Exercise 8.4 (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 8.4 Question 1. Multiply the binomials:
(i) $(2 x+5)$ and $(4 x-3)$
(ii) $(y-8)$ and $(3 y-4)$
(iii) $(2.5 l-0.5 m)$ and $(2.5 l+0.5 m)$
(iv) $(a+3 b)$ and $(x+5)$
(v) $\left(2 p q+3 q^2\right)$ and $\left(3 p q-2 q^2\right)$
(vi) $\left(\frac{3}{4} a^2+3 b^2\right)$ and $4\left(a^2-\frac{2}{3} b^2\right)$

Answer.
$
\text { (i) } \begin{aligned}
& (2 x+5) \times(4 x-3) \\
= & 2 x(4 x-3)+5(4 x-3) \\
= & 2 x \times 4 x-2 x \times 3+5 \times 4 x-5 \times 3 \\
= & 8 x^2-6 x+20 x-15 \\
= & 8 x^2+14 x-15
\end{aligned}
$

$\begin{aligned}
&=8 x^2+14 x-15\\
&\begin{aligned}
& \text { (ii) }(y-8) \times(3 y-4)=y(3 y-4)-8(3 y-4) \\
& =y \times 3 y-y \times 4-8 \times 3 y-8 \times-4 \\
&
\end{aligned}
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& =3 y^2-4 y-24 y+32 \\
& =3 y^2-28 y+32
\end{aligned}\\
&\begin{aligned}
& \text { (iii) }(2.5 l-0.5 m) \times(2.5 l+0.5 m) \\
& =2.5 l \times(2.5 l+0.5 m)-0.5 m \times(2.5 l+0.5 m) \\
& =2.5 l \times 2.5 l+2.5 l \times 0.5 m-0.5 m \times 2.5 l-0.5 m \times 0.5 m \\
& =6.25 l^2+1.25 l m-1.25 l m-0.25 m^2 \\
& =6.25 l^2-0.25 m^2
\end{aligned}\\
&\begin{aligned}
& \text { (iv) }(a+3 b) \times(x+5)=a(x+5)+3 b(x+5) \\
& =a \times x+a \times 5+3 b \times x+3 b \times 5 \\
& =a x+5 a+3 b x+15 b
\end{aligned}\\
&\begin{aligned}
& \text { (v) }\left(2 p q+3 q^2\right)\left(3 p q-2 q^2\right) \\
& =2 p q \times\left(3 p q-2 q^2\right)+3 q^2\left(3 p q-2 q^2\right) \\
& =2 p q \times 3 p q-2 p q \times 2 q^2+3 q^2 \times 3 p q-3 q^2 \times 2 q^2 \\
& =6 p^2 q^2-4 p q^3+9 p q^3-6 q^4
\end{aligned}
\end{aligned}$

$\begin{aligned}
& =6 p^2 q^2+5 p q^3-6 q^4 \\
& \text { (vi) }\left(\frac{3}{4} a^2+3 b^2\right) \times 4\left(a^2-\frac{2}{3} b^2\right) \\
& =\left(\frac{3}{4} a^2+3 b^2\right) \times\left(4 a^2-\frac{8}{3} b^2\right) \\
& =\frac{3}{4} a^2 \times\left(4 a^2-\frac{8}{3} b^2\right)+3 b^2 \times\left(4 a^2-\frac{8}{3} b^2\right)
\end{aligned}$

$
\begin{aligned}
& =\frac{3}{4} a^2 \times 4 a^2-\frac{3}{4} a^2 \times \frac{8}{3} b^2+3 b^2 \times 4 a^2-3 b^2 \times \frac{8}{3} b^2 \\
& =3 a^4-2 a^2 b^2+12 a^2 b^2-8 b^4 \\
& =3 a^4+10 a^2 b^2-8 b^4
\end{aligned}
$
Ex 8.4 Question 2.

Find the product:
(i) $(5-2 x)(3+x)$
(ii) $(x+7 y)(7 x-y)$
(iii) $\left(a^2+b\right)\left(a+b^2\right)$
(iv) $\left(p^2-q^2\right)(2 p+q)$

Answer.

(i) $(5-2 x)(3+x)$
$
\begin{aligned}
& =5 \times(3+x)-2 x(3+x) \\
& =5 \times 3+5 \times x-2 \times \times 3-2 x \times x \\
& =15+5 x-6 x-2 x^2=15-x-2 x^2
\end{aligned}
$
(ii) $(x+7 y)(7 x-y)$

$
\begin{aligned}
& =x(7 x-y)+7 y \times(7 x-y) \\
& =x \times 7 x-x \times y+7 y \times 7 x-7 y \times y \\
& =7 x^2-x y+49 x y-7 y^2 \\
& =7 x^2+48 x y-7 y^2
\end{aligned}
$
(iii) $\left(a^2+b\right)\left(a+b^2\right)$

$
\begin{aligned}
& =a^2 \times\left(a+b^2\right)+b \times\left(a+b^2\right) \\
& =a^2 \times a+a^2 \times b^2+b \times a+b \times b^2 \\
& =a^3+a^2 b^2+a b+b^3
\end{aligned}
$
$
\text { (iv) } \begin{aligned}
& \left(p^2-q^2\right)(2 p+q) \\
& =p^2 \times(2 p+q)-q^2(2 p+q) \\
& =p^2 \times 2 p+p^2 \times q-q^2 \times 2 p-q^2 \times q \\
& =2 p^3+p^2 q-2 p q^2-q^3
\end{aligned}
$
Ex 8.4 Question 3.

