# Use the identity (x + a)(x + b) = x^{2} + (a + b)x + ab to find the following products

(i) (x + 3)(x + 7) (ii) (4x + 5)(4x + 1) (iii) (4x - 5)(4x -1)

(iv) (4x + 5)(4x -1) (v) (2x + 5y)(2x + 3y)

(vi) (2a^{2} + 9)(2a^{2} + 5) (vii) (xyz - 4)(xyz - 2)

**Solution:**

We will be using the algebraic identity (x + a)(x + b) = x^{2} + (a + b)x + ab to solve the given questions.

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

= x^{2} + (3 + 7)x + (3)(7)

= x^{2} + 10x + 21

(ii) (4x + 5)(4x + 1)

= (4x)^{2} + (5 + 1)(4x) + (5)(1)

= 16x^{2} + 24x + 5

(iii) (4x - 5)(4x - 1)

= (4x)^{2} + [(-5) + (-1)](4x) + (-5) (-1)

= 16x^{2} - 24x + 5

(iv) (4x + 5)(4x -1)

= (4x)^{2} + [(5) + (-1)](4x) + (5)(-1)

= 16x^{2} + 16x - 5

(v) (2x + 5y)(2x + 3y)

= (2x)^{2} + (5y + 3y)(2x) + (5y)(3y)

= 4x^{2} + 16xy + 15y^{2}

(vi) (2a^{2} + 9)(2a^{2} + 5)

= (2a^{2})^{2} + (9 + 5)(2a^{2}) + (9)(5)

= 4a^{4} + 28a^{2} + 45

(vii) (xyz - 4)(xyz - 2)

= (xyz)^{2} + [(-4) + (-2)](xyz) + (-4)(-2)

= x^{2}y^{2}z^{2} - 6xyz + 8

**☛ Check: **NCERT Solutions for Class 8 Maths Chapter 9

**Video Solution:**

## Use the identity (x + a)(x + b) = x² + (a + b)x + ab to find the following products (i) (x + 3)(x + 7) (ii) (4x + 5)(4x + 1) (iii) (4x - 5)(4x -1) (iv) (4x + 5)(4x -1) (v) (2x + 5y)(2x + 3y) (vi) (2a^{2} + 9)(2a^{2} + 5) (vii) (xyz - 4)(xyz - 2)

NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5 Question 2

**Summary:**

Using the identity (x + a)(x + b) = x^{2} + (a + b)x + ab the following products (i) (x + 3)(x + 7) (ii) (4x + 5)(4x + 1) (iii) (4x - 5)(4x -1) (iv) (4x + 5)(4x -1) (v) (2x + 5y)(2x + 3y) (vi) (2a^{2} + 9)(2a^{2} + 5) (vii) (xyz - 4)(xyz - 2) are i) x^{2} + 10x + 21 ii) 16x^{2} + 24x + 5 iii) 16x^{2} - 24x + 5 iv) 16x^{2} + 16x - 5 v) 4x^{2} + 16xy + 15y^{2} vi) 4a^{4} + 28a^{2} + 45 vii) x^{2}y^{2}z^{2} - 6xyz + 8

**☛ Related Questions:**

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