(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