Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

Question 1.

Complete the table.

Solution:

We have:

1. a × (b + c + d) = a × b + a × c + a × d

= ab + ac + ad

2. (x + y – 5) × 5xy

= x × 5xy + y × 5xy + (-5) × 5xy

= 5x^{2}y + 5x^{2}y – 25xy

3. p × (6p^{2} – 7p + 5)

= p × 6p^{2} + p × (-7p) + p × 5

= 6p^{3} – 7p^{2} + 5p

4. 4p^{2}q^{2} × (p^{2} – q^{2})

= 4p^{2}q^{2} × p^{2} + 4p^{2}q^{2} × (-q^{2})

= (4 × 1) × p^{2}q^{2} × p^{2} + [4 x (-1)] × p^{2}q^{2} × (-q^{2})

= 4 × p^{4}q^{2} + (-4) × p^{2}q^{4} = 4p^{4}q^{2} – 4p^{2}q^{4}

5. (a + b + c) × abc

= a × abc + b × abc + c × abc

= (1 × 1) × a × abc + (1 × 1) × b × abc + (1 × 1) × c × abc

= (1) × a^{2}bc + (1) × ab^{2}c + abc^{2}

Question 2.

Find the product?

- (a
^{2}) × (2a^{22}) × (4a^{26}) - (\(\frac{2}{3}\)xy) × (\(\frac{-9}{10}\)x
^{2}y^{2}) - (-\(\frac{10}{3}\)pq
^{3}) × (\(\frac{6}{5}\)p^{3}q) - x × x
^{2}× x^{4}× x^{4}

Solution:

1. (a^{2}) × (2a^{22}) × (4a^{26})

= (1 × 2 × 4) × a^{2} × a^{22} × a^{26}

= 8 × a^{50} [∵2 + 22 + 26 = 50]

= 8a^{50}

2. (\(\frac{2}{3}\)xy) × (-\(\frac{9}{10}\)x^{2}y^{2}) = \(\frac{2}{3}\) × (-\(\frac{9}{10}\)) × xy × x^{2}y^{2}

= – \(\frac{2}{3}\) × \(\frac{9}{10}\) × x^{3}y^{3} = \(\frac{-3}{5}\)x^{3}y^{3}

3. (-\(\frac{10}{3}\)pq^{3}) × p^{3}q

= [(-\(\frac{10}{3}\)) × \(\frac{6}{5}\)] × pq^{3} × p^{3}q

= –\(\frac{10}{3}\) × \(\frac{6}{5}\) × p^{4}q^{4} = -4p^{4}q^{4}

4. x × x^{2} × x^{3} × x^{4}

= (1 × 1 × 1 × 1) × x × x^{2} × x^{3} × x^{4}

= (1) × x^{10} = x^{10}

Question 3.

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

(i) x = 3 and

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

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

(i) a = 0

(ii) a = 1 and

(iii) a = -1.

Solution:

(a) 3x(4x – 5) + 3 = 3x × 4x + 3x × (-5) + 3

= (3 × 4) × x × x + 3 × (-5) × x + 3

= 12 × x^{2} + (-15) × x + 3

= 12x^{2} – 15x + 3

(i) For x = 3, 12x^{2} – 15x + 3

= 12(3)^{2} – 15(3) + 3

= 12 × 9 – 45 + 3

= 108 + 3 – 45

= 111 – 45 = 66

(ii) For x = \(\frac{1}{2}\), 12x^{2} – 15x + 3

= 12(\(\frac{1}{2}\))^{2} – 15(\(\frac{1}{2}\)) + 3

= 12(\(\frac{1}{4}\)) – \(\frac{15}{2}\) + 3 = 3 – \(\frac{15}{2}\) + 3 = 6 – \(\frac{15}{2}\)

= \(\frac{12-15}{2}\) = \(\frac{-3}{2}\)

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

= a × a^{2} + a × a + a × 1 + 5

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

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

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

= 1 + 1 + 1 + 5 = 8

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

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

= 1 + 1 + 1 + 5 = 8

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

= (-1)^{3} + (-1)^{2} + (-1) + 5

= -1 + 1 – 1 + 5 = 4

Question 4.

(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) = p × p – p × q = p^{2} – pq

q(q – r) = q × q – q × r = q^{2} – qr

and r(r – p) = r × r – r × p = r^{2} – rp

∴ Adding the above products, we have

(p^{2} – pq) + (q^{2} – qr) + (r^{2} – rp)

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

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

= p^{2} + q^{2} + r^{2} – (pq + qr + rp)

(b) ∵ 2x(z – x – y) = 2x × z – 2x × x – 2x × y

= 2xz – 2x^{2} – 2xy and 2y(z – y – x)

= 2y × z – 2y × y – 2y × x

= 2yz – 2y^{2} – 2yx

(c) ∵ 3l(l – 4m + 5n)

= 3l × l – 3l × 4m + 3l × 5n

= 3l^{2} – 12lm + 15ln and 4l(10n – 3m + 2l)

= 4l × 10n – 4l × 3m + 4l × 2l

= 40ln – 12lm + 8l^{2}

(d) ∵ 3a(a + b + c) – 2b(a – b + c)

= [3a × a + 3a × b + 3a × b + 3a × c] – 2b(a × a + 2b × (-b) + 2b × c]

= [3a^{2} + 3ab + 3ac] – [2ab – 2b^{2} + 2abc]

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

= 3a^{2} + ab + 3ac + 2b^{2} – 2bc and 4c(-a + b + c)

= 4c × (-a) + 4c × b + 4c × c

= – 4ac + 4bc + 4c^{2}