# Algebraic Expressions And Identities : Exercise 9.3 (Mathematics NCERT Class 8th)

Q.1 Carry out the multiplication of the expressions in each of the following pairs.

(i) $4p,q + r$ (ii) $ab,a - b$ (iii) $a + b,7{a^2}{b^2}$ (iv) ${a^2} - 9,4a$

(v) $pq + qr + rp,0$

Sol. (i) $4p,q + r$

$4p \times (q + r) = 4p \times q + 4p \times r = 4pq + 4pr$

(ii) $ab,a - b$

$ab \times (a - b) = ab \times a - ab \times b = {a^2}b + a{b^2}$

(iii) $a + b,7{a^2}{b^2}$

$(a + b) \times 7{a^2}{b^2} = a \times 7{a^2}{b^2} + b \times 7{a^2}{b^2} = 7{a^3}{b^2} + 7{a^2}{b^3}$

(iv) ${a^2} - 9,4a$

$({a^2} - 9) \times 4a = {a^2} \times 4a - 4a \times 9 = 4{a^3} - 36a$

(v) $pq + qr + rp,0$

$(pq + qr + rp) \times 0 = pq \times 0 + qr \times 0 + rp \times 0 = 0 + 0 + 0 = 0$

Q.2 Complete the table.

 Â First expression Second expression Product (i) (ii) (iii) (iv) (v) $a$ $x + y - 5$ $p$ $4{p^2}{q^2}$ $a + b + c$ $b + c + d$ $5xy$ $6{p^2} - 7p + 5$ ${p^2} - {q^2}$ $abc$ ... ... ... ... ...

Sol. The complete table is as follows:

 Â First expression Second expression Product (i) (ii) (iii) (iv) (v) $a$ $x + y - 5$ $p$ $4{p^2}{q^2}$ $a + b + c$ $b + c + d$ $5xy$ $6{p^2} - 7p + 5$ ${p^2} - {q^2}$ $abc$ $ab + ac + ad$ $5{x^2}y + 5x{y^2} - 25xy$ $6{p^3} - 7{p^2} + 5p$ $4{p^4}{q^2} - 4{p^2}{q^4}$ ${a^2}bc + a{b^2}c + ab{c^2}$

Q.3 Find the product.

(i) $({a^2}) \times (2{a^{22}}) \times (4{a^{26}})$ (ii)$\left( {{2 \over 3}xy} \right) \times \left( {{{ - 9} \over {10}}{x^2}{y^2}} \right)$

(iii) $\left( { - {{10} \over 3}p{q^3}} \right) \times \left( {{6 \over 5}{p^3}q} \right)$ (iv)$x \times {x^2} \times {x^3} \times {x^4}$

Sol. (i) $({a^2}) \times (2{a^{22}}) \times (4{a^{26}})$

= $2 \times 4 \times {a^2} \times {a^{22}} \times {a^{26}} = 8{a^{50}}$

(ii) $\left( {{2 \over 3}xy} \right) \times \left( {{{ - 9} \over {10}}{x^2}{y^2}} \right)$

= $\left( {{2 \over 3}} \right) \times \left( {{{ - 9} \over {10}}} \right) \times x \times y \times {x^2} \times {y^2} = {{ - 3} \over 5}{x^3}{y^3}$

(iii) $\left( { - {{10} \over 3}p{q^3}} \right) \times \left( {{6 \over 5}{p^3}q} \right)$

= $\left( {{{ - 10} \over 3}} \right) \times \left( {{6 \over 5}} \right) \times p{q^3} \times {p^3}q = - 4{p^4}{q^4}$

(iv) $x \times {x^2} \times {x^3} \times {x^4}$

=$x \times {x^2} \times {x^3} \times {x^4} = {x^{10}}$

Q.4 (a) Simplify 3x (4x â€“ 5) + 3 and find its values for (i) x = 3 (ii) x = ${1 \over 2}$.

(b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = â€“ 1.

Sol. (a) $3x(4x - 5) + 3$

$3x(4x - 5) + 3 = 12{x^2} - 15x + 3$

For (i) x = 3

$12{x^2} - 15x + 3$= $12{(3)^2} - 15(3) + 3$

= 108 â€“ 45 + 3

= 66

(ii) x = ${1 \over 2}$

$12{x^2} - 15x + 3$= $12{\left( {{1 \over 2}} \right)^2} - 15\left( {{1 \over 2}} \right) + 3$

= $3 - {{15} \over 2} + 3$

=$6 - {{15} \over 2}$

= ${{12 - 15} \over 2}$=${{ - 3} \over 2}$

(b) $a({a^2} + a + 1) + 5$

$a({a^2} + a + 1) + 5$= ${a^3} + {a^2} + a + 5$

(i) a = 0

${a^3} + {a^2} + a + 5$= ${(0)^3} + {(0)^2} + 0 + 5$

= 5

(ii) a = 1

${a^3} + {a^2} + a + 5$= ${(1)^3} + {(1)^2} + 1 + 5$

= 1 + 1 + 1 + 5

= 8

(iii) a = â€“ 1

${a^3} + {a^2} + a + 5$= ${( - 1)^3} + {( - 1)^2} + ( - 1) + 5$

= -1 + 1 - 1 + 5

= 4

Q.5 (a) Add: p ( p â€“ q), q ( q â€“ r) and r ( r â€“ p)

(b) Add: 2x (z â€“ x â€“ y) and 2y (z â€“ y â€“ x)

(c) Subtract: 3l (l â€“ 4 m + 5 n) from 4l ( 10 n â€“ 3 m + 2 l )

(d) Subtract: 3a (a + b + c ) â€“ 2 b (a â€“ b + c) from 4c ( â€“ a + b + c )

Sol. (a) $p(p - q) + q(q - r) + r(r - p)$

= ${p^2} - pq + {q^2} - qr + {r^2} - rp$

= ${p^2} + {q^2} + {r^2} - pq - qr - rp$

(b) $2x(z - x - y) + 2y(z - y - x)$

= $2xz - 2{x^2} - 2xy + 2yz - 2{y^2} - 2xy$

= $- 2{x^2} - 2{y^2} - 4xy + 2yz + 2xz$

(c)$4l(10n - 3m + 2l) - 3l(l - 4m + 5n)$

= $40ln - 12lm + 8{l^2} - 3{l^2} + 12lm - 15ln$

= $8{l^2} - 3{l^2} - 12lm + 12lm + 40ln - 15ln$

=$5{l^2} + 25ln$

(d) $4c( - a + b + c) - [3a(a + b + c) - 2b(a - b + c)]$

= $- 4ac + 4bc + 4{c^2} - [3{a^2} + 3ab + 3ac - 2ab + 2{b^2} - 2bc]$

= $- 4ac + 4bc + 4{c^2} - [3{a^2} + 2{b^2} + 3ab - 2bc + 3ac - 2ab]$

= $- 4ac + 4bc + 4{c^2} - [3{a^2} + 2{b^2} + ab - 2bc + 3ac]$

= $- 4ac + 4bc + 4{c^2} - 3{a^2} - 2{b^2} - ab + 2bc - 3ac$

= $- 3{a^2} - 2{b^2} + 4{c^2} - ab + 6bc - 7ac$