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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)and4\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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