Exercise 2.4 NCERT Class 9 Chapter 2 Polynomials Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4

Exercise 2.4 NCERT Class 9 contains a total of 16 questions. The questions are based on the topic: Algebraic Identities.

Algebraic Identities:
Identity I: ${(x + y)}^{2} = x^{2} + 2 x y + y^{2}$
Identity II: ${(x - y)}^{2} = x^{2} - 2 x y + y^{2}$
Identity III: $x^{2} - y^{2} = (x + y) (x - y)$
Identity IV: $(x + a) (x + b) = x^{2} + (a + b) x + a b$
Identity V: ${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x$
Identity VI: ${(x + y)}^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}$
Identity VII: ${(x - y)}^{3} = x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3}$
Identity VIII: $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

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Exercise 2.4 NCERT Class 9 Polynomials Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16

The first question of Exercise 2.4 NCERT Class 9 asks us to find products using suitable identities.

1. Use suitable identities to find the following products:

(i) $(x + 4) (x + 10)$

Answer

We have
$(x + 4) (x + 10)$

$= x^{2} + (10 + 4) x + 4 \times 10$ ... $[∵ (x + a) (x + b) = x^{2} + (a + b) x + a b]$

= $x^{2} + 14 x + 40$ .

(ii) $(x + 8) (x - 10)$

Answer

We have
$(x + 8) (x - 10)$

$= x^{2} + {8 + (- 10)} x + 8 \times (- 10)$ ... $[∵ (x + a) (x + b) = x^{2} + (a + b) x + a b]$

$= x^{2} + {8 - 10} x - 80$

$= x^{2} + {- 2} x - 80$

= $x^{2} - 2 x - 80$ .

(iii) $(3 x + 4) (3 x - 5)$

Answer

We have
$(3 x + 4) (3 x - 5)$

= ${(3 x)}^{2} + {4 + (- 5)} x + 4 \times (- 5)$ ... $[∵ (x + a) (x + b) = x^{2} + (a + b) x + a b]$

= $9 x^{2} + {4 - 5} x - 20$

= $9 x^{2} + {- 1} x - 20$

= $9 x^{2} - x - 20$ .

(iv) $(y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2})$

Answer

We have
$(y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2})$

= ${(y^{2})}^{2} - {(\frac{3}{2})}^{2}$ ... $[∵ x^{2} - y^{2} = (x + y) (x - y)]$

= $y^{4} - \frac{3^{2}}{2^{2}}$

= $y^{4} - \frac{9}{4}$

(iv) $(3 - 2 x) (3 + 2 x)$

Answer

We have
$(3 - 2 x) (3 + 2 x)$

= $3^{2} - {(2 x)}^{2}$ ... $[∵ x^{2} - y^{2} = (x + y) (x - y)]$

= $9 - 4 x^{2}$

The second question of Exercise 2.4 NCERT Class 9 asks us to find the product of numbers using suitable identities.

2. Evaluate the following products without multiplying directly:

(i) $103 \times 107$

Answer

We have
$103 \times 107$

= $(100 + 3) (100 + 7)$
Applying $(x + a) (x + b) = x^{2} + (a + b) x + a b$ ,
where, $x = 100, a = 3$ and $b = 7$

= ${(100)}^{2} + (3 + 7) \times 100 + 3 \times 7$

= $10000 + 10 \times 100 + 21$

= $10000 + 1000 + 21$

= $11021$

(ii) $95 \times 96$

Answer

We have
$95 \times 96$

= $(100 - 5) (100 - 4)$
Applying $[(x + a) (x + b) = x^{2} + (a + b) x + a b]$
where $x = 100, a = - 5$ and $b = - 4$

= ${(100)}^{2} + {(- 5) + (- 4)} \times 100 + (- 5) \times (- 4)$

= $10000 + {- 5 - 4} \times 100 + 20$

= $10000 + {- 9} \times 100 + 20$

= $10000 - 900 + 20$

= $10020 - 900$

= $9120$

(iii) $104 \times 96$

Answer

We have
$104 \times 96$

= $(100 + 4) (100 - 4)$

= ${(100)}^{2} - 4^{2}$

= $10000 - 16$

= $9984$

The third question of Exercise 2.4 NCERT Class 9 asks us to factorise the given algebraic expressions using appropriate identities:

3. Factorise the following using appropriate identities:

(i) $9 x^{2} + 6 x y + y^{2}$

Answer

We have
$9 x^{2} + 6 x y + y^{2}$

= ${(3 x)}^{2} + 2 \times (3 x) \times y + y^{2}$
Applying ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ ,
where $a = 3 x, b = y$ .

