# Algebraic Expressions And Identities : Exercise 9.4 (Mathematics NCERT Class 8th)

Q.1 Multiply the binomials.

(i) $(2x + 5)$ and $(4x - 3)$ (ii) $(y - 8)$ and $(3y - 4)$

(iii) $(2.5l - 0.5m)$and $(2.5l + 0.5m)$ (iv) $(a + 3b)$ and $(x + 5)$

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

(vi) $\left( {{3 \over 4}{a^2} + 3{b^2}} \right)$and$4\left( {{a^2} - {2 \over 3}{b^2}} \right)$

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

= $2x(4x - 3) + 5(4x - 3)$

= $2x \times 4x - 2x \times 3 + 5 \times 4x - 5 \times 3$

= $8{x^2} - 6x + 20x - 15$

= $8{x^2} + 14x - 15$

(ii) $(y - 8) \times (3y - 4)$

= $y(3y - 4) - 8(3y - 4)$

= $y \times 3y - y \times 4 - 8 \times 3y + 8 \times 4$

= $3{y^2} - 4y - 24y + 32$

= $3{y^2} - 28y + 32$

(iii) $(2.5l - 0.5m) \times (2.5l + 0.5m)$

= $2.5l \times (2.5l + 0.5m) - 0.5m \times (2.5l + 0.5m)$

= $2.5l \times 2.5l + 2.25l \times 0.5m - 0.5m \times 2.5l - 0.5m \times 0.5m$

= $6.25{l^2} + 1.25lm - 1.25lm - 0.25{m^2}$

= $6.25{l^2} - 0.25{m^2}$

(iv) $(a + 3b) \times (x + 5)$

= $a \times (x + 5) + 3b(x + 5)$

= $a \times x + a \times 5 + 3b \times x + 3b \times 5$

=$ax + 5a + 3bx + 15b$

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

= $2pq \times (3pq - 2{q^2}) + 3{q^2} \times (3pq - 2{q^2})$

= $6{p^2}{q^2} - 4p{q^3} + 9p{q^3} - 6{q^4}$

= $6{p^2}{q^2} + 5p{q^3} - 6{q^4}$

(vi) $\left( {{3 \over 4}{a^2} + 3{b^2}} \right) \times 4\left( {{a^2} - {2 \over 3}{b^2}} \right)$

= $\left( {{3 \over 4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - {8 \over 3}{b^2}} \right)$

= ${3 \over 4}{a^2} \times \left( {4{a^2} - {8 \over 3}{b^2}} \right) + 3{b^2} \times \left( {4{a^2} - {8 \over 3}{b^2}} \right)$

= $3{a^4} - 2{a^2}{b^2} + 12{a^2}{b^2} - 8{b^4}$

= $3{a^4} + {120^2}{b^2} - 8{b^4}$

Q.2 Find the product.

(i)$(5 - 2x)(3 + x)$ (ii)$(x + 7y)(7x - y)$

(iii) $({a^2} + b)(a + {b^2})$ (iv) $({p^2} - {q^2})(2p + q)$

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

= $5 \times (3 + x) - 2x \times (3 + x)$

= $15 + 5x - 6x + 2{x^2}$

= $15 - x + 2{x^2}$

(ii)$(x + 7y) \times (7x - y)$

= $x \times (7x - y) + 7y \times (7x - y)$

= $7{x^2} - xy + 49xy - 7{y^2}$

= $7{x^2} + 48xy - 7{y^2}$

(iii) $({a^2} + b) \times (a + {b^2})$

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

= ${a^3} + {a^2}{b^2} + ab + {b^3}$

(iv) $({p^2} - {q^2}) \times (2p + q)$

= ${p^2} \times (2p + q) - {q^2} \times (2p + q)$

= $2{p^3} + {p^2}q - 2p{q^2} - {q^3}$

Q.3 Simplify.

(i) $({x^2} - 5)(x + 5) + 25$

(ii) $({a^2} + 5)({b^3} + 3) + 5$

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

(iv)$(a + b)(c - d) + (a - b)(c + d) + 2(ac + bd)$

(v)$(x + y)(2x + y) + (x + 2y)(x - y)$

(vi)$(x + y)({x^2} - xy + {y^2})$

(vii)$(1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y$

(viii)$(a + b + c)(a + b - c)$

Sol. (i) $({x^2} - 5)(x + 5) + 25$

= ${x^2} \times (x + 5) - 5 \times (x + 5) + 25$

= ${x^3} + 5{x^2} - 5x - 25 + 25$

= ${x^3} + 5{x^2} - 5x$

(ii) $({a^2} + 5)({b^3} + 3) + 5$

= ${a^2} \times ({b^3} + 3) + 5 \times ({b^3} + 3) + 5$

= ${a^2}{b^3} + 3{a^2} + 5{b^3} + 15 + 5$

= ${a^2}{b^3} + 3{a^2} + 5{b^3} + 20$

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

= $t \times {t^2} - t \times s + {s^2} \times {t^2} - {s^2} \times s$

= ${t^3} - st + {s^2}{t^2} - {s^3}$

(iv)$(a + b)(c - d) + (a - b)(c + d) + 2(ac + bd)$

= $a(c - d) + b(c - d) + a(c + d) - b(c + d) + 2ac + 2bd$

= $ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd$

= $2ac - 2bd + 2ac + 2bd$

= $4ac$

(v)$(x + y)(2x + y) + (x + 2y)(x - y)$

= $x(2x + y) + y(2x + y) + x(x - y) + 2y(x - y)$

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

= $3{x^2} + 4xy - {y^2}$

(vi)$(x + y)({x^2} - xy + {y^2})$

= $x({x^2} - xy + {y^2}) + y({x^2} - xy + {y^2})$

= ${x^3} - {x^2}y + x{y^2} + {x^2}y - x{y^2} + {y^3}$

= ${x^3} + {y^3}$

(vii)$(1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y$

= $1.5x(1.5x + 4y + 3) - 4y(1.5x + 4y + 3) - 4.5x + 12y$

= $2.25{x^2} + 6xy + 4.5 - 6xy - 16{y^2} - 12y - 4.5x + 12y$

= $2.25{x^2} - 16{y^2}$

(viii)$(a + b + c)(a + b - c)$

= $a(a + b - c) + b(a + b - c) + c(a + b - c)$

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

= ${a^2} + {b^2} - {c^2} + 2ab$

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