Simplify:
(i) $\left(x^2-5\right)(x+5)+25$
(ii) $\left(a^2+5\right)\left(b^2+3\right)+5$
(iii) $\left(t+s^2\right)\left(t^2-s\right)$
(iv) $(a+b)(c-d)+(a-b)(c+d)+2(a c+b d)$

(v) $(x+y)(2 x+y)+(x+2 y)(x-y)$
(vi) $(x+y)\left(x^2-x y+y^2\right)$
(vii) $(1.5 x-4 y)(1.5 x+4 y+3)-4.5 x+12 y$
(viii) $(a+b+c)(a+b-c)$

Answer.

(i) $\left(x^2-5\right)(x+5)+25$
$=x^2(x+5)-5(x+5)+25$

$\begin{aligned}
&\begin{aligned}
& =x^2 \times x+x^2 \times 5-5 \times x-5 \times 5+25 \\
& =x^3+5 x^2-5 x-25+25 \\
& =x^3+5 x^2-5 x
\end{aligned}\\
&\begin{aligned}
& \text { (ii) }\left(a^2+5\right)\left(b^3+3\right)+5 \\
& =a^2\left(b^3+3\right)+5\left(b^3+3\right)+5 \\
& =a^2 \times b^3+a^2 \times 3+5 \times b^3+5 \times 3+5 \\
& =a^2 b^3+3 a^2+5 b^3+15+5 \\
& =a^2 b^3+3 a^2+5 b^3+20
\end{aligned}\\
&\begin{aligned}
& \text { (iii) }\left(t+s^2\right)\left(t^2-s\right)=t\left(t^2-s\right)+s^2\left(t^2-s\right) \\
& =t \times t^2-t \times s+s^2 \times t^2-s^2 \times s \\
& =t^3-s t+s^2 t^2-s^3
\end{aligned}\\
&\begin{aligned}
& \text { (iv) }(a+b)(c-d)+(a-b)(c+d)+2(a c+b d) \\
& =a(c-d)+b(c-d)+a(c+d)-b(c+d)+2 a c+2 b d \\
& =a c-a d+b c-b d+a c+a d-b c-b d+2 a c+2 b d
\end{aligned}
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& =a c+a c-a d+a d+b c-b c-b d-b d+2 a c+2 b d \\
& =2 a c-2 b d+2 a c+2 b d \\
& =4 a c
\end{aligned}\\
&\begin{aligned}
& \text { (v) }(x+y)(2 x+y)+(x+2 y)(x-y) \\
& =x(2 x+y)+y(2 x+y)+x(x-y)+2 y(x-y) \\
& =2 x^2+x y+2 x y+y^2+x^2-x y+2 x y-2 y^2
\end{aligned}
\end{aligned}$

$
\begin{aligned}
& =2 x^2+x^2+x y+2 x y-x y+2 x y+y^2-2 y^2 \\
& =3 x^2+4 x y-y^2
\end{aligned}
$
$
\begin{aligned}
& \text { (vi) }(x+y)\left(x^2-x y+y^2\right) \\
& =x\left(x^2-x y+y^2\right)+y\left(x^2-x y+y^2\right) \\
& =x^3-x^2 y+x y^2+x^2 y-x y^2+y^3 \\
& =x^3-x^2 y+x^2 y+x y^2-x y^2+y^3 \\
& =x^3+y^3
\end{aligned}
$
$
\begin{aligned}
& \text { (vii) }(1.5 x-4 y)(1.5 x+4 y+3)-4.5 x+12 y \\
& =1.5 x(1.5 x+4 y+3)-4 y(1.5 x+4 y+3)-4.5 x+12 y \\
& =2.25 x^2+6.0 x y+4.5 x-6.0 x y-16 y^2-12 y-4.5 x+12 y \\
& =2.25 x^2+6.0 x y-6.0 x y+4.5 x-4.5 x-16 y^2-12 y+12 y \\
& =2.25 x^2-16 y^2
\end{aligned}
$
(viii) $(a+b+c)(a+b-c)$
$
=a(a+b-c)+b(a+b-c)+c(a+b-c)
$

$\begin{aligned}
& =a^2+a b-a c+a b+b^2-b c+a c+b c-c^2 \\
& =a^2+a b+a b-a c+a c-b c+b c+b^2-c^2 \\
& =a^2+b^2-c^2+2 a b
\end{aligned}$