Chapter 2 of 12

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Updated 11 June 2026

NCERT Solutions

Polynomials

Madhya Pradesh Board · Class 9 · Mathematics

NCERT Solutions for Polynomials — Madhya Pradesh Board Class 9 Mathematics.

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30 Questions Solved · 4 Sections

Exercise 2.1

1Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i)

4x^{2} - 3x + 7

(ii)

y^{2} + \sqrt{2}

(iii)

3\sqrt{t} + t\sqrt{2}

(iv)

y + \frac{2}{y}

(v)

x^{10} + y^3 + t^{50}

Show solution

Concept: A polynomial in one variable is an expression of the form

a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0

where all exponents of the variable are whole numbers (non-negative integers).

(i) $4x^{2} - 3x + 7$
All exponents of

x

are whole numbers (2, 1, 0). It is a polynomial in one variable

x

.

(ii) $y^{2} + \sqrt{2}$
All exponents of

y

are whole numbers (2, 0). It is a polynomial in one variable

y

.

(iii) $3\sqrt{t} + t\sqrt{2}$

3\sqrt{t} = 3t^{1/2}

. The exponent

\frac{1}{2}

is not a whole number. Hence it is not a polynomial.

(iv) $y + \frac{2}{y}$

\frac{2}{y} = 2y^{-1}

. The exponent

-1

is not a whole number. Hence it is not a polynomial.

(v) $x^{10} + y^3 + t^{50}$
This expression contains three variables

x

y

, and

t

. Hence it is not a polynomial in one variable (it is a polynomial in three variables).

2Write the coefficients of

x^2

in each of the following:
(i)

2 + x^{2} + x

(ii)

2 - x^{2} + x^{3}

(iii)

\frac{\pi}{2} x^2 + x

(iv)

\sqrt{2} x - 1

Show solution

Concept: The coefficient of

x^2

is the number multiplied with

x^2

in the expression.

(i) $2 + x^{2} + x$
The term containing

x^2

1 \cdot x^2

.
Coefficient of

x^2

\mathbf{1}

(ii) $2 - x^{2} + x^{3}$
The term containing

x^2

-1 \cdot x^2

.
Coefficient of

x^2

\mathbf{-1}

(iii) $\frac{\pi}{2} x^2 + x$
The term containing

x^2

\frac{\pi}{2} x^2

.
Coefficient of

x^2

\dfrac{\boldsymbol{\pi}}{\mathbf{2}}

(iv) $\sqrt{2} x - 1$
There is no

x^2

term in this expression.
Coefficient of

x^2

\mathbf{0}

3Give one example each of a binomial of degree 35, and of a monomial of degree 100.Show solution

Concept:
- A binomial has exactly two terms.
- A monomial has exactly one term.
- The degree is the highest power of the variable.

Binomial of degree 35:

x^{35} + 1

This has two terms and the highest power is 35.

Monomial of degree 100:

x^{100}

This has one term and the highest power is 100.

4Write the degree of each of the following polynomials:
(i)

5x^{3} + 4x^{2} + 7x

(ii)

4 - y^{2}

(iii)

5t - \sqrt{7}

(iv) 3Show solution

Concept: The degree of a polynomial is the highest power of the variable in the polynomial.

(i) $5x^{3} + 4x^{2} + 7x$
Highest power of

x

is 3.
Degree =

\mathbf{3}

(ii) $4 - y^{2}$
Highest power of

y

is 2.
Degree =

\mathbf{2}

(iii) $5t - \sqrt{7}$
Highest power of

t

is 1.
Degree =

\mathbf{1}

(iv) $3$

3 = 3 \cdot x^0

. This is a non-zero constant polynomial.
Degree =

\mathbf{0}

5Classify the following as linear, quadratic and cubic polynomials:
(i)

x^{2} + x

(ii)

x - x^3

(iii)

y + y^2 + 4

(iv)

1 + x

(v)

3t

(vi)

r^2

(vii)

7x^3

Show solution

Concept:
- Linear polynomial: degree 1
- Quadratic polynomial: degree 2
- Cubic polynomial: degree 3

