# Exercise 9.4 Algebraic Expressions and Identities- NCERT Solutions Class 8

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## Chapter 9 Ex.9.4 Question 1

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}

### 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}

## Chapter 9 Ex.9.4 Question 2

Find the product.

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

ii) $$\left( {x{\rm{ }} + {\rm{ }}7y} \right)$$$$\left( {7x{\rm{ }} - {\rm{ }}y} \right)$$

iii) $$\left( {{a^2}{\rm{ }} + {\rm{ }}b} \right)$$ $$\left( {a{\rm{ }} + {\rm{ }}{b^2}} \right)$$

iv) $$\left( {{p^2}{\rm{ }} - {\rm{ }}{q^2}} \right)$$ $$\left( {2p{\rm{ }} + {\rm{ }}q} \right)$$

### Solution

What is known?

Expressions

What is unknown?

Simplification

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) $$\left( {5{\rm{ }} - {\rm{ }}2x} \right)$$ $$\left( {3{\rm{ }} + {\rm{ }}x} \right)$$

\begin{align}&= 5{\rm{ }}\left( {3 + x} \right){\rm{ }} - 2x\left( {3 + x} \right)\\&= {\rm{ }}15 + 5x{\rm{ }} - 6x - 2{x^2}\\&= {\rm{ }}15 - x - 2{x^2}\end{align}

ii) $$\left( {x{\rm{ }} + {\rm{ }}7y} \right)$$$$\left( {7x{\rm{ }} - {\rm{ }}y} \right)$$

\begin{align}&= {\rm{ }}x{\rm{ }}\left( {7x{\rm{ }} - {\rm{ }}y} \right){\rm{ }} + {\rm{ }}7y{\rm{ }}\left( {7x{\rm{ }} - {\rm{ }}y} \right)\\&= {\rm{ }}7{x^2} - xy{\rm{ }} + 49xy{\rm{ }} - 7{y^2}\\&= {\rm{ }}7{x^2} + 48xy{\rm{ }} - 7{y^2}\end{align}

iii) $$\left( {{a^2}{\rm{ }} + {\rm{ }}b} \right)$$ $$\left( {a{\rm{ }} + {\rm{ }}{b^2}} \right)$$

\begin{align}&= {\rm{ }}{a^2}\left( {a + {b^2}} \right){\rm{ }} + {\rm{ }}b\left( {a + {b^2}} \right)\\&= {\rm{ }}{a^3} + {a^2}{b^2} + ab + {b^3}\end{align}

iv) $$\left( {{p^2}{\rm{ }} - {\rm{ }}{q^2}} \right)$$ $$\left( {2p{\rm{ }} + {\rm{ }}q} \right)$$

\begin{align}&= {\rm{ }}{p^2}{\rm{ }}\left( {2p + {\rm{ }}q} \right){\rm{ }} - {\rm{ }}{q^2}\left( {2p{\rm{ }} + {\rm{ }}q} \right)\\&= {\rm{ }}2{p^{3{\rm{ }}}} + {p^2}q - 2p{q^2} - {q^3}\end{align}

## Chapter 9 Ex.9.4 Question 3

Simplify.

i) $$\left( {{x^2} - 5} \right){\rm{ }}\left( {x{\rm{ }} + 5} \right) + 25$$

ii) $$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right){\rm{ }} + 5$$

iii) $$\left( {t + {s^2}} \right)\left( {{t^2} - s} \right)$$

iv) $$\begin{bmatrix} \left( {a{\rm{ }} + {\rm{ }}b} \right)\left( {c{\rm{ }} - {\rm{ }}d} \right) \\+ {\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right)\left( {c{\rm{ }} + {\rm{ }}d} \right)\\+ {\rm{ }}2{\rm{ }}\left( {ac{\rm{ }} + {\rm{ }}bd} \right)\end{bmatrix}$$

v) $$\begin{bmatrix}\left( {x{\rm{ }} + {\rm{ }}y} \right)\left( {2x{\rm{ }} + {\rm{ }}y} \right)\\+ {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}2y} \right)\left( {x{\rm{ }} - {\rm{ }}y} \right)\end{bmatrix}$$

vi) $$\left( {x{\rm{ }} + {\rm{ }}y} \right)\left( {{x^2}{\rm{ }} - {\rm{ }}xy{\rm{ }} + {\rm{ }}{y^2}} \right)$$

vii) $$\begin{bmatrix}\left( {1.5x{\rm{ }} - {\rm{ }}4y} \right)\left( {1.5x{\rm{ }} + {\rm{ }}4y{\rm{ }} + {\rm{ }}3} \right){\rm{ }}\\ - {\rm{ }}4.5x{\rm{ }} + {\rm{ }}12y \end{bmatrix}$$

viii) $$\left( {a{\rm{ }} + b + {\rm{ }}c} \right)\left( {a{\rm{ }} + b{\rm{ }} - {\rm{ }}c} \right)$$

### Solution

What is known?

Expressions

What is unknown?

