# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

**NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4**

Ex 9.4 Class 8 Maths 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) (\(\frac { 3 }{ 4 }\)a^{2} + 3b^{2}) and 4(a^{2} – \(\frac { 2 }{ 3 }\) b^{2})

Solution:

(i) (2x + 5) × (4x – 3)

= 2x × (4x – 3) + 5 × (4x – 3)

= (2x × 4x) – (3 × 2x) + (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} – 28y + 32

(iii) (2.5l – 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} + 1.25ml – 1.25ml – 0.25m^{2}

= 6.25l^{2} + 0 – 0.25m^{2}

= 6.25l^{2} – 0.25m^{2}

(iv) (a + 3b) × (x + 5)

= a × (x + 5) + 36 × (x + 5)

= (a × x) + (a × 5) + (36 × x) + (36 × 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q^{2}) × (3pq – 2q^{2})

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

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

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

= 6p^{2}q^{2} + 5pq^{3} – 6q^{4}

Ex 9.4 Class 8 Maths 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)

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

= 15 + 5x – 6x – 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} + a^{2}b^{2} + ab + b^{3}

(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}

Ex 9.4 Class 8 Maths 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

= x^{3} + 5x^{2} – 5x + 0

= 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} + 5b^{3} + 20

(iii) (t + s^{2}) (t^{2} – s)

= t(t^{2} – s) + s^{2}(t^{2} – s)

= t^{3} – st + s^{2}t^{2} – s^{3}

= t^{3} + s^{2}t^{2} – st – 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

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

= 4ac + 0 + 0 + 0

= 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}

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

= x^{3} – 0 + 0 + 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} + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y^{2}

= 2.25x^{2} + 0 + 0 + 0 – 16y^{2}

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

= a^{2} + 2ab + b^{2} – c^{2} + 0 + 0

= a^{2} + 2ab + b^{2} – c^{2}