Mathematics Solutions Solutions for Class 8 Maths Chapter 6 - Factorisation Of Algebraic Expressions

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Page No 30:

Question 1:

Factorise.
(1) x²+ 9x + 18
(2) x²− 10x + 9
(3) y²+ 24y + 144
(4) 5y²+ 5y − 10
(5) p²− 2p − 35
(6) p²− 7p − 44
(7) m²− 23m + 120
(8) m²− 25m + 100
(9) 3x²+ 14x + 15
(10) 2x²+ x − 45
(11) 20x²− 26x + 8
(12) 44x²− x − 3

Answer:

$(1) x^{2} + 9 x + 18 = x^{2} + 6 x + 3 x + 18 = x (x + 6) + 3 (x + 6) = (x + 3) (x + 6)$
$(2) x^{2} - 10 x + 9 = x^{2} - 9 x - x + 9 = x (x - 9) - 1 (x - 9) = (x - 1) (x - 9)$
$(3) y^{2} + 24 y + 144 = y^{2} + 12 y + 12 y + 144 = y (y + 12) + 12 (y + 12) = (y + 12) (y + 12)$
$(4) 5 y^{2} + 5 y - 10 = 5 y^{2} + 10 y - 5 y - 10 = 5 y (y + 2) - 5 (y + 2) = (5 y - 5) (y + 2) = 5 (y - 1) (y + 2)$
$(5) p^{2} - 2 p - 35 = p^{2} - 7 p + 5 p - 35 = p (p - 7) + 5 (p - 7) = (p + 5) (p - 7)$
$(6) p^{2} - 7 p - 44 = p^{2} - 11 p + 4 p - 44 = p (p - 11) + 4 (p - 11) = (p + 4) (p - 11)$
$(7) m^{2} - 23 m + 120 = m^{2} - 15 m - 8 m + 120 = m (m - 15) - 8 (m - 15) = (m - 8) (m - 15)$
$(8) m^{2} - 25 m + 100 = m^{2} - 20 m - 5 m + 100 = m (m - 20) - 5 (m - 20) = (m - 5) (m - 20)$
$(9) 3 x^{2} + 14 x + 15 = 3 x^{2} + 9 x + 5 x + 15 = 3 x (x + 3) + 5 (x + 3) = (3 x + 5) (x + 3)$
$(10) 2 x^{2} + x - 45 = 2 x^{2} + 10 x - 9 x - 45 = 2 x (x + 5) - 9 (x + 5) = (2 x - 9) (x + 5)$
$(11) 20 x^{2} - 26 x + 8 = 20 x^{2} - 16 x - 10 x + 8 = 4 x (5 x - 4) - 2 (5 x - 4) = (4 x - 2) (5 x - 4) = 2 (2 x - 1) (5 x - 4)$
$(12) 44 x^{2} - x - 3 = 44 x^{2} + 11 x - 12 x - 3 = 11 x (4 x + 1) - 3 (4 x + 1) = (11 x - 3) (4 x + 1)$

Page No 31:

Question 1:

Factorise.
(1) x³+ 64y³
(2) 125p³+ q³
(3) 125k³+ 27m³
(4) 2l³+ 432m³
(5) 24a³+ 81b³
(6) y³+ $\frac{1}{8 y^{3}}$
(7) a³+ $\frac{8}{a^{3}}$
(8) 1 + $\frac{q^{3}}{125}$

Answer:

