myhelper: RD Sharma 2020 solution class 9 chapter 5 Factorization of Algebraic Expressions FBQS

FBQS

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Question 1:

The factorized form of the expression y² + (x – 1)y – x is ____________.

Answer 1:

$y^{2} + (x - 1) y - x = y^{2} + x y - y - x = y (y + x) - 1 (y + x) = (y - 1) (y + x)$

Hence, the factorized form of the expression y² + (x – 1)y – x is (y – 1)(y + x).

Question 2:

The factorized form of a³ + (b – a)³ – b³ is ____________.

Answer 2:

$a^{3} + {(b - a)}^{3} - b^{3} = a^{3} + (b^{3} - a^{3} - 3 b a (b - a)) - b^{3} (Using the identity : {(x - y)}^{3} = x^{3} - y^{3} - 3 x y (x - y)) = a^{3} + b^{3} - a^{3} - 3 b a (b - a) - b^{3} = - 3 b a (b - a) = 3 a b (a - b)$

Hence, the factorized form of a³ + (b – a)³ – b³ is 3ab(a – b).

Question 3:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c =____________.$

Answer 3:

$Given : \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 . . . (1) a b c = 2 . . . (2) Now, a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = a^{2} b^{2} c^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(a b c)}^{2} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = {(2)}^{2} (1) (From (1) and (2)) = 4$

Hence, if

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c = 4 .$

Question 4:

Factorization of the polynomial

$11 x^{2} - 10 \sqrt{3} x - 3$ gives ____________.

Answer 4:

$11 x^{2} - 10 \sqrt{3} x - 3 = 11 x^{2} - 11 \sqrt{3} x + \sqrt{3} x - 3 = 11 x (x - \sqrt{3}) + \sqrt{3} (x - \sqrt{3}) = (11 x + \sqrt{3}) (x - \sqrt{3})$

Hence, factorization of the polynomial

$11 x^{2} - 10 \sqrt{3} x - 3$ gives

$(11 x + \sqrt{3}) (x - \sqrt{3})$ .

Question 5:

The polynomial x² + y² – z² – 2xy on factorization gives _____________.

Answer 5:

$x^{2} + y^{2} - z^{2} - 2 x y = (x^{2} + y^{2} - 2 x y) - z^{2} = {(x - y)}^{2} - {(z)}^{2} (Using the identity : {(a - b)}^{2} = a^{2} + b^{2} - 2 a b) = (x - y + z) (x - y - z) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the polynomial x² + y² – z² – 2xy on factorization gives

$(x - y + z) (x - y - z)$ .

Question 6:

The factors of the expression

$a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are ____________.

Answer 6:

$a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a} = {(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} + {(- \sqrt{c})}^{2} + 2 \sqrt{a} \sqrt{b} + 2 \sqrt{b} (- \sqrt{c}) + 2 \sqrt{a} (- \sqrt{c}) = {(\sqrt{a} + \sqrt{b} - \sqrt{c})}^{2} (Using the identity : a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c = {(a + b + c)}^{2}) = (\sqrt{a} + \sqrt{b} - \sqrt{c}) (\sqrt{a} + \sqrt{b} - \sqrt{c})$

Hence, the factors of the expression

$a + b + c + 2 \sqrt{a b} - 2 \sqrt{b c} - 2 \sqrt{c a}$ are

$(\sqrt{a} + \sqrt{b} - \sqrt{c}) and (\sqrt{a} + \sqrt{b} - \sqrt{c}) .$ .

Question 7:

The polynomial , x⁶ + 64y⁶ on factorization gives _____________.

Answer 7:

$x^{6} + 64 y^{6} = {(x^{2})}^{3} + {(4 y^{2})}^{3} = (x^{2} + 4 y^{2}) ({(x^{2})}^{2} + {(4 y^{2})}^{2} - (x^{2}) (4 y^{2})) (Using the identity : a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)) = (x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$

Hence, the polynomial x⁶ + 64y⁶ on factorization gives

$(x^{2} + 4 y^{2}) (x^{4} + 16 y^{4} - 4 x^{2} y^{2})$ .

Question 8:

The factorization form of a⁴ + b⁴– a²b² is _____________.

Answer 8:

$a^{4} + b^{4} - a^{2} b^{2} Adding and subtracting 2 a^{2} b^{2}, we get = {(a^{2})}^{2} + {(b^{2})}^{2} + 2 a^{2} b^{2} - 2 a^{2} b^{2} - a^{2} b^{2} = {(a^{2} + b^{2})}^{2} - 3 a^{2} b^{2} (Using the identity : a^{2} + b^{2} + 2 a b = {(a + b)}^{2}) = {(a^{2} + b^{2})}^{2} - {(\sqrt{3} a b)}^{2} = (a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b) (Using the identity : a^{2} - b^{2} = (a + b) (a - b))$

Hence, the factorization form of a⁴ + b⁴– a²b² is

$(a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b)$ .

Question 9:

$3 x - \frac{y}{5} = 10 and x y = 5$ , then the value of

$27 x^{3} - \frac{y^{3}}{125}$ is ____________.

Answer 9:

$Given : 3 x - \frac{y}{5} = 10 . . . (1) x y = 5 . . . (2) Now, 3 x - \frac{y}{5} = 10 Taking cube on both sides, we get \Rightarrow {(3 x - \frac{y}{5})}^{3} = 10^{3} \Rightarrow {(3 x)}^{3} - {(\frac{y}{5})}^{3} - 3 (3 x) (\frac{y}{5}) (3 x - \frac{y}{5}) = 1000 (Using the identity : {(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a - b)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} x y (3 x - \frac{y}{5}) = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - \frac{9}{5} \times 5 \times 10 = 1000 (From (1) and (2)) \Rightarrow 27 x^{3} - \frac{y^{3}}{125} - 90 = 1000 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1000 + 90 \Rightarrow 27 x^{3} - \frac{y^{3}}{125} = 1090$

Hence, the value of

$27 x^{3} - \frac{y^{3}}{125}$ is 1090.

Question 10:

The factorized form of

$\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is _____________.

Answer 10:

$\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \frac{(x^{2} + y^{2} + z^{2})}{x y z} + 2 (\frac{y z + x z + x y}{x y x}) = \frac{x^{2} + y^{2} + z^{2} + 2 (y z + x z + x y)}{x y z} = \frac{{(x + y + z)}^{2}}{x y z} (Using the identity : {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c) = \frac{1}{x y z} {(x + y + z)}^{2}$

Hence, the factorized form of

$\frac{1}{x y z} (x^{2} + y^{2} + z^{2}) + 2 (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is

$\frac{1}{x y z} {(x + y + z)}^{2} .$

Question 11:

The factorized form of a³ + b³ + 3ab – 1 is ____________.

Answer 11:

$a^{3} + b^{3} + 3 a b - 1 = a^{3} + b^{3} - 1 + 3 a b = a^{3} + b^{3} + {(- 1)}^{3} - 3 a b (- 1) = (a + b - 1) (a^{2} + b^{2} + {(- 1)}^{2} - a b - b (- 1) - a (- 1)) (Using the identity : a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)) = (a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a)$

Hence, the factorized form of a³ + b³ + 3ab – 1 is

$(a + b - 1) (a^{2} + b^{2} + 1 - a b + b + a) .$

RD Sharma 2020 solution class 9 chapter 5 Factorization of Algebraic Expressions FBQS