Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorize x2 – x then we can do it by taking x common.

x(x – 1) so that x and x –1 are the factors of x2 – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x2 + bx + c = x2 + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x2 + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5

= 6x2 + 2x + 15x + 5

= 2x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Example: 2

Factorize 8x3 + 27y3 + 36x2y + 54xy2

Solution:

The given expression can be written as

= (2x)+ (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)

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

= (2x + 3y)3 

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.

Solution:

4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)+ (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]2  = (2x – y + z)2 = (2x – y + z) (2x – y + z)

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