= ${(3 x + y)}^{2}$

= $(3 x + y) (3 x + y)$ .

(ii) $4 y^{2} + 4 y + 1$

Answer

We have
$4 y^{2} + 4 y + 1$

= ${(2 y)}^{2} + 2 \times (2 y) \times 1 + 1^{2}$
Applying ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ ,
where $a = 2 y, b = 1$ .

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

= $(2 y + 1) (2 y + 1)$

(iii) $x^{2} - \frac{y^{2}}{100}$

Answer

We have
$x^{2} - \frac{y^{2}}{100}$

= $x^{2} - {(\frac{y}{10})}^{2}$
Applying $a^{2} - b^{2} = (a + b) (a - b)$ ,
where $a = x, b = \frac{y}{10}$ .

= $(x + \frac{y}{10}) (x - \frac{y}{10})$

The fourth question of Exercise 2.4 NCERT Class 9 asks us to expand the given algebraic expressions using suitable identities:

4. Expand each of the following, using suitable identities:

(i) ${(x + 2 y + 4 z)}^{2}$

Answer

We have
${(x + 2 y + 4 z)}^{2}$

= $x^{2} + {(2 y)}^{2} + {(4 z)}^{2} + 2 \cdot x \cdot 2 y + 2 \cdot 2 y \cdot 4 z + 2 \cdot 4 z \cdot x$ ... $[∵ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a]$

= $x^{2} + 2^{2} \cdot y^{2} + 4^{2} \cdot z^{2} + 4 x y + 16 y z + 8 z x$

= $x^{2} + 4 y^{2} + 16 z^{2} + 4 x y + 16 y z + 8 x z$

(ii) ${(2 x - y + z)}^{2}$

Answer

We have
${(2 x - y + z)}^{2}$

$= {(2 x)}^{2} + {(- y)}^{2} + z^{2} + 2 \cdot 2 x \cdot (- y) + 2 \cdot (- y) \cdot z + 2 \cdot z \cdot (2 x)$ ... $[∵ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a]$

= $2^{2} \cdot x^{2} + y^{2} + z^{2} - 4 x y - 2 y z + 4 z x$

= $4 x^{2} + y^{2} + z^{2} - 4 x y - 2 y z + 4 z x$

(iii) ${(- 2 x + 3 y + 2 z)}^{2}$

Answer

We have
${(- 2 x + 3 y + 2 z)}^{2}$

$= {(- 2 x)}^{2} + {(3 y)}^{2} + {(2 z)}^{2} + 2 \cdot (- 2 x) \cdot (3 y) + 2 \cdot (3 y) \cdot (2 z) + 2 \cdot (2 z) \cdot (- 2 x)$ ... $[∵ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a]$

= ${(- 2)}^{2} \cdot x^{2} + 3^{2} \cdot y^{2} + 2^{2} \cdot z^{2} - 12 x y + 12 y z - 8 z x$

= $4 x^{2} + 9 y^{2} + 4 z^{2} - 12 x y + 12 y z - 8 z x$

(iv) ${(3 a - 7 b - c)}^{2}$

Answer

We have
${(3 a - 7 b - c)}^{2}$

$= {(3 a)}^{2} + {(- 7 b)}^{2} + {(- c)}^{2} + 2 \cdot (3 a) \cdot (- 7 b) + 2 \cdot (- 7 b) \cdot (- c) + 2 \cdot (- c) \cdot (3 a)$ ... $[∵ {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a]$