(i) $x^{2} + x$ — Highest degree = 2 → Quadratic polynomial

(ii) $x - x^3$ — Highest degree = 3 → Cubic polynomial

(iii) $y + y^2 + 4$ — Highest degree = 2 → Quadratic polynomial

(iv) $1 + x$ — Highest degree = 1 → Linear polynomial

(v) $3t$ — Highest degree = 1 → Linear polynomial

(vi) $r^2$ — Highest degree = 2 → Quadratic polynomial

(vii) $7x^3$ — Highest degree = 3 → Cubic polynomial

Exercise 2.2

1Find the value of the polynomial

5x - 4x^2 + 3

at
(i)

x = 0

(ii)

x = -1

(iii)

x = 2

Show solution

Let

p(x) = 5x - 4x^2 + 3

(i) At $x = 0$ :

p(0) = 5(0) - 4(0)^2 + 3 = 0 - 0 + 3 = \mathbf{3}

(ii) At $x = -1$ :

p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = \mathbf{-6}

(iii) At $x = 2$ :

p(2) = 5(2) - 4(2)^2 + 3 = 10 - 4(4) + 3 = 10 - 16 + 3 = \mathbf{-3}

2Find

p(0), p(1)

and

p(2)

for each of the following polynomials:
(i)

p(y) = y^2 - y + 1

(ii)

p(t) = 2 + t + 2t^2 - t^3

(iii)

p(x) = x^3

(iv)

p(x) = (x - 1)(x + 1)

Show solution

(i) $p(y) = y^2 - y + 1$

p(0) = 0 - 0 + 1 = \mathbf{1}

p(1) = 1 - 1 + 1 = \mathbf{1}

p(2) = 4 - 2 + 1 = \mathbf{3}

(ii) $p(t) = 2 + t + 2t^2 - t^3$

p(0) = 2 + 0 + 0 - 0 = \mathbf{2}

p(1) = 2 + 1 + 2(1) - 1 = 2 + 1 + 2 - 1 = \mathbf{4}

p(2) = 2 + 2 + 2(4) - 8 = 2 + 2 + 8 - 8 = \mathbf{4}

(iii) $p(x) = x^3$

p(0) = 0^3 = \mathbf{0}

p(1) = 1^3 = \mathbf{1}

p(2) = 2^3 = \mathbf{8}

(iv) $p(x) = (x-1)(x+1)$

p(0) = (0-1)(0+1) = (-1)(1) = \mathbf{-1}

p(1) = (1-1)(1+1) = (0)(2) = \mathbf{0}

p(2) = (2-1)(2+1) = (1)(3) = \mathbf{3}

3Verify whether the following are zeroes of the polynomial, indicated against them.
(i)

p(x) = 3x + 1,\ x = -\frac{1}{3}

(ii)

p(x) = 5x - \pi,\ x = \frac{4}{5}

(iii)

p(x) = x^2 - 1,\ x = 1, -1

(iv)

p(x) = (x + 1)(x - 2),\ x = -1, 2

(v)

p(x) = x^2,\ x = 0

(vi)

p(x) = lx + m,\ x = -\frac{m}{l}

(vii)

p(x) = 3x^2 - 1,\ x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}

(viii)

p(x) = 2x + 1,\ x = \frac{1}{2}

Show solution

Concept:

x = a

is a zero of

p(x)

if and only if

p(a) = 0

.

(i) $p(x) = 3x + 1$ at $x = -\frac{1}{3}$ :

p\!\left(-\tfrac{1}{3}\right) = 3\left(-\tfrac{1}{3}\right) + 1 = -1 + 1 = 0

Yes,

x = -\dfrac{1}{3}

is a zero of

p(x)

.

(ii) $p(x) = 5x - \pi$ at $x = \frac{4}{5}$ :

p\!\left(\tfrac{4}{5}\right) = 5\left(\tfrac{4}{5}\right) - \pi = 4 - \pi \neq 0

No,

x = \dfrac{4}{5}

is not a zero of

p(x)

.

