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.
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Factorization (Continued):

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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.

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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.

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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)

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

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

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