= $3^{2} \cdot a^{2} + {(- 7)}^{2} b^{2} + c^{2} - 42 a b + 14 b c - 6 c a$

= $9 a^{2} + 49 b^{2} + c^{2} - 42 a b + 14 b c - 6 c a$

(v) ${(- 2 x + 5 y - 3 z)}^{2}$

Answer

We have
${(- 2 x + 5 y - 3 z)}^{2}$

= $3^{2} \cdot a^{2} + {(- 7)}^{2} b^{2} + c^{2} - 42 a b + 14 b c - 6 c a$

= $9 a^{2} + 49 b^{2} + c^{2} - 42 a b + 14 b c - 6 c a$

(vi) ${[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}$

Answer

We have
${[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}$

$= {(\frac{1}{4} a)}^{2} + {(- \frac{1}{2} b)}^{2} + 1^{2} + 2 \cdot (\frac{1}{4} a) \cdot (- \frac{1}{2} b) + 2 \cdot (- \frac{1}{2} b) \cdot 1 + 2 \cdot 1 \cdot (\frac{1}{4} a)$ ... $[∵ {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x]$

= ${(\frac{1}{4})}^{2} a^{2} + {(- \frac{1}{2})}^{2} b^{2} + 1^{2} - \frac{a b}{4} - b + \frac{1}{2} a$

= $\frac{a^{2}}{16} + \frac{b^{2}}{4} + 1 - \frac{a b}{4} - b + \frac{a}{2}$ .

The fifth question of Exercise 2.4 NCERT Class 9 asks us to factorise the given algebraic expressions.

5. Factorise:

(i) $4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z$

Answer

Since the terms $- 24 y z$ and $- 16 x z$ are negative and $z$ is common in both, we write the term $16 z^{2} = {(- 4 z)}^{2}$ .

We have
$4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z$

$= {(2 x)}^{2} + {(3 y)}^{2} + {(- 4 z)}^{2} + 2 \cdot 2 x \cdot 3 y + 2 \cdot 3 y \cdot (- 4 z) + 2 \cdot (- 4 z) \cdot 2 x$

$= {[2 x + 3 y + (- 4 z)]}^{2}$

$= {(2 x + 3 y - 4 z)}^{2}$

$= (2 x + 3 y - 4 z) (2 x + 3 y - 4 z)$

(ii) $2 x^{2} + y^{2} + 8 z^{2} - 2 \sqrt{2} x y + 4 \sqrt{2} y z - 8 x z$

Answer

Since the terms $- 2 \sqrt{2} x y$ and $- 8 x z$ are negative and $x$ is common in both, we write the term $2 x^{2} = {(- \sqrt{2} x)}^{2}$ .

We have
$2 x^{2} + y^{2} + 8 z^{2} - 2 \sqrt{2} x y + 4 \sqrt{2} y z - 8 x z$

$= {(- \sqrt{2} x)}^{2} + y^{2} + {(2 \sqrt{2} z)}^{2} + 2 \cdot (- \sqrt{2} x) \cdot y + 2 \cdot y \cdot (2 \sqrt{2} z) + 2 \cdot (2 \sqrt{2} z) \cdot (- \sqrt{2} x)$

$= {[- \sqrt{2} x + y + 2 \sqrt{2} z]}^{2}$

$= {(- \sqrt{2} x + y + 2 \sqrt{2} z)}^{2}$

= $(- \sqrt{2} x + y + 2 \sqrt{2} z) (- \sqrt{2} x + y + 2 \sqrt{2} z)$

The sixth question of Exercise 2.4 NCERT Class 9 asks us to write the cubes in expanded form:

6. Write the following cubes in expanded form:

(i) ${(2 x + 1)}^{3}$

Answer

We have
${(2 x + 1)}^{3}$

= ${(2 x)}^{3} + 3 \cdot {(2 x)}^{2} \cdot 1 + 3 \cdot (2 x) \cdot 1^{2} + 1^{3}$ ... $[∵ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}]$
where $a = 2 x, b = 1$

= $2^{3} \cdot x^{3} + 3 \cdot 2^{2} \cdot x^{2} \cdot 1 + 3 \cdot 2 \cdot x \cdot 1 + 1$

= $8 x^{3} + 3 \cdot 4 \cdot x^{2} \cdot 1 + 6 x + 1$

= $8 x^{3} + 12 x^{2} + 6 x + 1$

(ii) ${(2 a - 3 b)}^{3}$

Answer

We have
${(2 a - 3 b)}^{3}$

= ${(2 a)}^{3} - 3 \cdot {(2 a)}^{2} \cdot (3 b) + 3 \cdot (2 a) \cdot {(3 b)}^{2} - {(3 b)}^{3}$ ... $[∵ {(x + y)}^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}]$ ,
where $x = 2 a, y = 3 b$ .

