## X PLUS A X PLUS B FORMULA

1. (x+a) (x+b)= x² + x (a+b) + ab

Let us see the explanation of the above formula. For that let us use distributive property.

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

= x² + xb + ax +ab

=  x² + x(a+b) + ab [by combining﻿ like terms﻿]

2. (x-a)(x-b) = x² - x(a+b) + ab

Like the above formula we can explain this using distributive property.

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

= x² - x(a+b) + ab [by combining like terms]

### x plus a x plus b formula -examples

The following examples are based on algebraic identity.

Question 1 :

Expand (3x+5) (3x+8)

Solution:

Here the question is in the form of (x+a)(x+b) . Instead of a we have"5" and instead of b we have "8" . Now we need to apply the formula (x ² + (a+b)x + ab ) and we need to apply those values instead of a and b

(3x+5) (3x+8)    = (3x)² + (5 + 8) x + (5 x 8)

= 9 x² + 13 x + 40

Question 2 :

Expand (x + 1) (x - 3)

Solution:

Here the question is in the form of (x+a)(x+b) . Instead of a we have"1" and instead of b we have "-3" . Now we need to apply the formula (x² + (a+b)x + ab ) and we need to apply those values instead of a and b

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

= x² - 2 x - 3

Question 3 :

Expand (x-2)(x-3)

Solution:

Here the question is in the form of (x-a)(x-b) . Instead of a we have "2" and instead of b we have "3" . Now we need to apply the formula x² - (a + b) x + ab and we need to apply those values instead of a and b .

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

= x² - 5 x + 6 (a - b)² = a² - 2 ab + b²

a² - b² = (a + b) (a - b)

(x+a)(x+b)=x²+(a+b)x+ab

(a+b)³=a³+3a²b+3ab²+b³

(a-b)³=a³-3a²b+3ab²-b³

(a³+b³)= (a+b)(a²-ab+b²)

(a³-b³)=(a-b)(a²+ab+ b²)

(a+b+c)²= a²+b²+c²+2ab+2bc+2ca