It is known that,
$a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)$
$(1) x^{3} + 64 y^{3} = {(x)}^{3} + {(4 y)}^{3} = (x + 4 y) \{{(x)}^{2} + {(4 y)}^{2} - (x) \times (4 y)\} = (x + 4 y) (x^{2} + 16 y^{2} - 4 x y)$
$(2) 125 p^{3} + q^{3} = {(5 p)}^{3} + {(q)}^{3} = (5 p + q) \{{(5 p)}^{2} + {(q)}^{2} - (5 p) \times (q)\} = (5 p + q) (25 p^{2} + q^{2} - 5 p q)$
$(3) 125 k^{3} + 27 m^{3} = {(5 k)}^{3} + {(3 m)}^{3} = (5 k + 3 m) \{{(5 k)}^{2} + {(3 m)}^{2} - (5 k) \times (3 m)\} = (5 k + 3 m) (25 k^{2} + 9 m^{2} - 15 k m)$
$(4) 2 l^{3} + 432 m^{3} = 2 [l^{3} + 216 m^{3}] = 2 [{(l)}^{3} + {(6 m)}^{3}] = 2 [(l + 6 m) \{{(l)}^{2} + {(6 m)}^{2} - (l) \times (6 m)\}] = 2 (l + 6 m) (l^{2} + 36 m^{2} - 6 l m)$
$(5) 24 a^{3} + 81 b^{3} = 3 [8 a^{3} + 27 b^{3}] = 3 [{(2 a)}^{3} + {(3 b)}^{3}] = 3 [(2 a + 3 b) \{{(2 a)}^{2} + {(3 b)}^{2} - (2 a) \times (3 b)\}] = 3 (2 a + 3 b) (4 a^{2} + 9 b^{2} - 6 a b)$
$(6) y^{3} + \frac{1}{8 y^{3}} = {(y)}^{3} + {(\frac{1}{2 y})}^{3} = (y + \frac{1}{2 y}) \{{(y)}^{2} + {(\frac{1}{2 y})}^{2} - (y) \times (\frac{1}{2 y})\} = (y + \frac{1}{2 y}) (y^{2} + \frac{1}{4 y^{2}} - \frac{1}{2})$
$(7) a^{3} + \frac{8}{a^{3}} = {(a)}^{3} + {(\frac{2}{a})}^{3} = (a + \frac{2}{a}) \{{(a)}^{2} + {(\frac{2}{a})}^{2} - (a) \times (\frac{2}{a})\} = (a + \frac{2}{a}) (a^{2} + \frac{4}{a^{2}} - 2)$
$(8) 1 + \frac{q^{3}}{125} = {(1)}^{3} + {(\frac{q}{5})}^{3} = (1 + \frac{q}{5}) \{{(1)}^{2} + {(\frac{q}{5})}^{2} - (1) \times (\frac{q}{5})\} = (1 + \frac{q}{5}) (1 + \frac{q^{2}}{25} - \frac{q}{5})$

Page No 32:

Question 1:

Factorise:
(1) y³− 27
(2) x³− 64y³
(3) 27m³− 216n³
(4) 125y³− 1
(5) 8p³− $\frac{27}{p^{3}}$
(6) 343a³ − 512b³
(7) 64x³− 729y³
(8) 16a³− $\frac{128}{b^{3}}$

Answer:

It is known that,
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
$(1) y^{3} - 27 = {(y)}^{3} - {(3)}^{3} = (y - 3) \{{(y)}^{2} + {(3)}^{2} + (y) \times (3)\} = (y - 3) (y^{2} + 9 + 3 y)$
$(2) x^{3} - 64 y^{3} = {(x)}^{3} - {(4 y)}^{3} = (x - 4 y) \{{(x)}^{2} + {(4 y)}^{2} + (x) \times (4 y)\} = (x - 4 y) (x^{2} + 16 y^{2} + 4 x y)$
$(3) 27 m^{3} - 216 n^{3} = 27 [m^{3} - 8 n^{3}] = 27 [{(m)}^{3} - {(2 n)}^{3}] = 27 [(m - 2 n) \{{(m)}^{2} + {(2 n)}^{2} + (m) \times (2 n)\}] = 27 (m - 2 n) (m^{2} + 4 n^{2} + 2 m n)$
$(4) 125 y^{3} - 1 = {(5 y)}^{3} - {(1)}^{3} = (5 y - 1) \{{(5 y)}^{2} + {(1)}^{2} + (5 y) \times (1)\} = (5 y - 1) (25 y^{2} + 1 + 5 y)$
$(5) 8 p^{3} - \frac{27}{p^{3}} = {(2 p)}^{3} - {(\frac{3}{p})}^{3} = (2 p - \frac{3}{p}) \{{(2 p)}^{2} + {(\frac{3}{p})}^{2} + (2 p) \times (\frac{3}{p})\} = (2 p - \frac{3}{p}) (4 p^{2} + \frac{9}{p^{2}} + 6)$
$(6) 343 a^{3} - 512 b^{3} = {(7 a)}^{3} - {(8 b)}^{3} = (7 a - 8 b) \{{(7 a)}^{2} + {(8 b)}^{2} + (7 a) \times (8 b)\} = (7 a - 8 b) (49 a^{2} + 64 b^{2} + 56 a b)$
$(7) 64 x^{3} - 729 y^{3} = {(4 x)}^{3} - {(9 y)}^{3} = (4 x - 9 y) \{{(4 x)}^{2} + {(9 y)}^{2} + (4 x) \times (9 y)\} = (4 x - 9 y) (16 x^{2} + 81 y^{2} + 36 x y)$
$(8) 16 a^{3} - \frac{128}{b^{3}} = 16 [a^{3} - \frac{8}{b^{3}}] = 16 [{(a)}^{3} - {(\frac{2}{b})}^{3}] = 16 [(a - \frac{2}{b}) \{{(a)}^{2} + {(\frac{2}{b})}^{2} + (a) \times (\frac{2}{b})\}] = 16 (a - \frac{2}{b}) (a^{2} + \frac{4}{b^{2}} + \frac{2 a}{b})$

Page No 32:

Question 2:

