Question

USE THE IDENTITY
$(x + a)(x + b) = {x }^{2} + (a + b)x + ab$
TO FIND THE FOLLOWING PRODUCTS.

CLASS 8TH
CHAPTER 9= ALGEBRAIC EXPRESSIONS AND IDENTITIES​

Answer 1

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Answer 2

Answer:

$(i) \: a = 3 \: \: and \: \: b = 7 \\ \\ = > {x}^{2} + 10x + 21 = 0 \\ \\ (ii) \: a = 5 \: \: \: and \: \: \: b = 1 \\ \\ x \: \: replaced \: \: by \: \: 4x \\ \\ = > 16 {x}^{2} + 24x + 5 \\ \\ (iii) \: \: a = - 5 \: \: and \: \: b = - 1 \\ \\ x \: \: replaced \: \: by \: \: 4x \\ \\ = > 16 {x }^{2} - 24x + 5 = 0 \\ \\ (iv) \: \: a = 5 \: \: b = - 1 \\ \\ x \: \: \: \: replaced \: \: by \: \: 4x \\ \\ = > 16 {x}^{2} - 16x - 5 = 0 \\ \\ (v) \: \: a = 5y \: \: and \: \: a = 3y \\ \\ x \: \: replaced \: \: by \: \: 2x \\ \\ = > 4 {x}^{2} + 16xy + 15 {y}^{2} \\ \\ (vi) \: \: a = 9 \: \: and \: \: b = 5 \\ \\ x \: \: replaced \: \: by \: \: 2 {a}^{2} \\ \\ 4 {a}^{4} + 28 {a}^{2} + 45 = 0 \\ \\ (vii) \: \: a = - 4 \: \: and \: \: a = - 2 \\ \\ x \: \: replaced \: \: by \: \: xyz \\ \\ = > {(xyz)}^{2} - 6xyz + 8 = 0$