Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 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 3(pq – 2q^{2})

(vi)

Solution:

(i) (2x + 5) × (4x – 3)

= 8x^{2} – 6x + 20x – 15

= 8x^{2} + 14x – 15.

(ii) (y – 8) × (3y – 4) = 3y^{2} – 4y – 24y + 32

= 3y^{2} – 28y + 32.

(iii) (2.5l – 0.5m) × (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)

= ax + 5a + 3bx + 15b

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

(v) (2pq + 3q^{2}) × (3pq – 2q^{2})

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

= 6p^{2}a^{2} + 5pq^{3} – 6q^{4}.

(vi)

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)

= 15 + 5x – 6x – 2x^{2}

= 15 – x – 2x^{2}.

(ii) (x + 7y) × (7x -y)

= 7x^{2} – xy + 49xy – 7y^{2}

= 7x^{2} – 7y^{2} + 48xy.

(iii) (a^{2} + b) × (a + b^{2})

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

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

(iv) (p^{2} – q^{2}) × (2p + q)

= 2p^{3} + p^{2}q – 2pq^{2} – q^{3}

= 2p^{3} – q^{3} + p^{2}q – 2pq^{2}.

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^{3} + 5x^{2} – 5x – 25 + 25

= x^{3} + 5x^{2} – 5x.

(ii) (a^{2} + 5) × (b^{3} + 3) + 5

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

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

(iii) (t + s^{2}) (t^{2} – s)

= t^{3} – st + s^{2}t^{2} – s^{3}

= t^{3} – s^{3} + s^{2} + t^{2} – st.

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac +bd)

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

= 4ac.

(v) (x + y) × (2x + y) + (x + 2y) x (x – y)

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

= 3x^{2} – y^{2} + 4xy.

(vi) (x + 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

= 2.25x^{2} + 60xy + 4.5x – 60xy – 16y^{2} – 12y – 4.5x + 12y

= 2.25x^{2}.

(viii) (a + b + c) × (a + b – c)

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

= a^{2} + b^{2} – c^{2} + 2ab.