(iii) $p(x) = x^2 - 1$ at $x = 1$ and $x = -1$ :

p(1) = 1^2 - 1 = 0

p(-1) = (-1)^2 - 1 = 1 - 1 = 0

Yes, both

x = 1

and

x = -1

are zeroes of

p(x)

.

(iv) $p(x) = (x+1)(x-2)$ at $x = -1$ and $x = 2$ :

p(-1) = (-1+1)(-1-2) = (0)(-3) = 0

p(2) = (2+1)(2-2) = (3)(0) = 0

Yes, both

x = -1

and

x = 2

are zeroes of

p(x)

.

(v) $p(x) = x^2$ at $x = 0$ :

p(0) = 0^2 = 0

Yes,

x = 0

is a zero of

p(x)

.

(vi) $p(x) = lx + m$ at $x = -\frac{m}{l}$ :

p\!\left(-\tfrac{m}{l}\right) = l\left(-\tfrac{m}{l}\right) + m = -m + m = 0

Yes,

x = -\dfrac{m}{l}

is a zero of

p(x)

.

(vii) $p(x) = 3x^2 - 1$ at $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{2}{\sqrt{3}}$ :

p\!\left(-\tfrac{1}{\sqrt{3}}\right) = 3\left(\tfrac{1}{3}\right) - 1 = 1 - 1 = 0

x = -\dfrac{1}{\sqrt{3}}

is a zero.

p\!\left(\tfrac{2}{\sqrt{3}}\right) = 3\left(\tfrac{4}{3}\right) - 1 = 4 - 1 = 3 \neq 0

x = \dfrac{2}{\sqrt{3}}

is not a zero of

p(x)

.

(viii) $p(x) = 2x + 1$ at $x = \frac{1}{2}$ :

p\!\left(\tfrac{1}{2}\right) = 2\left(\tfrac{1}{2}\right) + 1 = 1 + 1 = 2 \neq 0

No,

x = \dfrac{1}{2}

is not a zero of

p(x)

4Find the zero of the polynomial in each of the following cases:
(i)

p(x) = x + 5

(ii)

p(x) = x - 5

(iii)

p(x) = 2x + 5

(iv)

p(x) = 3x - 2

(v)

p(x) = 3x

(vi)

p(x) = ax,\ a \neq 0

(vii)

p(x) = cx + d,\ c \neq 0

Show solution

Concept: To find the zero, set

p(x) = 0

and solve for

x

.

(i) $p(x) = x + 5$ :

x + 5 = 0 \implies x = -5

Zero is

\mathbf{-5}

.

(ii) $p(x) = x - 5$ :

x - 5 = 0 \implies x = 5

Zero is

\mathbf{5}

.

(iii) $p(x) = 2x + 5$ :

2x + 5 = 0 \implies 2x = -5 \implies x = -\dfrac{5}{2}

Zero is

\mathbf{-\dfrac{5}{2}}

.

(iv) $p(x) = 3x - 2$ :

3x - 2 = 0 \implies 3x = 2 \implies x = \dfrac{2}{3}

Zero is

\mathbf{\dfrac{2}{3}}

.

(v) $p(x) = 3x$ :

3x = 0 \implies x = 0

Zero is

\mathbf{0}

.

(vi) $p(x) = ax,\ a \neq 0$ :

ax = 0 \implies x = 0

Zero is

\mathbf{0}

.

(vii) $p(x) = cx + d,\ c \neq 0$ :

cx + d = 0 \implies cx = -d \implies x = -\dfrac{d}{c}

Zero is

\mathbf{-\dfrac{d}{c}}

Exercise 2.3

1Determine which of the following polynomials has

(x + 1)

a factor:
(i)

x^3 + x^2 + x + 1

(ii)

x^4 + x^3 + x^2 + x + 1

(iii)

x^4 + 3x^3 + 3x^2 + x + 1

(iv)

x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}

Show solution

Concept (Factor Theorem):

(x+1)

is a factor of

p(x)

if and only if

p(-1) = 0

.

(i) $p(x) = x^3 + x^2 + x + 1$ :

p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0

Since

p(-1) = 0

(x+1)

is a factor.

