# Ex.9.5 Q2 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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## Question

Use the identity

\begin{align} &\left( {x + a} \right)\left( {x + b} \right) \\ & = {x^2} + \left( {a + b} \right)x + ab\end{align}

to find the following products.

(i) \begin{align} \quad \left( {x + 3} \right)\left( {x + 7} \right)\end{align}

(ii) \begin{align}\quad {\rm}\left( {4x + 5} \right)\left( {4x + 1} \right)\end{align}

(iiii) \begin{align}\quad \left( {4x - 5} \right)\left( {4x - 1} \right)\end{align}

(iv) \begin{align} \quad {\rm}\left( {4x + 5} \right)\left( {4x - 1} \right)\end{align}

(v) \begin{align}\quad {\rm}\left( {2x + 5y} \right)\left( {2x + 3y} \right)\end{align}

(vi)    \begin{align} \,\left( 2{{a}^{2}}+9 \right)\left( 2{{a}^{2}}+5 \right)\end{align}

(vii) \begin{align} \quad \left( {xyz - 4} \right)\left( {xyz - 2} \right)\end{align}

Video Solution
Algebraic Expressions & Identities
Ex 9.5 | Question 2

## Text Solution

What is known?

\begin{align}& \left( {x + a} \right)\left( {x + b} \right) \\ &= {x^2} + \left( {a + b} \right)x + ab\end{align}

What is unknown?

Simplification

Steps:

The products will be as follows.

(i) \begin{align} \left( {x + 3} \right)\left( {x + 7} \right)\end{align}

\begin{align}&= {x^2} \! + \! \left( {3 + 7} \right)x \! +\! ( 3 )( 7 )\\& = {x^2} + 10x + 21\end{align}

(ii) \begin{align} {\rm}\left( {4x + 5} \right)\left( {4x + 1} \right)\end{align}

\begin{align}&= {\left( {4x} \right)^2} + \left( {5 + 1} \right)\left( {4x} \right) + \left( 5 \right)\left( 1 \right) \\&= 16{x^2} + 24x + 5\end{align}

(iiii) \begin{align} \left( {4x - 5} \right)\left( {4x - 1} \right)\end{align}

\begin{align}&=\begin{bmatrix} {\left( {4x} \right)^2}\! +\! \left[ {\left( { - 5} \right) \!+\! \left( { - 1} \right)} \right] \left( {4x} \right) \\\!+\! \left( { - 5} \right)\left( { - 1} \right)\end{bmatrix} \\&= 16{x^2} - 24x + 5\end{align}

(iv) \begin{align} {\rm}\left( {4x + 5} \right)\left( {4x - 1} \right)\end{align}

\begin{align}&= \begin{bmatrix} {( {4x})^2} \!+\! \left[ {( { + 5} ) \!+ \!( { - 1} )} \right] ( {4x} ) \\\! +\! ( { + 5} )\left( { - 1} \right) \end{bmatrix}\\&= 16{x^2} + 16x - 5\end{align}

(v) \begin{align} {\rm}\left( {2x + 5y} \right)\left( {2x + 3y} \right)\end{align}

\begin{align}&=\begin{bmatrix} {\left( {2x} \right)^{2}} + \left( {5y + 3y} \right)\left( {2x} \right) \\ + \left( {5y} \right)\left( {3y} \right) \end{bmatrix} \\& = 4{x^2} + 16xy + 15{y^2}\end{align}

(vi) \begin{align} \left( 2{{a}^{2}}+9 \right)\left( 2{{a}^{2}}+5 \right)\end{align}

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

(vii) \begin{align} \left( {xyz - 4} \right)\left( {xyz - 2} \right)\end{align}

\begin{align}&=\begin{bmatrix} {{\left( {xyz} \right)}^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right] \\ \left( {xyz} \right) + \left( { - 4} \right)\left( { - 2} \right) \end{bmatrix} \\&= {x^2}{y^2}{z^2} - 6xyz + 8\end{align}

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