Exercise 8.3 (Revised) - Chapter 9 - Algebraic Expressions & Identities - Ncert Solutions class 8 - Maths

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Chapter 8 - Algebraic Expressions & Identities - NCERT Solutions for Class 8 Maths

Ex 8.3 Question 1.

Carry out the multiplication of the expressions in each of the following pairs:
(i) $4 p: q+r$
(ii) $a b, a-b$
(iii) $a+b, 7 a^2 b^2$
(iv) $a^2-9=4 a$
(v) $p q+q r+r p=0$

Answer.
(i) $4 p \times(q+\gamma)=4 p \times q+4 p \times \gamma$
$
=4 p q+4 p r
$
(ii) $a b \times(a-b)=a b \times a-a b \times b$
$
=a^2 b-a b^2
$
(iii) $(a+b) \times 7 a^2 b^2=a \times 7 a^2 b^2+b \times 7 a^2 b^2=7 a^3 b^2+7 a^2 b^3$
(iv) $\left(a^2-9\right) \times 4 a=a^2 \times 4 a-4 a \times 9=4 a^3-36 a$
(v) $(p q+q r+r p) \times 0=p q \times 0+q r \times 0+r p \times 0$
$
=0+0+0=0
$

Ex 8.3 Question 2.

Complete the table: 

Answer:

Ex 8.3 Question 3.

Find the product:
(i) $\left(a^2\right) \times\left(2 a^{22}\right) \times\left(4 a^{26}\right)$
(ii) $\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)$
(iii) $\left(\frac{-10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)$
(iv) $x \times x^2 \times x^3 \times x^4$

Answer.
$
\begin{aligned}
& \text { (i) }\left(a^2\right) \times\left(2 a^{22}\right) \times\left(4 a^{26}\right) \\
& =(2 \times 4)\left(a^2 \times a^{22} \times a^{25}\right) \\
& =8 \times a^{2+22+26}=8 a^{50}
\end{aligned}
$
(ii) $\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)$
$
=\left(\frac{2}{3} \times \frac{-9}{10}\right)\left(x \times x^2 \times y \times y^2\right)
$

$\begin{aligned}
&=\frac{-3}{5} x^3 y^3\\
&\begin{aligned}
& \text { (iii) }\left(\frac{-10}{3} p q^3\right)\left(\frac{6}{5} p^3 q\right) \\
& =\left(\frac{-10}{3} \times \frac{6}{5}\right)\left(p \times p^3 \times q^3 \times q\right) \\
& =-4 p^4 q^4
\end{aligned}
\end{aligned}$

(iv) $x \times x^2 \times x^3 \times x^4=x^{1+2+3+4}=x^{10}$
Ex 8.3 Question 4.

(a) Simplify: $3 x(4 x-5)+3$ 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) $a=-1$.

Answer.

(a) $3 x(4 x-5)+3$
$
\begin{aligned}
& =3 x \times 4 x-3 x \times 5+3 \\
& =12 x^2-15 x+3
\end{aligned}
$
(i) For $x=3,12 x^2+15 x+3$
$
=12(3)^2-15 \times 3+3
$

$\begin{aligned}
&=12 \times 9-45+3=108-45+3=66\\
&\begin{aligned}
& \text { (ii) For } x=\frac{1}{2}=12 x^2-15 x+3 \\
& =12\left(\frac{1}{2}\right)^2-15 \times \frac{1}{2}+3 \\
& =12 \times \frac{1}{4}-\frac{15}{2}+3
\end{aligned}
\end{aligned}$

$
\begin{aligned}
&=6-\frac{15}{2}=\frac{12-15}{2}=\frac{-3}{2}\\
&\begin{aligned}
& \text { (b) } a\left(a^2+a+1\right)+5 \\
& =a \times a^2+a \times a+a \times 1+5 \\
& =a^3+a^2+a+5
\end{aligned}\\
&\begin{aligned}
& \text { (i) For } a=0: a^3+a^2+a+5 \\
& =(0)^3+(0)^2+(0)+5 \\
& =0+0+0+5=5 \\
& \text { (ii) For } a=1=a^3+a^2+a+5 \\
& =(1)^3+(1)^2+(1)+5 \\
& =1+1+1+5=8
\end{aligned}\\
&\begin{aligned}
& \text { (ii) For } a=1: a^3+a^2+a+5 \\
& =(1)^3+(1)^2+(1)+5 \\
& =1+1+1+5=8
\end{aligned}\\
&\begin{aligned}
& \text { (iii) For } a=-1 \cdot a^3+a^2+a+5 \\
& =(-1)^3+(-1)^2+(-1)+5 \\
& =-1+1-1+5=-2+6=4
\end{aligned}
\end{aligned}
$

Ex 8.3 Question 5.

(a) Add: $p(p-q), q(q-r)$ and $\mu(\mu-p)$.
(b) Add: $2 x(z-x-y)$ and $2 y(z-y-z x)$.
(c) Subtract: $3 l(l-4 m+5 n)$ from $4 l(10 n-3 m+2 l)$.
(d) Subtract: $3 a(a+b+c)-2 b(a-b+c)$ from $4 c(-a+b+c)$.

Answer.

(a) $p(p-q)+q(q-r)+\mu(\mu-p)$

$\begin{aligned}
&\begin{aligned}
& =p^2-p q+q^2-q r+r^2-r p \\
& =p^2+q^2+r^2-p q-q r-r p
\end{aligned}\\
&\begin{aligned}
& \text { (b) } 2 x(z-x-y)+2 y(z-y-x) \\
& =2 x z-2 x^2-2 x y+2 y z-2 y^2-2 x y \\
& =2 x z-2 x y-2 x y+2 y z-2 x^2-2 y^2 \\
& =-2 x^2-2 y^2-4 x y+2 y z+2 z x
\end{aligned}\\
&\begin{aligned}
& \text { (c) } 4 l(10 n-3 m+2 l)-3 l(l-4 m+5 n) \\
& =40 l n-12 l m+8 l^2-3 l^2+12 l m-15 l n \\
& =8 l^2-3 l^2-12 l m+12 m+40 l n-15 m \\
& =5 l^2+25 l n
\end{aligned}\\
&\begin{aligned}
& \text { (d) } 4 c(-a+b+c)-[3 a(a+b+c)-2 b(a-b+c)] \\
& =-4 a c+4 b c+4 c^2-\left[3 a^2+3 a b+3 a c-2 a b+2 b^2-2 b c\right] \\
& =-4 a c+4 b c+4 c^2-\left[3 a^2+2 b^2+3 a b-2 b c+3 a c-2 a b\right]
\end{aligned}
\end{aligned}$

$\begin{aligned}
& =-4 a c+4 b c+4 c^2-\left[3 a^2+2 b^2+a b+3 a c-2 b c\right] \\
& =-4 a c+4 b c+4 c^2-3 a^2-2 b^2-a b-3 a c+2 b c \\
& =-3 a^2-2 b^2+4 c^2-a b+4 b c+2 b c-4 a c-3 a c \\
& =-3 a^2-2 b^2+4 c^2-a b+6 b c-7 a c
\end{aligned}$