= $2^{3} \cdot a^{3} - 3 \cdot 2^{2} \cdot a^{2} \cdot 3 b + 3 \cdot 2 a \cdot 3^{2} \cdot b^{2} - 3^{3} \cdot b^{3}$

= $8 a^{3} - 3 \cdot 4 \cdot 3 \cdot a^{2} b + 3 \cdot 2 \cdot 9 \cdot a b^{2} - 27 b^{3}$

= $8 a^{3} - 36 a^{2} b + 54 a b^{2} - 27 b^{3}$

(iv) ${[\frac{3}{2} x + 1]}^{3}$

Answer

We have
${[\frac{3}{2} x + 1]}^{3}$

= ${(\frac{3}{2} x)}^{3} + 3 \cdot {(\frac{3}{2} x)}^{2} \cdot 1 + 3 \cdot (\frac{3}{2} x) \cdot 1^{2} + 1^{3}$ ... $[∵ {(a - b)}^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}]$
where $a = \frac{3}{2} x, b = 1$ .

= ${(\frac{3}{2})}^{3} \cdot x^{3} + 3 \cdot {(\frac{3}{2})}^{2} \cdot x^{2} \cdot 1 + 3 \cdot (\frac{3}{2}) \cdot x \cdot 1 + 1$

= $\frac{3^{3}}{2^{3}} \cdot x^{3} + 3 \cdot \frac{3^{2}}{2^{2}} \cdot x^{2} \cdot 1 + 3 \cdot \frac{3}{2} \cdot x \cdot 1 + 1$

= $\frac{27}{8} x^{3} + \frac{27}{4} x^{2} + \frac{9}{2} x + 1$

(iv) ${[x - \frac{2}{3} y]}^{3}$

Answer

We have
${[x - \frac{2}{3} y]}^{3}$

= ${(x)}^{3} - 3 \cdot x^{2} \cdot (\frac{2}{3} y) + 3 \cdot x \cdot {(\frac{2}{3} y)}^{2} - {(\frac{2}{3} y)}^{3}$ ... $[∵ {(a - b)}^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}]$ ,
where $a = x, b = \frac{2}{3} y$ .

= $x^{3} - 3 \cdot x^{2} \cdot \frac{2}{3} y + 3 \cdot x \cdot \frac{2^{2}}{3^{2}} \cdot y^{2} - \frac{2^{3}}{3^{3}} \cdot y^{3}$

= $x^{3} - 2 x^{2} y + \frac{4}{3} x y^{2} - \frac{8}{27} y^{3}$

The seventh question of Exercise 2.4 NCERT Class 9 asks us to evaluate the given numbers using suitable identities.

7. Evaluate the following using suitable identities:

(i) ${(99)}^{3}$

Answer

We have
${(99)}^{3}$

= ${(100 - 1)}^{3}$ ... $[∵ 99 = 100 - 1]$

= ${(100)}^{3} - 3 \cdot {(100)}^{2} \cdot 1 + 3 \cdot 100 \cdot 1^{2} - 1^{3}$ ... $[∵ {(a - b)}^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}]$ ,
where $a = 100, b = 1$ .

= $1000000 - 3 \times 10000 + 300 - 1$

= $1000000 - 30000 + 300 - 1$

= $1000300 - 30001$

= $970299$ .

(ii) ${(102)}^{3}$

Answer

We have
${(102)}^{3}$

= ${(100 + 2)}^{3}$ ... $[∵ 102 = 100 + 2]$

Applying ${(x + y)}^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}$ ,
where $x = 100, y = 2$ .

