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) (2a² + 9)(2a² + 5)   (vii) (xyz - 4)(xyz - 2)

Solution:

We will be using the algebraic identity (x + a)(x + b) = x² + (a + b)x + ab to solve the given questions.

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

= x² + (3 + 7)x + (3)(7) 

= x² + 10x + 21

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

= (4x)² + (5 + 1)(4x) + (5)(1)

= 16x² + 24x + 5

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

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

= 16x² - 24x + 5

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

= (4x)² + [(5) + (-1)](4x) + (5)(-1)

= 16x² + 16x - 5

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

= (2x)² + (5y + 3y)(2x) + (5y)(3y)

= 4x² + 16xy + 15y²

(vi) (2a² + 9)(2a² + 5)

= (2a²)² + (9 + 5)(2a²) + (9)(5)

= 4a⁴ + 28a² + 45

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

= (xyz)² + [(-4) + (-2)](xyz) + (-4)(-2)

= x²y²z² - 6xyz + 8

☛ Check: NCERT Solutions for Class 8 Maths Chapter 9

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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) (2a² + 9)(2a² + 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) = x² + (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) (2a² + 9)(2a² + 5) (vii) (xyz - 4)(xyz - 2) are i) x² + 10x + 21 ii) 16x² + 24x + 5 iii) 16x² - 24x + 5 iv) 16x² + 16x - 5 v) 4x² + 16xy + 15y² vi) 4a⁴ + 28a² + 45 vii) x²y²z² - 6xyz + 8

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