(ii) $p(x) = x^4 + x^3 + x^2 + x + 1$ :

p(-1) = 1 - 1 + 1 - 1 + 1 = 1 \neq 0

Since

p(-1) \neq 0

(x+1)

is not a factor.

(iii) $p(x) = x^4 + 3x^3 + 3x^2 + x + 1$ :

p(-1) = 1 + 3(-1) + 3(1) + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1 \neq 0

Since

p(-1) \neq 0

(x+1)

is not a factor.

(iv) $p(x) = x^3 - x^2 - (2+\sqrt{2})x + \sqrt{2}$ :

p(-1) = (-1)^3 - (-1)^2 - (2+\sqrt{2})(-1) + \sqrt{2}

= -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = 0 + 2\sqrt{2} = 2\sqrt{2} \neq 0

Since

p(-1) \neq 0

(x+1)

is not a factor.

2Use the Factor Theorem to determine whether

g(x)

is a factor of

p(x)

in each of the following cases:
(i)

p(x) = 2x^3 + x^2 - 2x - 1,\ g(x) = x + 1

(ii)

p(x) = x^3 + 3x^2 + 3x + 1,\ g(x) = x + 2

(iii)

p(x) = x^3 - 4x^2 + x + 6,\ g(x) = x - 3

Show solution

Concept:

g(x) = x - a

is a factor of

p(x)

iff

p(a) = 0

.

(i) $g(x) = x + 1$ , zero is $x = -1$ :

p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0

Since

p(-1) = 0

g(x)

is a factor of

p(x)

.

(ii) $g(x) = x + 2$ , zero is $x = -2$ :

p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 \neq 0

Since

p(-2) \neq 0

g(x)

is not a factor of

p(x)

.

(iii) $g(x) = x - 3$ , zero is $x = 3$ :

p(3) = 3^3 - 4(3)^2 + 3 + 6 = 27 - 36 + 3 + 6 = 0

Since

p(3) = 0

g(x)

is a factor of

p(x)

3Find the value of

k

, if

x - 1

is a factor of

p(x)

in each of the following cases:
(i)

p(x) = x^2 + x + k

(ii)

p(x) = 2x^{2} + kx + \sqrt{2}

(iii)

p(x) = kx^2 - \sqrt{2}x + 1

(iv)

p(x) = kx^2 - 3x + k

Show solution

Concept: If

(x-1)

is a factor of

p(x)

, then by Factor Theorem,

p(1) = 0

.

(i) $p(x) = x^2 + x + k$ :

p(1) = 1 + 1 + k = 0 \implies 2 + k = 0 \implies \mathbf{k = -2}

(ii) $p(x) = 2x^2 + kx + \sqrt{2}$ :

p(1) = 2 + k + \sqrt{2} = 0 \implies k = -2 - \sqrt{2} \implies \mathbf{k = -(2 + \sqrt{2})}

(iii) $p(x) = kx^2 - \sqrt{2}x + 1$ :

p(1) = k - \sqrt{2} + 1 = 0 \implies k = \sqrt{2} - 1 \implies \mathbf{k = \sqrt{2} - 1}

(iv) $p(x) = kx^2 - 3x + k$ :

p(1) = k - 3 + k = 0 \implies 2k = 3 \implies \mathbf{k = \dfrac{3}{2}}

4Factorise:
(i)

12x^{2} - 7x + 1

(ii)

2x^{2} + 7x + 3

(iii)

6x^{2} + 5x - 6

(iv)

3x^{2} - x - 4

Show solution

Method: Splitting the middle term.