= ${(100)}^{3} + 3 \cdot {(100)}^{2} \cdot 2 + 3 \cdot 100 \cdot 2^{2} + 2^{3}$

= $1000000 + 60000 + 1200 + 8$

= $1061208$

(ii) ${(998)}^{3}$

Answer

We have
${(998)}^{3}$

= ${(1000 - 2)}^{3}$

= ${(1000)}^{3} - 3 \cdot {(1000)}^{2} \cdot 2 + 3 \cdot 1000 \cdot 2^{2} - 2^{3}$

= $1000000000 - 3 \cdot 1000000 \cdot 2 + 3 \cdot 1000 \cdot 4 - 8$

= $1000000000 - 6000000 + 12000 - 8$

= $1000012000 - 6000008$

= $994011992$

The eighth question of Exercise 2.4 NCERT Class 9 asks us to factorise the given algebraic identities using suitable identities.

8. Factorise each of the following:

(i) $8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}$

Answer

We have
$8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}$

= ${(2 a)}^{3} + b^{3} + 3 \cdot {(2 a)}^{2} \cdot b + 3 \cdot 2 a \cdot b^{2}$
Applying $(x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2}) = {(x + y)}^{3}$ ,
where $x = 2 a, y = b$ .

= ${(2 a + b)}^{3}$

= $(2 a + b) (2 a + b) (2 a + b)$

(ii) $8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}$

Answer

We have
$8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}$

= ${(2 a)}^{3} - b^{3} - 3 \cdot {(2 a)}^{2} \cdot b + 3 \cdot 2 a \cdot b^{2}$
Applying $(x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}) = {(x - y)}^{3}$ ,
where $x = 2 a, y = b$ .

= ${(2 a - b)}^{3}$

= $(2 a - b) (2 a - b) (2 a - b)$

(iii) $27 - 125 a^{3} - 135 a + 225 a^{2}$

Answer

We have
$27 - 125 a^{3} - 135 a + 225 a^{2}$

= ${(3)}^{3} - {(5 a)}^{3} - 3 \cdot {(3)}^{2} \cdot 5 a + 3 \cdot 3 \cdot {(5 a)}^{2}$
Applying $(x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}) = {(x - y)}^{3}$
where $x = 3, y = 5 a$

= ${(3 - 5 a)}^{3}$

= $(3 - 5 a) (3 - 5 a) (3 - 5 a)$ .

(iv) $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

Answer

We have
$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

= ${(4 a)}^{3} - {(3 b)}^{3} - 3 \cdot {(4 a)}^{2} \cdot 3 b + 3 \cdot 4 a \cdot {(3 b)}^{2}$
Applying $(x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}) = {(x - y)}^{3}$
where $x = 4 a, y = 3 b$ .

= ${(4 a - 3 b)}^{3}$

= $(4 a - 3 b) (4 a - 3 b) (4 a - 3 b)$

(v) $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

Answer

We have
$27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

= ${(3 p)}^{3} - {(\frac{1}{6})}^{3} - 3 \cdot {(3 p)}^{2} \cdot \frac{1}{6} + 3 \cdot 3 p \cdot {(\frac{1}{6})}^{2}$
Applying $(x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2}) = {(x - y)}^{3}$
where $x = 3 p, y = \frac{1}{6}$

= ${(3 p - \frac{1}{6})}^{3}$

= $(3 p - \frac{1}{6}) (3 p - \frac{1}{6}) (3 p - \frac{1}{6})$ .

The ninth question of Exercise 2.4 NCERT Class 9 asks us to verify the given algebraic identities.

9. Verify:

(i) $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$

Answer

To Verify: $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$

Taking RHS, we have
$(x + y) (x^{2} - x y + y^{2})$

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

= $x^{3} - x^{2} y + x y^{2} + y x^{2} - x y^{2} + y^{3}$

= $x^{3} + y^{3}$ ... $[∵ - x^{2} y + y x^{2} = 0, x y^{2} - x y^{2} = 0]$

= LHS
Hence, verified.

(ii) $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

Answer

To Verify: $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

Taking RHS, we have
$(x - y) (x^{2} + x y + y^{2})$

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

= $x^{3} + x^{2} y + x y^{2} - y x^{2} - x y^{2} - y^{3}$

= $x^{3} - y^{3}$ ... $[∵ x^{2} y - y x^{2} = 0, x y^{2} - x y^{2} = 0]$

= LHS
Hence, verified.

The eleventh question of Exercise 2.4 NCERT Class 9 asks us to factorise the given algebraic expression.

