## MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4

Question 1.

Multiply the binomials.

(i) (2x + 5) and (4x – 3)

(ii) (y – 8) and (3y – 4)

(iii) (2.5l – 0.5 m) and (2.5l +0.5 m)

(iv) (a + 3b) and (x + 5)

(v) (2pq + 3q^{2}) and (3pq – 2q^{2})

Solution:

(i) (2x + 5) × (4x – 3) = 2x (4x – 3) + 5(4x – 3) = 8x^{2} – 6x + 20x – 15 = 8x^{2} + 14x – 15

(ii) (y – 8) × (3y – 4) = y(3y – 4) – 8(3y – 4)

= 3 y^{2} – 4y – 24y + 32 = 3y^{2} – 28y + 32

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)

= 2.5l (2.5l + 0.5m) – 0.5 m (2.5l + 0.5m)

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

(iv) (a + 3b) × (x + 5) = a(x + 5) + 3b(x + 5)

= ax + 5a + 3 bx + 15 b

(v) (2pq + 3q^{2}) × (3pq – 2q^{2})

= 2pq (3pq – 2q^{2}) + 3q^{2} (3pq – 2q^{2})

= 6p^{2}q^{2} – 4pq^{3} + 9pq^{3} – 6q^{4}

= 6 p^{2}q^{2} + 5pq^{3} – 6q^{4}

Question 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)

Solution:

(i) (5 – 2x) × (3 + x) = 5(3 + x) – 2x(3 + x)

= 15 + 5x – 6x – 2x^{2} – 15 – x – 2x^{2}

(ii) (x + 7y)(7x – y) = x(7x – y) + 7y(7x – y)

=7x^{2} – xy + 49xy – 7y^{2} = 7x^{2} + 48xy – 7y^{2}

(iii) (a^{2} + b) (a + b^{2}) = a^{2} (a + b^{2}) + b (a + b^{2})

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

(iv) (p^{2} – q^{2}) (2p + q) = p^{2} (2p + q) – q^{2} (2p + q)

= 2p^{3} + p^{2}q – 2q^{2}p – q^{3}

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Question 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)

Solution:

(i) (x^{2} – 5) (x + 5) + 25

= x^{2}(x +5) – 5(x + 5) +25

=x^{3}+ 5x^{2} – 5x – 25 + 25 = r^{3}+ 5r^{2} – 5x

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

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

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

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

(iii) (t + s^{2}) (t^{2} – s) = t(t^{2} – s) + s^{2}(t^{2} – s)

= t^{3} – ts +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) + 2a + 2bd

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)

=2x^{2} + xy + 2xy + y^{2} + x^{2} – xy + 2xy – 2y^{2}

= 3x^{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 + xy^{2} + x^{2}y – xy^{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

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

= 2.25x^{2} – 16y^{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} + 2 ab + b^{2} – c^{2}