# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities

Question 3 Exercise 9.5

Use the identity (x+a)(x+b)=x^{2}+(a+b) x+a b 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) \left(2 a^{2}+9\right)\left(2 a^{2}+5\right)

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

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

\begin{array}{l} =x^{2}+(3+7) x+21 \\ =x^{2}+10 x+21 \end{array}

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

\begin{array}{l} =16 \mathrm{x}^{2}+(5+1) 4 \mathrm{x}+5 \\ =16 \mathrm{x}^{2}+24 \mathrm{x}+5 \end{array}

iii) (4x – 5) (4x – 1)

\begin{array}{l} =16 x^{2}+(-5-1) 4 x+5 \\ =16 x^{2}-20 x+5 \end{array}

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

\begin{array}{l} =16 \mathrm{x}^{2}+(5-1) 4 \mathrm{x}-5 \\ =16 \mathrm{x}^{2}+16 \mathrm{x}-5 \end{array}

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

\begin{array}{l} =4 \mathrm{x}^{2}+(5 \mathrm{y}+3 \mathrm{y}) 2 \mathrm{x}+15 \mathrm{y}^{2} \\ =4 \mathrm{x}^{2}+16 \mathrm{xy}+15 \mathrm{y}^{2} \end{array}

\begin{aligned} &\text { vi) }\left(2 a^{2}+9\right)\left(2 a^{2}+5\right)\\ &=4 a^{4}+(9+5) 2 a^{2}+45\\ &=4 a^{4}+28 a^{2}+45 \end{aligned}

\begin{aligned} &\text { vii) }(x y z-4)(x y z-2)\\ &=x^{2} y^{2} z^{2}+(-4-2) x y z+8\\ &=x^{2} y^{2} z^{2}-6 x y z+8 \end{aligned}

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