Question

Using the identity (x + a)(x + b) – x² + x(a + b) + ab, find the following product.

(i) (x + 3) (x + 7)

(ii) (6a + 9) (6a – 5)

(iii) (4x + 3y) (4x + 5y)

(iv) (8 + pq) (pq + 7)

Answer 1

(i) (x + 3) (x + 7)

Let a = 3; b = 7, then

(x + 3) (x + 7) is of the form x² + x (a + b) + ab

(x + 3) (x + 7) = x² + x (3 + 7) + (3 x 7) 

= x² + 10x + 21

(ii) (6a + 9) (6a – 5)

Substituting x = 6a; a = 9 and b = -5

In (x + a) (x + b) = x² + x (a + b) + ab, we get

(6a + 9)(6a – 5) = (6a)² + 6a (9 + (-5)) + (9 x (-5))

6²a² + 6a (4) + (-45) = 36a² + 24a – 45

(6a + 9) (6a – 5) = 36a² + 24a – 45

(iii) (4x + 3y) (4x + 5y)

Substituting x = 4x; a = 3y and b = 5y in

(x + a) (x + b) = x² + x (a + b) + ab, we get

(4x + 3y) (4x – 5y) = (4x)² + 4x (3y + 5y) + (3y) (5y)

= 4² x² + 4x (8y) + 15y² = 16x² + 32xy + 15y²

(4x + 3y) (4x + 5y) = 16x² + 32xy + 15y²

(iv) (8 + pq) (pq + 7)

Substituting x = pq; a = 8 and b = 7 in

(x + a) (x + b) = x² + x (a + b) + ab, we get

(pq + 8) (pq + 7) = (pq)² + pq (8 + 7) + (8) (7)

= p² q² + pq (15) + 56

(8 + pq) (pq + 7) = p² q² + 15pq + 56