KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

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KSEEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

Question 1.
Carry out the multiplication of the expression in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) 4p(q + r)
= 4p × q + 4p × r
= 4pq + 4pr

(ii) ab × (a – b)
= ab × a – ab × b
= a²b – ab²

(iii) (a + b) (7a²b²)
= a × 7a²b² + b × 7a²b²
= 7a³b² + 7a²b³

(iv) (a² – 9) (4a)
= a² × 4a – 9 × 4a
= 4a³ – 36a

(v) (pq +qr + rp) (0) = 0

Question 2.
Complete the table.

Solution:
(i) (a) (b + c + d)
= (ab + ac + ad)

(ii) (x + y – 5)(5xy)
= 5xy(x + y – 5)
= 5yx × x + 5xy × y – 5 × 5xy
= 5x²y + 5xy² × -25xy

(iii) p(6p² – 7p + 5)
= p × 6p² – 7p × p + p × 5)
= 6p³ – 7p² + 5p

(iv) 4p²q²(p² – q²)
= 4p²q² × p² – 4p²q² × q²
= 4p⁴q² – 4p²q⁴

(v) (a + b + c) (abc)
= a × abc + b × abc + c × abc
= a²bc + ab²c + abc²

Question 3.
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)\)
(iii) \(\left(-\frac{10}{3} p q^{3}\right) \times\left(\frac{6}{5} p^{3} q\right)\)
(iv) x × x² × x³ × x⁴
Solution:
(i) (a²) × (2a²²) × (4a²⁶)
= a² × 2a²² × 4a²⁶
= 1 × 2 × 4 × a² × a²² × a²⁶
= 8a⁵⁰

(ii) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)\)
= \(\frac{2}{3} \times \frac{-9}{10} x y \times x^{2} y^{2}\)
= \(\frac{-3}{5} x^{3} y^{3}\)

(iii) \(\left(-\frac{10}{3} p q^{3}\right) \times\left(\frac{6}{5} p^{3} q\right)\)
= \(-\frac{10}{3} \times \frac{6}{5} p q^{3} \times p^{3} q\)
= -4p⁴q⁴

(iv) x × x² × x³ × x⁴ = x¹⁰

Question 4.
(a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3
(ii) x = \(\frac{1}{2}\)
(b) Simplify a(a² + a + 1) + 5 and find its values for
(i) a = 0
(ii) a = 1
(iii) a = -1
Solution:
(a) 3x(4x – 5) + 3
= 3x × 4x – 3x × 5 + 3
= 12x² – 15x + 3
when (i) x = 3
12x² – 15x + 3
= 12(3)² – 15(3) + 3
= 12(9) – 45 + 3
= 108 – 45 + 3
= 66

(ii) when x = \(\frac{1}{2}\)
12x² – 15x + 3
= \(12\left(\frac{1}{2}\right)^{2}-15\left(\frac{1}{2}\right)+3\)
= \(12 \times \frac{1}{4}-\frac{15}{2}+3\)
= 3 – \(\frac{15}{2}\) + 3
= 6 – \(\frac{15}{2}\)
= \(\frac{12-15}{2}\)
= \(\frac{-3}{2}\)

(b) a(a² + a + 1) + 5
= a × a² + a × a + a × 1 + 5
= a³ × a² + a + 5

(i) when a = 0
(0)³ + (0)² + 0 + 5
= 0 + 0 + 0 + 5
= 5

(ii) when a = 1
= (1)³ + (1)² + (1) + 5
= 1 + 1 + 1 + 5
= 8

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: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) p(p – q), q(q – r) and r(r – p)
= p × p – p × q + q × q – q × r + r × r – r × p
= p × p – pq + q × q – qr + r × r – rp
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – pq – qr – rp

(b) 2x(z – x – y) and 2y(z – y – x)
= 2xz – 2x × x – 2x × y and 2y × z – 2y × y – 2y × x)
= 2xz – 2x² – 2xy and 2yz – 2y² – 2yx
= 2xz – 2x² – 2xy – 2yx – 2y² + 2yz
= 2xz – 2x² – 4xy – 2y² + 2yz

(c) 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
= 3l × l – 3l × 4m + 3l × 5n from 4l × 10n – 4l × 3m + 4l × 2l
= 3l² – 12lm + 15ln from 40ln – 12lm + 8l²
= 40ln – 12lm + 8l² – 15ln + 12lm + 3l²
= 25ln + 0 + 5l²

(d) 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
= 3a × a + 3a × b + 3a × c – 2b × a + 2b × b × b – 2b × c
from -4ca + 4cb + 4c × c = 3a² + 3ab + 3ac
-2ab + 2b² × 2bc
from -4ca + 4bc + 4c² = -4ca + 4bc + 4c²
= -3ac + 2bc + 3a² + 2b² + 3ab
= -7ac – 6bc + 4a² – 3a² – 2b² + 3ab