# Ex.9.4 Q1 Algebraic Expressions and Identities - NCERT Maths Class 8

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## Question

Multiply the binomials.

(i) $$\left( {2x + 5} \right)$$ and $$\left( {4x - 3} \right)$$

(ii) $$\left( {y - 8} \right)$$ and $$\left( {3y - 4} \right)$$

(iii) $$\left( {2.5l{\rm{ }} - {\rm{ }}0.5 m} \right)$$ and  $$\left( {2.5l{\rm{ }} + {\rm{ }}0.5m} \right)$$

(iv) $$\left( {a + 3b} \right)$$ and $${\rm{ }}\left( {x + 5} \right)$$

(v) $$\left( {2pq + 3{q^2}} \right)$$ and $$\left( {3pq - 2{q^2}} \right)$$

(vi) \begin{align}\left( \frac{3}{4}{{a}^{2}}+3{{b}^{2}} \right)\end{align} and \begin{align}\left[ 4\left( {{a}^{2}}-\frac{2}{3}{{b}^{2}} \right) \right]\end{align}

Video Solution
Algebraic Expressions & Identities
Ex 9.4 | Question 1

## Text Solution

What is known?

Expressions

What is unknown?

Multiplication

Reasoning:

i) By using the distributive law, we can carry out the multiplication term by term.

ii) In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.

Steps:

(i)$$( 2x+5 ) \times ( 4x-3 )$$

\begin{align} &= 2x \!\times \! \left( 4x-3 \right) \! + \!5\!\times\! \left( 4x\!-\!3 \right) \\ &=8{{x}^{2}}-6x+20x-15 \\&=8{{x}^{2}}+14x-15\\ &\quad \left[ \text{By adding } \text{like terms} \right]\\\end{align}

(ii) $$\left( {y - 8} \right)$$  $$\times$$  $$\left( {3y - 4} \right)$$

\begin{align} &= y \!\times(\! 3y-4 ) \! -8 \!\times \! (3y-4 ) \\ &=3{{y}^{2}}-4y\text{ }-24y+32 \\&=3{{y}^{2}}-28y+32 \\& \quad [ \text{By adding } \text{like terms}] \\\end{align}

(iii) $$\left( {2.5l{\rm{ }} - {\rm{ }}0.5 m} \right)$$  $$\times$$   $$\left( {2.5l{\rm{ }} + {\rm{ }}0.5m} \right)$$

\begin{align}& = \begin{bmatrix} 2.5l\times \left( 2.5l+0.5m \right)-\\ 0.5m\left( 2.5l+0.5m \right) \end{bmatrix} \\ &= \begin{bmatrix} 6.25{{l}^{2}}+1.25lm - \\ 1.25lm-0.25{{m}^{2}} \end{bmatrix} \\&=6.25{{l}^{2}}-0.25{{m}^{2}} \\\end{align}

(iv) $$\left( {a + 3b} \right)$$ $$\times$$ $${\rm{ }}\left( {x + 5} \right)$$

\begin{align}&= a \!\times\! \left( x \!+5 \right) \! +3b \! \times\! \left( x\! + \!5 \right) \\&=ax+5a+3bx+15b \\\end{align}

(v) $$\left( {2pq + 3{q^2}} \right)$$ $$\times$$ $$\left( {3pq - 2{q^2}} \right)$$

\begin{align}& = \begin{bmatrix} 2pq\times \left( 3pq-2{{q}^{2}} \right)\\ +3{{q}^{2}}\times \left( 3pq-2{{q}^{2}} \right) \end{bmatrix} \\ &= 6{{p}^{2}}{{q}^{2}}\!-\!\text{ }4p{{q}^{3}} \!+\!9p{{q}^{3}}\!-\!6{{q}^{4}}\\&=6{{p}^{2}}{{q}^{2}}\text{ }+\text{ }5p{{q}^{3}}-6{{q}^{4}} \\\end{align}

(vi) \begin{align}\left( \frac{3}{4}{{a}^{2}}+3{{b}^{2}} \right)\times \left[ 4\left( {{a}^{2}}-\frac{2}{3}{{b}^{2}} \right) \right]\end{align}

\begin{align}&= \left( {\frac{3}{4}{a^2} \!+\! 3{b^2}} \right)\! \! \times \!\!\left( {4{a^2} \!- \!\frac{8}{3}{b^2}} \right) \\&=\begin{bmatrix} \frac{3}{4}{a^2} \times \left( {4{a^2} - \frac{8}{3}{b^2}} \right)+ \\ 3{b^2} \times \left( {4{a^2} - \frac{8}{3}{b^2}} \right) \end{bmatrix} \\&= \begin{bmatrix} \left( {\frac{3}{\not{4}}{a^2} \times {\not{4}}{a^2}} \right)-\\ \left( {\frac{\not{3}}{\not{4}}{a^2} \times \frac{{{\not{8}^2}}}{\not{3}}{b^2}} \right)+ \\ \left( {3{b^2} \times 4{a^2}} \right)- \\ \left( \not3{{b^2} \times \frac{8}{\not{3}}{b^2}} \right) \end{bmatrix} \\&=\begin{bmatrix} 3{a^4} - 2{b^2}{a^2} +\\ 12{b^2}{a^2} - 8{b^4} \end{bmatrix} \\&= 3{a^4} + 10{a^2}{b^2} - 8{b^4}\end{align}

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