+2ca
= a2+4b2+25c2+4ab+20bc+10ca
- (2a-5b-7c)2
= (2a)2+(-5b)2+(-7c)2+(2x2ax-5b)+(2x-5bx-7c)+(2x-7cx2a)
= 4a2+25b2+49c2-20ab+70bc-28ca
- (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
- (5a-3b)3
= (5a)3-(3b)3+3x(5a)2x(-3b)+3x(5a)x(-3b)2
=125a3-27b3-225a2b+135ab2
- 64a3+343
We know that a3+b3= (a+b)(a2+b2-ab)
= (4a)3+73
= (4a+7)(16a2+49-28a)
- 32a4-500a
= 4a(8a3-125)
= 4a [(2a)3-53]
= 4a (2a-5)(4a2+25+10a)
- 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)
- (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
- x2+11x+30
= x2+5x+6x+30
= (x2+5x) + (6x+30)
= x(x+5) + 6(x+5)
= (x+5)(x+6)
- 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
- 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)
- 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)
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