Simplify:
(1) (x + y)³− (x − y)³
(2) (3a + 5b)³− (3a − 5b)³
(3) (a + b)³− a³− b³
(4) p³− (p + 1)³
(5) (3xy − 2ab)³ − (3xy + 2ab)³

Answer:

It is known that,
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
$(1) {(x + y)}^{3} - {(x - y)}^{3} = \{(x + y) - (x - y)\} \{{(x + y)}^{2} + {(x - y)}^{2} + (x + y) (x - y)\} = 2 y (x^{2} + y^{2} + 2 x y + x^{2} + y^{2} - 2 x y + x^{2} - y^{2}) = 2 y (3 x^{2} + y^{2})$
$(2) {(3 a + 5 b)}^{3} - {(3 a - 5 b)}^{3} = \{(3 a + 5 b) - (3 a - 5 b)\} \{{(3 a + 5 b)}^{2} + {(3 a - 5 b)}^{2} + (3 a + 5 b) (3 a - 5 b)\} = 10 b (9 a^{2} + 25 b^{2} + 30 a b + 9 a^{2} + 25 b^{2} - 30 a b + 9 a^{2} - 25 b^{2}) = 10 b (27 a^{2} + 25 b^{2})$
$(3) {(a + b)}^{3} - a^{3} - b^{3} = {(a + b)}^{3} - (a^{3} + b^{3}) = {(a + b)}^{3} - \{(a + b) (a^{2} + b^{2} - a b)\} = (a + b) \{{(a + b)}^{2} - (a^{2} + b^{2} - a b)\} = (a + b) (a^{2} + b^{2} + 2 a b - a^{2} - b^{2} + a b) = (a + b) (3 a b) = 3 a^{2} b + 3 a b^{2}$
$(4) p^{3} - {(p + 1)}^{3} = \{p - (p + 1)\} \{{(p)}^{2} + {(p + 1)}^{2} + (p) \times (p + 1)\} = (- 1) (p^{2} + p^{2} + 1 + 2 p + p^{2} + p) = (- 1) (3 p^{2} + 3 p + 1) = - 3 p^{2} - 3 p - 1$
$(5) {(3 x y - 2 a b)}^{3} - {(3 x y + 2 a b)}^{3} = \{(3 x y - 2 a b) - (3 x y + 2 a b)\} \{{(3 x y - 2 a b)}^{2} + {(3 x y + 2 a b)}^{2} + (3 x y - 2 a b) (3 x y + 2 a b)\} = (- 4 a b) (9 x^{2} y^{2} + 4 a^{2} b^{2} - 12 a b x y + 9 x^{2} y^{2} + 4 a^{2} b^{2} + 12 a b x y + 9 x^{2} y^{2} - 4 a^{2} b^{2}) = (- 4 a b) (27 x^{2} y^{2} + 4 a^{2} b^{2}) = - 108 x^{2} y^{2} a b - 16 a^{3} b^{3}$

Page No 33:

Question 1:

Simplify:
(1) $\frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}}$

(2) $\frac{a^{2} + 10 a + 21}{a^{2} + 6 a - 7} \times \frac{a^{2} - 1}{a + 3}$

(3) $\frac{8 x^{3} - 27 y^{3}}{4 x^{2} - 9 y^{2}}$

(4) $\frac{x^{2} - 5 x - 24}{(x + 3) (x + 8)} \times \frac{x^{2} - 64}{{(x - 8)}^{2}}$

(5) $\frac{3 x^{2} - x - 2}{x^{2} - 7 x + 12} \div \frac{3 x^{2} - 7 x - 6}{x^{2} - 4}$

(6) $\frac{4 x^{2} - 11 x + 6}{16 x^{2} - 9}$

(7) $\frac{a^{3} - 27}{5 a^{2} - 16 a + 3} \div \frac{a^{2} + 3 a + 9}{25 a^{2} - 1}$

(8) $\frac{1 - 2 x + x^{2}}{1 - x^{3}} \times \frac{1 + x + x^{2}}{1 + x}$

Answer:

