In continuation of

**Blog-97**, we will see all the important formulas and useful statements which are to be used in Math test of GRE.### Factorization (Continued):

#### b) Factorization of Polynomial:

Here following 4 possibilities can be studied to understand the factorization in a better way.1) (x + a) (x + b) = x (x + b) + a (x + b)

= x

^{2}+ b x + a x + ab

= x

^{2}+ a x + b x + ab

= x

^{2}+ (a + b) x + ab

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

= x

^{2}- b x + a x - ab

= x

^{2}+ a x - b x - ab

= x

^{2}+ (a - b) x - ab

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

= x

^{2}+ b x - a x - ab

= x

^{2}- a x + b x - ab

= x

^{2}- (a - b) x - ab

Note: Here formulae 2 and 3 are of the same type. the coefficient of the middle term is the difference of the Constance and the sign is to be taken of the greater Constance.

4) (x - a) (x - b) = x (x - b) - a (x - b)

= x

^{2}- b x - a x + ab

= x

^{2}- a x - b x + ab

= x

^{2}- (a + b) x + ab

Now we will study these types in detail:

1) (x + a) (x + b) = x (x + b) + a (x + b)

= x

^{2}+ b x + a x + ab

= x

^{2}+ a x + b x + ab

= x

^{2}+ (a + b) x + ab

Generally we call

**x**

^{2 }as the first term,

**(a + b) x**as the middle term and

**ab**as the last term.

### Basic concept:

a) Step-1: See the sign of the last term.b) Step-2: Here it is "+" so factorize the product of the coefficient of the first term (here it is 1) and the last term in such a way that the

**SUM**of these two factors must be the coefficient of the middle term.

c) Step-3: Get the factors.

### Example-1:

Factorize: x^{2}+ 10 x + 21.

a) Step-1: Here sign of the last term 21 is "+"

b) Step-2: The coefficient of the first term is 1 and the last term is 21, so the product of 1 and 21 is 21. Now the factors of 21 are 3 and 7 and as the sign of the last term is "+", their addition is 3 + 7 = 10 which is the coefficient of the middle term.

= x

^{2}+ 10 x + 21

= x

^{2}+ (3 + 7) x + (3 x 7)

=

__x__

__+__

^{2}+ 3 x__7 x__

__+ (3 x 7)__

=

__x__

__(x + 3)__+

__7 (x__

__+ 3)__

= (x + 3) (x + 7)

c) Step-3: So the factors of x

^{2}+ 10 x + 21 are (x + 3) and (x + 7)

### Example-2:

Factorize: 8 x^{2}+ 14 x + 5.

a) Step-1: Here sign of the last term 5 is "+"

b) Step-2: The coefficient of the first term is 8 and the last term is 5, so the product of 8 and 5 is 8 X 5. Now the factors of 8 X 5 are 2, 2, 2 and 5 and as the sign of the last term is "+", so, we take two factor in such a way that their sum will be 14. Here 2, 2, 2 and 5 will give us 4 and 10. So, here addition is 4 + 10 = 14 which is the coefficient of the middle term.

= 8 x

^{2}+ 14 x + 5

= 8 x

^{2}+ (4 + 10) x + 5

=

__8 x__

__+__

^{2}+ 4 x__10 x__

__+ 5__

=

__4 x__

__(2 x + 1)__+

__5 (2 x__

__+ 1)__

= (4 x + 5) (2 x + 1)

c) Step-3: So the factors of 8 x

^{2}+ 14 x + 5 are (2 x + 1) and (4 x + 5).

In the next part, we will see the remaining 3 types in detail. These 3 types are given below.

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

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

4) (x - a) (x - b) = x

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

^{2}+ (a - b) x - ab3) (x - a) (x + b) = x

^{2}- (a - b) x - ab4) (x - a) (x - b) = x

^{2}- (a + b) x + ab
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