(i) $12x^2 - 7x + 1$ :
We need two numbers whose product =

12 \times 1 = 12

and sum =

-7

.
Numbers:

-3

and

-4

12x^2 - 7x + 1 = 12x^2 - 4x - 3x + 1

= 4x(3x - 1) - 1(3x - 1)

= \mathbf{(4x - 1)(3x - 1)}

(ii) $2x^2 + 7x + 3$ :
Product =

2 \times 3 = 6

, sum =

7

. Numbers:

6

and

1

2x^2 + 7x + 3 = 2x^2 + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= \mathbf{(2x + 1)(x + 3)}

(iii) $6x^2 + 5x - 6$ :
Product =

6 \times (-6) = -36

, sum =

5

. Numbers:

9

and

-4

6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6

= 3x(2x + 3) - 2(2x + 3)

= \mathbf{(3x - 2)(2x + 3)}

(iv) $3x^2 - x - 4$ :
Product =

3 \times (-4) = -12

, sum =

-1

. Numbers:

-4

and

3

3x^2 - x - 4 = 3x^2 - 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= \mathbf{(x + 1)(3x - 4)}

5Factorise:
(i)

x^{3} - 2x^{2} - x + 2

(ii)

x^{3} - 3x^{2} - 9x - 5

(iii)

x^{3} + 13x^{2} + 32x + 20

(iv)

2y^{3} + y^{2} - 2y - 1

Show solution

Method: Factor Theorem — find a zero by trial, then divide/group.

(i) $p(x) = x^3 - 2x^2 - x + 2$ :

p(1) = 1 - 2 - 1 + 2 = 0

, so

(x-1)

is a factor.

x^3 - 2x^2 - x + 2 = x^2(x-1) - (x-1)(x+2)

Let us group:

= x^2(x - 1) - (x - 1) - x(x-1)

Better: divide

p(x)

(x-1)

x^3 - 2x^2 - x + 2 = (x-1)(x^2 - x - 2)

Now factorise

x^2 - x - 2 = (x-2)(x+1)

\boxed{(x-1)(x-2)(x+1)}

(ii) $p(x) = x^3 - 3x^2 - 9x - 5$ :

p(5) = 125 - 75 - 45 - 5 = 0

, so

(x-5)

is a factor.
Divide:

x^3 - 3x^2 - 9x - 5 = (x-5)(x^2 + 2x + 1) = (x-5)(x+1)^2

\boxed{(x-5)(x+1)^2}

(iii) $p(x) = x^3 + 13x^2 + 32x + 20$ :

p(-1) = -1 + 13 - 32 + 20 = 0

, so

(x+1)

is a factor.
Divide:

x^3 + 13x^2 + 32x + 20 = (x+1)(x^2 + 12x + 20)

Factorise

x^2 + 12x + 20 = (x+2)(x+10)

\boxed{(x+1)(x+2)(x+10)}

(iv) $p(y) = 2y^3 + y^2 - 2y - 1$ :

p(1) = 2 + 1 - 2 - 1 = 0

, so

(y-1)

is a factor.
Group:

2y^3 + y^2 - 2y - 1 = y^2(2y+1) - 1(2y+1) = (2y+1)(y^2-1) = (2y+1)(y-1)(y+1)

\boxed{(2y+1)(y-1)(y+1)}

Exercise 2.4

1Use suitable identities to find the following products:
(i)

(x + 4)(x + 10)

(ii)

(x + 8)(x - 10)

(iii)

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

(iv)

\left(y^{2} - \frac{3}{2}\right)\left(y^{2} - \frac{3}{2}\right)

(v)

(3 - 2x)(3 + 2x)

Show solution

Identity used:

(x+a)(x+b) = x^2 + (a+b)x + ab

and

(x+y)(x-y) = x^2 - y^2

(x-y)^2 = x^2 - 2xy + y^2

.

(i) $(x+4)(x+10)$ : Using Identity IV with

a=4, b=10

= x^2 + (4+10)x + (4)(10) = \mathbf{x^2 + 14x + 40}

(ii) $(x+8)(x-10)$ : Using Identity IV with

a=8, b=-10

= x^2 + (8-10)x + (8)(-10) = \mathbf{x^2 - 2x - 80}

(iii) $(3x+4)(3x-5)$ : Using Identity IV with

x \to 3x

a=4

b=-5

= (3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 - 3x - 20 = \mathbf{9x^2 - 3x - 20}