11. Factorise : $27 x^{3} + y^{3} + z^{3} - 9 x y z$

Answer

We have
$27 x^{3} + y^{3} + z^{3} - 9 x y z$

= ${(3 x)}^{3} + y^{3} + z^{3} - 3 \cdot (3 x) \cdot y \cdot z$
Applying $a^{3} + b^{3} + c^{3} - 3 a b c =$ $(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
where $a = 3 x, b = y, c = z$

= $(3 x + y + z) {{(3 x)}^{2} + y^{2} + z^{2} - (3 x) \cdot y - y \cdot z - z \cdot 3 x}$

= $(3 x + y + z) (9 x^{2} + y^{2} + z^{2} - 3 x y - y z - 3 z x)$

The twelfth question of Exercise 2.4 NCERT Class 9 asks us to verify the given algebraic equation.

12. Verify that $x^{3} + y^{3} + z^{3} - 3 x y z$ $= \frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]$

Answer

To verify:
$x^{3} + y^{3} + z^{3} - 3 x y z$ $= \frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]$

Taking LHS, we have
$x^{3} + y^{3} + z^{3} - 3 x y z$

$= (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ ... [Using identities]

Multiplying and dividing by 2, we get.

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

$= \frac{1}{2} (x + y + z) \times 2 \times (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

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

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

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

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

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

= RHS.
Hence, verified.

You can do this by simplifying RHS also.

The thirteenth question of Exercise 2.4 NCERT Class 9 asks us to prove an equality at a given condition.

13. If $x + y + z = 0$ , show that $x^{3} + y^{3} + z^{3} = 3 x y z$ .

Answer

We have,
$x + y + z = 0$

Cubing on both sides, we get
${(x + y + z)}^{3} = 0$

$\Rightarrow {(x + y + z)}^{2} (x + y + z) = 0$

$\Rightarrow (x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x) (x + y + z) = 0$ ... $[∵ {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x]$

$\Rightarrow x (x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x) + y (x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x) + z (x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x) = 0$

$\Rightarrow x^{3} + x y^{2} + x z^{2} + 2 x^{2} y + 2 x y z + 2 z x^{2} + y x^{2} + y^{3} + y z^{2} + 2 x y^{2} + 2 y^{2} z + 2 z x y + z x^{2} + z y^{2} + z^{3} + 2 x y z + 2 y z^{2} + 2 z^{2} x = 0$
Adding like terms, we get

$\Rightarrow x^{3} + y^{3} + z^{3} + 6 x y z + 3 x y^{2} + 3 x z^{2} + 3 x^{2} y + 3 z x^{2} + 3 y z^{2} + 3 y^{2} z = 0$
Adding and subtracting $3 x y z$ , we get

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + 9 x y z + 3 x y^{2} + 3 x z^{2} + 3 x^{2} y + 3 z x^{2} + 3 y z^{2} + 3 y^{2} z = 0$
Now, regrouping the terms to get $(x + y + z)$ as common, we get

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + (3 x^{2} y + 3 x y^{2} + 3 x y z) + (3 z x^{2} + 3 x y z + 3 x z^{2}) + (3 x y z + 3 y^{2} z + 3 y z^{2}) = 0$
Taking common as much as possible from each group, we get

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + 3 x y (x + y + z) + 3 z x (x + y + z) + 3 y z (x + y + z) = 0$

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + (3 x y + 3 z x + 3 y z) (x + y + z) = 0$

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + (3 x y + 3 z x + 3 y z) \times 0 = 0$ ... $[∵ (x + y + z) = 0]$

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z + 0 = 0$

$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z = 0$

$\Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z$
Hence proved!

The fourteenth question of Exercise 2.4 NCERT Class 9 asks us to find the value of the given expressions.

14. Without actually calculating the cubes, find the value of each of the following:

(i) ${(- 12)}^{3} + 7^{3} + 5^{3}$

Answer

We have
${(- 12)}^{3} + 7^{3} + 5^{3}$

$∵ (- 12 + 7 + 5) = 0$

$∴ {(- 12)}^{3} + 7^{3} + 5^{3} = 3 \times (- 12) \times 7 \times 5 = - 1260$ ... $[∵ (x + y + z) = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z]$

(i) ${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3}$

Answer

We have
${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3}$

$∵ 28 + (- 15) + (- 13) = 0$

$∴ {(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3} = 3 \times 28 \times (- 15) \times (- 13) = 16380$ ... $[∵ (x + y + z) = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z]$

The fifteenth question of Exercise 2.4 NCERT Class 9 is a word problem based on algebraic identities.