It is known that,
$a^{2} - b^{2} = (a + b) (a - b); a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
$(1) \frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}} = \frac{(m + n) (m - n)}{(m + n) (m + n)} \times \frac{(m^{2} + m n + n^{2})}{(m - n) (m^{2} + m n + n^{2})} = \frac{1}{m + n}$
$(2) \frac{a^{2} + 10 a + 21}{a^{2} + 6 a - 7} \times \frac{a^{2} - 1}{a + 3} = \frac{a^{2} + 7 a + 3 a + 21}{a^{2} + 7 a - a - 7} \times \frac{(a + 1) (a - 1)}{(a + 3)} = \frac{a (a + 7) + 3 (a + 7)}{a (a + 7) - 1 (a + 7)} \times \frac{(a + 1) (a - 1)}{(a + 3)} = \frac{(a + 7) (a + 3)}{(a - 1) (a + 7)} \times \frac{(a + 1) (a - 1)}{(a + 3)} = (a + 1)$
$(3) \frac{8 x^{3} - 27 y^{3}}{4 x^{2} - 9 y^{2}} = \frac{{(2 x)}^{3} - {(3 y)}^{3}}{{(2 x)}^{2} - {(3 y)}^{2}} = \frac{(2 x - 3 y) [{(2 x)}^{2} + (2 x) \times (3 y) + {(3 y)}^{2}]}{(2 x + 3 y) (2 x - 3 y)} = \frac{(2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2})}{(2 x + 3 y) (2 x - 3 y)} = \frac{4 x^{2} + 6 x y + 9 y^{2}}{(2 x + 3 y)}$
$(4) \frac{x^{2} - 5 x - 24}{(x + 3) (x + 8)} \times \frac{x^{2} - 64}{{(x - 8)}^{2}} = \frac{x^{2} - 8 x + 3 x - 24}{(x + 3) (x + 8)} \times \frac{{(x)}^{2} - {(8)}^{2}}{{(x - 8)}^{2}} = \frac{x (x - 8) + 3 (x - 8)}{(x + 3) (x + 8)} \times \frac{(x + 8) (x - 8)}{(x - 8) (x - 8)} = \frac{(x + 3) (x - 8)}{(x + 3) (x + 8)} \times \frac{(x + 8) (x - 8)}{(x - 8) (x - 8)} = 1$
$(5) \frac{3 x^{2} - x - 2}{x^{2} - 7 x + 12} \div \frac{3 x^{2} - 7 x - 6}{x^{2} - 4} = \frac{3 x^{2} - 3 x + 2 x - 2}{x^{2} - 4 x - 3 x + 12} \div \frac{3 x^{2} - 9 x + 2 x - 6}{{(x)}^{2} - {(2)}^{2}} = \frac{3 x (x - 1) + 2 (x - 1)}{x (x - 4) - 3 (x - 4)} \div \frac{3 x (x - 3) + 2 (x - 3)}{(x + 2) (x - 2)} = \frac{(3 x + 2) (x - 1)}{(x - 3) (x - 4)} \div \frac{(3 x + 2) (x - 3)}{(x + 2) (x - 2)} = \frac{(3 x + 2) (x - 1)}{(x - 3) (x - 4)} \times \frac{(x + 2) (x - 2)}{(3 x + 2) (x - 3)} = \frac{(x - 1) (x + 2) (x - 2)}{(x - 4) (x - 3) (x - 3)}$
$(6) \frac{4 x^{2} - 11 x + 6}{16 x^{2} - 9} = \frac{4 x^{2} - 8 x - 3 x + 6}{{(4 x)}^{2} - {(3)}^{2}} = \frac{4 x (x - 2) - 3 (x - 2)}{(4 x + 3) (4 x - 3)} = \frac{(4 x - 3) (x - 2)}{(4 x + 3) (4 x - 3)} = \frac{x - 2}{4 x + 3}$
$(7) \frac{a^{3} - 27}{5 a^{2} - 16 a + 3} \div \frac{a^{2} + 3 a + 9}{25 a^{2} - 1} = \frac{{(a)}^{3} - {(3)}^{3}}{5 a^{2} - 15 a - a + 3} \div \frac{a^{2} + 3 a + 9}{{(5 a)}^{2} - {(1)}^{2}} = \frac{(a - 3) \{{(a)}^{2} + (a) \times (3) + {(3)}^{2}\}}{5 a (a - 3) - 1 (a - 3)} \div \frac{a^{2} + 3 a + 9}{(5 a + 1) (5 a - 1)} = \frac{(a - 3) (a^{2} + 3 a + 9)}{(5 a - 1) (a - 3)} \div \frac{(a^{2} + 3 a + 9)}{(5 a + 1) (5 a - 1)} = \frac{(a - 3) (a^{2} + 3 a + 9)}{(5 a - 1) (a - 3)} \times \frac{(5 a + 1) (5 a - 1)}{(a^{2} + 3 a + 9)} = 5 a + 1$
$(8) \frac{1 - 2 x + x^{2}}{1 - x^{3}} \times \frac{1 + x + x^{2}}{1 + x} = \frac{1 - x - x + x^{2}}{{(1)}^{3} - {(x)}^{3}} \times \frac{1 + x + x^{2}}{1 + x} = \frac{1 (1 - x) - x (1 - x)}{(1 - x) \{{(1)}^{2} + (1) \times (x) + {(x)}^{2}\}} \times \frac{(1 + x + x^{2})}{1 + x} = \frac{(1 - x) (1 - x)}{(1 - x) (1 + x + x^{2})} \times \frac{(1 + x + x^{2})}{(1 + x)} = \frac{1 - x}{1 + x}$

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