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Home >> Standard Identities & their applications >> (x + a) (x + b) = x2 + x(a + b) + ab >>

(X + A) (X + B)

(a + b)2 = a2 + b2 + 2ab (a - b)2 = a2 + b2 - 2ab a2 - b2 = (a + b) (a b) (x + a) (x + b) = x2 + x(a + b) + ab (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 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Before you understand (x + a) (x + b) = x2 + x(a + b) + ab, you are advised to read:

How to Multiply Variables ?
How to Multiply Polynomials ?
What is Exponential Form ?

How this identity of (x + a) (x + b) = x2 + x(a + b) + ab is obtained

Taking LHS of the identity:
(x + a) (x + b)

Multiply as we do multiplication of two binomials and we get:
= x(x + b) + a(x + b)
= x2 + bx + ax + ab

Since, bring out x as common from bx and ax & we get:
= x2 + x(b + a) + ab

Hence, in this way we obtain the identity i.e. (x + a) (x + b) = x2 + x(a + b) + ab

Following are few applications of identity third:


Example 1: Solve (p + 10) (p + 5)
Solution: This proceeds as:
Given polynomial (p + 10) (p + 5) represents identity third i.e. (x + a) (x + b)
Where x = p, a = 10 and b = 5

Now apply values of x, a and b on the identity i.e. (x + a) (x + b) = x2 + x(a + b) + ab and we get:
(p + 10) (p + 5) = (p)2 + p(10 + 5) + (10 X 5)
= p2 + p(15) + 50
= p2 + 15p + 50

Hence, (p + 10) (p + 5) = p2 + 15p + 50



Example 2: Solve (5c2 + 3) (5c2 + 2)
Solution: This proceeds as:
Given polynomial (5c2 + 3) (5c2 + 2) represents identity third i.e. (x + a) (x + b)
Where x = 5c2, a = 3 and b = 2

Now apply values of x, a and b on the identity i.e. (x + a) (x + b) = x2 + x(a + b) + ab and we get:
(5c2 + 3) (5c2 + 2) = (5c2)2 + 5c2 (3 + 2) + (3 X 2)

= 5c4 + 5c2(5) + 6
= 5c4 + 25c2 + 6

Hence, (5c2 + 3) (5c2 + 2) = 5c4 + 25c2 + 6

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