15. Give possible expressions for the length and breadth of each of the following rectangles, in which thier areas are given:

(i) Area : $25 a^{2} - 35 a + 12$

Answer

Let the length of the rectangle be $l$ , and breadth be $b$ .

∵ Area of the rectangle = $25 a^{2} - 35 a + 12$
$\Rightarrow l \times b = 25 a^{2} - 35 a + 12$

$\Rightarrow l \times b = 25 a^{2} - 20 a - 15 + 12$ ... $[∵ 35 = 20 + 15, 25 \times 12 = 20 \times 15]$

$\Rightarrow l \times b = 5 a (5 a - 4) - 3 (5 a - 4)$

$\Rightarrow l \times b = (5 a - 3) (5 a - 4)$
Comparing on both sides we get

$l = (5 a - 3), b = (5 a - 4)$

This is one possibility. There may be many other values.

(ii) Area : $35 y^{2} + 13 y - 12$

Answer

Let the length of the rectangle be $l$ , and breadth be $b$ .

∵ Area of the rectangle = $35 y^{2} + 13 y - 12$
$\Rightarrow l \times b = 35 y^{2} + 13 a - 12$

$\Rightarrow l \times b = 35 y^{2} + (28 - 15) y - 12$ ... $[∵ 13 = 28 - 15, 35 \times 12 = 28 \times 15]$

$\Rightarrow l \times b = 35 y^{2} + 28 y - 15 y - 12$

$\Rightarrow l \times b = 7 y (5 y + 4) - 3 (5 y + 4)$

$\Rightarrow l \times b = (7 y - 3) (5 y + 4)$

On comparing on both sides,
$l = (7 y - 3), b = (5 y + 4)$

This is one possibility. There may be many other values.

The sixteenth question of Exercise 2.4 NCERT Class 9 is a word problem based on algebraic identities.

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume : $3 x^{2} - 12 x$

Answer

Let the length, breadth and height of the cuboid be $l, b$ and $h$ respectively.

∵ Volume = $3 x^{2} - 12 x$

∴ $l \times b \times h = 3 x^{2} - 12 x$

$\Rightarrow l \times b \times h = 3 x (x - 4)$

$\Rightarrow l \times b \times h = 3 \times x \times (x - 4)$

On comparing, we get
$l = 3, b = x, h = (x - 4)$

Hence, the one possible answer is $3, x, x - 4$ .

(ii) Volume : $12 k y^{2} + 8 k y - 20 k$

Answer

Let the length, breadth and height of the cuboid be $l, b$ and $h$ respectively.

∵ Volume = $12 k y^{2} + 8 k y - 20 k$

$∴ l \times b \times h = 12 k y^{2} + 8 k y - 20 k$

$\Rightarrow l \times b \times h = 4 k (3 y^{2} + 2 y - 5)$

$\Rightarrow l \times b \times h = 4 k (3 y^{2} + 5 y - 3 y - 5)$ ... $[∵ 2 = 5 - 3, 15 = 5 \times 3]$

$\Rightarrow l \times b \times h = 4 k {y (3 y + 5) - 1 (3 y - 5)}$

$\Rightarrow l \times b \times h = 4 k (y - 1) (3 y - 5)$

On comparing both sides, we get
$l = 4 k, b = (y - 1), h = (3 y - 5)$

Hence, one possible answer is : $4 k, (y - 1), (3 y - 5)$ .

