These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths 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 – 0.5m) and (2.5l + 0.5m)

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

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

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

Answer:

(2x + 5) (4x – 3) = 2x (4x – 3) + 5 (4x – 3)

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

= 8x^{2} – 6x + 20x – 15

= 8x^{2} + 14x – 15

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

= y × 3y + y (-4) – [8 × 3y + 8 (-4)]

= 3y^{2} – 4y – (24y – 32)

= 3y^{2} – 4y – 24y + 32

= 3y^{2} – 28y + 32

(iii) (2.5l – 0.5m) (2.5l + 0.5m)

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

= 2.5l × 2.5l + 2.5l × 0.5m – [0.5m × 2.5l + 0.5m × 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 + 3bx + 15b

= ax + 5a + 3bx + 15b

(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}

= 6p^{2}q^{2} + 5pq^{3} – 6q^{4}

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

= 3a^{4} + 10a^{2}b^{2} – 8b^{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)

Answer:

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

= 5 × 3 + 5 × x – (2x) × 3 – 2x (x)

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

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

= x × 7x – x × y + 7y × 7x – 7y × 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^{2} × a + a^{2} × b^{2} + b × a + b × b^{2}

= a^{3</sup + a2b2 + ab + b3}

(iv) (p^{2} – q^{2}) (2p + q)

= p^{2} × (2p + q) – q^{2} (2p + q)

= p^{2} × 2p + p^{2} × q – q^{2} × 2p – q^{2} × q

= 2p^{3} + p^{2}q – 2pq^{2} – q^{3}

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) – 45x + 12y

(viii) (a + b + c) (a + b – c)

Answer:

(i) (x^{2} – 5) (x + 5) + 25

= x^{2} × (x + 5) – 5 (x + 5) + 25

= x^{2}× x + x^{2} × 5 – 5 × x – 5 × 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} + 3) + 5 × (b^{3} + 3) + 5

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

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

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

= t × t^{2} – t × s + s^{2} × t^{2} – S^{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 ) + 2 × ac + 2 × bd

= a × c – a × d+ b × c – b × d + a × c + a × d – b × c – b × d + 2ac + 2bd

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

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

= 4ac – 2bd + 2bd = 4ac

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

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

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

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

= 3x^{2} + 5xy – xy – y^{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 × x^{2} + x × (-xy) + x × y^{2} + y × x^{2} + y × (-xy) + y × y^{2}

= x^{3} – x^{2}y + xy^{2} + x^{2}y – xy^{2} + y^{3}

= x^{3} – x^{2}y + x^{2}y + xy^{2} – 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

= 1.5x × 1.5x + 1.5x × 4y + 1.5x × 3 – 4y × 1.5x – 4y × 4y – 4y × 3 – 4.5x + 12y

= 2.25xz + 6xy + 4.5x – 6xy – 16y2

– 12y – 4.5x + 12y

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

= 2.25x^{2} + 0 + 0 – 16y^{2} + 0

= 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 × a + a × b + a × -c + b × a + b × b – b × c + c × a + b × c – c × c

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

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

= a^{2} + 2ab + o + o + b^{2} – c^{2}

= a^{2} + b^{2} – c^{2} + 2ab