# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.3 in Chapter 9 - Algebraic Expressions and Identities

Question 7 Exercise 9.3

Q4) (a) Simplify 3x\left(4x-5\right)+3 and find values for

(i) x = 3

(ii) x = \frac{1}{2}

(b) Simplify a\left(a^2+a+1\right)+5 and find its value for

(i) a = 0

(ii) a = 1

(iii) a = -1

Solution:

(a) 3x\left(4x-5\right)+3

= 3x\times4x-3x\times5+3

= 12x^2-15x+3

(i) For x = 3, 12x^2-15x+3

= 12\left(3\right)^2- 15 x 3 + 3

= 12 x 9 - 45 + 3

= 108 - 45 + 3

= 66

(ii) For x=\frac{1}{2},12x^2-15x+3

= 12\left(\frac{1}{2}\right)^2-15\times\frac{1}{2}+3

= 12\times\frac{1}{4}-\frac{15}{2}+3

= 6-\frac{15}{2}=\frac{12-15}{2}=\frac{-3}{2}

(b) a\left(a^2+a+1\right)+5

= a\times a^2+a\times a+a\times1+5

= a^3+a^2+a+5

(i) For a = 0, a^3+a^2+a+5

= \left(0\right)^3+\left(0\right)^2+\left(0\right)+5

= 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a^3+a^2+a+5

= \left(1\right)^3+\left(1\right)^2+\left(1\right)+5

= 1 + 1 + 1 + 5 = 8

(iii) For a = -1, a^3+a^2+a+5

= \left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+5

= -1+1-1+5

= -2+6

= 4

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