## MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3

Question 1.

Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r

(ii) ab, a – b

(iii) a + b, 7a^{2}b^{2}

(iv) a^{2} – 9, 4a

(v) pq + qr + rp, 0

Solution:

(i) 4p × (q + r) = 4p × q + 4p × y = 4pq + 4pr

(ii) ab × (a – b) = a^{2}b – ab^{2}

(iii) (a + b) × 7a^{2}b^{2} = 7a^{3}b^{2} + 7a^{2}b^{3}

(iv) (a^{2} – 9) × (4a) = 4a^{3} – 36a

(v) (pq + qr + rp) × 0 = 0

Question 2.

Complete the table.

Solution:

Question 3.

Find the product.

(i) (a^{2}) × (2a^{22}) × (4a^{26})

(iv) x × x^{2} × x^{3} × x^{4}

Solution:

Question 4.

(a) Simplify 3x(4x – 5) + 3 and find its values for

(i) x = 3

(ii) x = \(\frac{1}{2}\)

(b) Simplify a(a^{2} + a + 1) + 5 and find its value for

(i) o = 0

(ii) a = 1

(iii) a = – 1

Solution:

(a) 3x(4x – 5) + 3 = 12x^{2} – 15x + 3

(i) For x = 3,

12(3)^{2} – 15(3) + 3 = 108 – 45 + 3 = 66

(b) a(a^{2} + a + 1) + 5 = a^{3} + a^{2} + a + 5

(i) For a = 0, (0)^{3} + 0^{2} + 0 + 5 = 5

(ii) For a = 1, 1^{3} + 1^{2} + 1 + 5 = 8

(iii) For a = -1, (-1)^{3} + (-1)^{2} + (-1) + 5 = -1 + 1 – 1 + 5 = 4

Question 5.

(a) Add: p(p – q), q(q – r) and r(r – p)

(b) Add : 2x(z – x – y) and 2y(z – y – x)

(c) Subtract: 3l(l -4m + 5n) from 4l(10n – 3m + 2l)

(d) Subtract: 3o(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)

Solution:

(a) First expression

= p(p – q) = p^{2} – pq

Second expression = q(q- r) = q^{2} – qr… (ii)

Third expression = r(r – p) = r^{2} – pr ….(iii)

Adding (i), (ii) and (iii), we get

p^{2} – pq + q^{2} – qr + r^{2} – pr = p^{2} + q^{2} + r^{2} – pq – qr – pr

(b) First expression = 2x(z -x-y)

= 2xz – 2x^{2} – 2xy

Second expression = 2y(z -y-x)

= 2yz – 2y^{2} – 2xy

Adding the two expressions, we get

(c) First expression = 3l(l – 4m + 5n)

= 3l^{2} – 121m + 15In

Second expression = 4l(10n – 3m + 2l)

= 40ln – 12lm + 8l^{2}

Subtracting the two expressions, we get 40/n –

(d)

First expression= 3a(a + b + c) – 2b(a – b + c)

= 3a^{2}+3ab + 3ac – 2ab + 2b^{2} – 2bc

= 3a^{2} + ab + 2b^{2} + 3ac – 2bc

Second expression = 4c(- a + b + c)

= – 4ac + 4bc + 4c^{2}

Subtracting the two expressions, we get