Simplification

Steps:

i) $$\left( {{x^2} - 5} \right){\rm{ }}\left( {x{\rm{ }} + 5} \right) + 25$$

\begin{align}&= \begin{bmatrix} {x^2}\left( {x + 5} \right) - \\ 5\left( {x + 5} \right) + 25 \end{bmatrix} \\ &= \begin{bmatrix}{x^3} + 5{x^2} - \\ 5x - 25 + 25 \end{bmatrix} \\&={x^3} + 5{x^2} - 5x\end{align}

ii) $$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right){\rm{ }} + 5$$

\begin{align}&=\begin{bmatrix} {a^2}\left( {{b^3} + 3} \right)+ \\ 5\left( {{b^3} + 3} \right) + 5 \end{bmatrix} \\&= \begin{bmatrix} {a^2}{b^3} + 3{a^2} + \\ 5{b^3} + 15 + 5 \end{bmatrix} \\&= \begin{bmatrix} {a^2}{b^3}+ 3{a^2} + \\ 5{b^3} + 20 \end{bmatrix} \end{align}

iii) $$\left( {t + {s^2}} \right)\left( {{t^2} - s} \right)$$

\begin{align}&= {\rm{ }}t{\rm{ }}\left( {{t^2} - s} \right) + {s^2}\left( {{t^2} - s} \right)\\&= {\rm{ }}{t^3} - st + {\rm{ }}{s^2}{t^2} - {s^3}\end{align}

iv)

$$\begin{bmatrix} \left( {a{\rm{ }} + {\rm{ }}b} \right)\left( {c{\rm{ }} - {\rm{ }}d} \right) \\+ {\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right)\left( {c{\rm{ }} + {\rm{ }}d} \right)\\+ {\rm{ }}2{\rm{ }}\left( {ac{\rm{ }} + {\rm{ }}bd} \right)\end{bmatrix}$$

\begin{align}&=\begin{bmatrix}a\left( {c - d} \right)+ b\left( {c - d} \right) \\ + a\left( {c + d} \right) - b\left( {c + d} \right)\\ + 2\left( {ac + bd} \right)\end{bmatrix} \\ &= \begin{bmatrix}ac - ad + bc - bd \\ + ac + ad - bc \\ - bd + 2ac + 2bd \end{bmatrix}\\ &=\begin{bmatrix}\left( {ac + ac+2ac} \right) \\ + \left( {ad - ad} \right) +\left( {bc - bc} \right) \\ + \left( {2bd - bd - bd} \right) \end{bmatrix}\\&= 4ac\end{align}

v)

$$\begin{bmatrix}\left( {x{\rm{ }} + {\rm{ }}y} \right)\left( {2x{\rm{ }} + {\rm{ }}y} \right)\\+ {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}2y} \right)\left( {x{\rm{ }} - {\rm{ }}y} \right)\end{bmatrix}$$

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

vi) $$\left( {x{\rm{ }} + {\rm{ }}y} \right)\left( {{x^2}{\rm{ }} - {\rm{ }}xy{\rm{ }} + {\rm{ }}{y^2}} \right)$$

\begin{align}&=\begin{bmatrix} x\left( {{x^2} -xy + {y^2}} \right) \\ + y \left( {{x^2} - xy + {y^2}} \right)\end{bmatrix} \\&=\begin{bmatrix} {x^3} - {x^2}y + x{y^2} \\ + {x^2}y - x{y^2} + {y^3}\end{bmatrix} \\&=\begin{bmatrix} {x^3} + {y^3} + \\ \left( {x{y^2}- x{y^2}} \right) \\ + \left( {{x^2}y - {x^2}y} \right)\end{bmatrix} \\&= {\rm{ }}{x^3}{\rm{ }} + {y^3}\end{align}

vii)

$$\begin{bmatrix}\left( {1.5x{\rm{ }} - {\rm{ }}4y} \right)\left( {1.5x{\rm{ }} + {\rm{ }}4y{\rm{ }} + {\rm{ }}3} \right){\rm{ }}\\ - {\rm{ }}4.5x{\rm{ }} + {\rm{ }}12y \end{bmatrix}$$

\begin{align}&=\begin{bmatrix}1.5x \left( {1.5x + 4y + 3} \right) \\ - 4y \left( {1.5x + 4y + 3} \right) \\ - 4.5x + 12y\end{bmatrix} \\ &= \begin{bmatrix} 2.25 {x^2} + 6xy + 4.5x \\ - 6xy - 16{y^2} - 12y \\ - 4.5x + 12y \end{bmatrix} \\&=\begin{bmatrix} 2.25 {x^2} + \left( {6xy - 6xy} \right) \\ + \left( {4.5x - 4.5x} \right)- 16{y^2} \\ + \left( {12y - 12y} \right) \end{bmatrix} \\&= 2.25{x^2} - 16{y^2}\end{align}

viii) $$\left( {a{\rm{ }} + b + {\rm{ }}c} \right)\left( {a{\rm{ }} + b{\rm{ }} - {\rm{ }}c} \right)$$

\begin{align}&= \begin{bmatrix} a \left( {a + b - c} \right) \\ + b\left( {a +b - c} \right) \\ + c\left( {a + b - c} \right)\end{bmatrix} \\&= \begin{bmatrix} {a^2} + ab - ac \\ + ab + {b^2} - bc \\ + ca + bc - {c^2}\end{bmatrix}\\&= \begin{bmatrix}{a^2} + {b^2} - {c^2} \\+ \left( {ab + ab} \right) + \left( {bc- bc} \right) \\+ \left( {ca - ca} \right)\end{bmatrix}\\&= {a^{2}} + {b^2} - {c^2} + 2ab\end{align}

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