Download NCERT Class 9 Book

Exercise 2.4 NCERT Class 9 Polynomials Mathematics Solutions

1. Use suitable identities to find the following products:

(i) (x+4)(x+10)

Answer

(ii) (x+8)(x−10)

Answer

(iii) (3x+4)(3x−5)

Answer

(iv) (y2+32)(y2−32)

Answer

(iv) (3−2x)(3+2x)

Answer

2. Evaluate the following products without multiplying directly:

(i) 103×107

Answer

(ii) 95×96

Answer

(iii) 104×96

Answer

3. Factorise the following using appropriate identities:

(i) 9x2+6xy+y2

Answer

(ii) 4y2+4y+1

Answer

(iii) x2−y2100

Answer

4. Expand each of the following, using suitable identities:

(i) (x+2y+4z)2

Answer

(ii) (2x−y+z)2

Answer

(iii) (−2x+3y+2z)2

Answer

(iv) (3a−7b−c)2

Answer

(v) (−2x+5y−3z)2

Answer

(vi) [14a−12b+1]2

Answer

5. Factorise:

(i) 4x2+9y2+16z2+12xy−24yz−16xz

Answer

(ii) 2x2+y2+8z2−22xy+42yz−8xz

Answer

6. Write the following cubes in expanded form:

(i) (2x+1)3

Answer

(ii) (2a−3b)3

Answer

(iv) [32x+1]3

Answer

(iv) [x−23y]3

Answer

7. Evaluate the following using suitable identities:

(i) (99)3

Answer

(ii) (102)3

Answer

(ii) (998)3

Answer

8. Factorise each of the following:

(i) 8a3+b3+12a2b+6ab2

Answer

(ii) 8a3−b3−12a2b+6ab2

Answer

(iii) 27−125a3−135a+225a2

Answer

(iv) 64a3−27b3−144a2b+108ab2

Answer

(v) 27p3−1216−92p2+14p

Answer

9. Verify:

(i) x3+y3=(x+y)(x2−xy+y2)

Answer

(ii) x3−y3=(x−y)(x2+xy+y2)

Answer

11. Factorise : 27x3+y3+z3−9xyz

Answer

12. Verify that x3+y3+z3−3xyz=12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]

(i) $(x + 4) (x + 10)$

(ii) $(x + 8) (x - 10)$

(iii) $(3 x + 4) (3 x - 5)$

(iv) $(y^{2} + \frac{3}{2}) (y^{2} - \frac{3}{2})$

(iv) $(3 - 2 x) (3 + 2 x)$

(i) $103 \times 107$

(ii) $95 \times 96$

(iii) $104 \times 96$

(i) $9 x^{2} + 6 x y + y^{2}$

(ii) $4 y^{2} + 4 y + 1$

(iii) $x^{2} - \frac{y^{2}}{100}$

(i) ${(x + 2 y + 4 z)}^{2}$

(ii) ${(2 x - y + z)}^{2}$

(iii) ${(- 2 x + 3 y + 2 z)}^{2}$

(iv) ${(3 a - 7 b - c)}^{2}$

(v) ${(- 2 x + 5 y - 3 z)}^{2}$

(vi) ${[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}$

(i) $4 x^{2} + 9 y^{2} + 16 z^{2} + 12 x y - 24 y z - 16 x z$

(ii) $2 x^{2} + y^{2} + 8 z^{2} - 2 \sqrt{2} x y + 4 \sqrt{2} y z - 8 x z$

(i) ${(2 x + 1)}^{3}$

(ii) ${(2 a - 3 b)}^{3}$

(iv) ${[\frac{3}{2} x + 1]}^{3}$

(iv) ${[x - \frac{2}{3} y]}^{3}$

(i) ${(99)}^{3}$

(ii) ${(102)}^{3}$

(ii) ${(998)}^{3}$

(i) $8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}$

(ii) $8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}$

(iii) $27 - 125 a^{3} - 135 a + 225 a^{2}$

(iv) $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

(v) $27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p$

(i) $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$

(ii) $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

11. Factorise : $27 x^{3} + y^{3} + z^{3} - 9 x y z$

12. Verify that $x^{3} + y^{3} + z^{3} - 3 x y z$ $= \frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]$

13. If $x + y + z = 0$ , show that $x^{3} + y^{3} + z^{3} = 3 x y z$ .

(i) ${(- 12)}^{3} + 7^{3} + 5^{3}$

(i) ${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3}$

(i) Area : $25 a^{2} - 35 a + 12$

(ii) Area : $35 y^{2} + 13 y - 12$

(i) Volume : $3 x^{2} - 12 x$

(ii) Volume : $12 k y^{2} + 8 k y - 20 k$