(iv) $\left(y^2 - \dfrac{3}{2}\right)^2$ : Using Identity II

(x-y)^2 = x^2 - 2xy + y^2

with

x = y^2

y = \dfrac{3}{2}

= y^4 - 2 \cdot y^2 \cdot \frac{3}{2} + \frac{9}{4} = \mathbf{y^4 - 3y^2 + \dfrac{9}{4}}

(v) $(3-2x)(3+2x)$ : Using Identity III

(a-b)(a+b) = a^2 - b^2

with

a=3

b=2x

= 9 - 4x^2 = \mathbf{9 - 4x^2}

2Evaluate the following products without multiplying directly:
(i)

103 \times 107

(ii)

95 \times 96

(iii)

104 \times 96

Show solution

(i) $103 \times 107$ :

103 \times 107 = (100 + 3)(100 + 7)

Using

(x+a)(x+b) = x^2 + (a+b)x + ab

with

x=100, a=3, b=7

= 10000 + (3+7)(100) + 21 = 10000 + 1000 + 21 = \mathbf{11021}

(ii) $95 \times 96$ :

95 \times 96 = (100 - 5)(100 - 4)

Using

(x+a)(x+b)

with

x=100, a=-5, b=-4

= 10000 + (-5-4)(100) + (-5)(-4) = 10000 - 900 + 20 = \mathbf{9120}

(iii) $104 \times 96$ :

104 \times 96 = (100 + 4)(100 - 4)

Using

(a+b)(a-b) = a^2 - b^2

with

a=100, b=4

= 10000 - 16 = \mathbf{9984}

3Factorise the following using appropriate identities:
(i)

9x^{2} + 6xy + y^{2}

(ii)

4y^{2} - 4y + 1

(iii)

x^{2} - \dfrac{y^{2}}{100}

Show solution

(i) $9x^2 + 6xy + y^2$ :

= (3x)^2 + 2(3x)(y) + y^2 = (3x + y)^2

Using Identity I:

(a+b)^2 = a^2 + 2ab + b^2

\boxed{(3x + y)^2}

(ii) $4y^2 - 4y + 1$ :

= (2y)^2 - 2(2y)(1) + 1^2 = (2y - 1)^2

Using Identity II:

(a-b)^2 = a^2 - 2ab + b^2

\boxed{(2y - 1)^2}

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

= x^2 - \left(\frac{y}{10}\right)^2 = \left(x + \frac{y}{10}\right)\left(x - \frac{y}{10}\right)

Using Identity III:

a^2 - b^2 = (a+b)(a-b)

\boxed{\left(x + \dfrac{y}{10}\right)\left(x - \dfrac{y}{10}\right)}

4Expand each of the following, using suitable identities:
(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

(vi)

\left[\frac{1}{4}a - \frac{1}{2}b + 1\right]^{2}

Show solution

Identity used:

(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

(i) $(x + 2y + 4z)^2$ :

= x^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

= \mathbf{x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz}

(ii) $(2x - y + z)^2$ :
Take

a=2x, b=-y, c=z

= 4x^2 + y^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)

= \mathbf{4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz}

(iii) $(-2x + 3y + 2z)^2$ :
Take

a=-2x, b=3y, c=2z

= 4x^2 + 9y^2 + 4z^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)

= \mathbf{4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz}

(iv) $(3a - 7b - c)^2$ :
Take

a=3a, b=-7b, c=-c

= 9a^2 + 49b^2 + c^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)

= \mathbf{9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac}

(v) $(-2x + 5y - 3z)^2$ :
Take

a=-2x, b=5y, c=-3z

= 4x^2 + 25y^2 + 9z^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)

= \mathbf{4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz}

(vi) $\left[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\right]^2$ :
Take

x=\frac{a}{4}, y=-\frac{b}{2}, z=1

= \frac{a^2}{16} + \frac{b^2}{4} + 1 + 2\cdot\frac{a}{4}\cdot\left(-\frac{b}{2}\right) + 2\cdot\left(-\frac{b}{2}\right)\cdot 1 + 2\cdot 1\cdot\frac{a}{4}

= \mathbf{\dfrac{a^2}{16} + \dfrac{b^2}{4} + 1 - \dfrac{ab}{4} - b + \dfrac{a}{2}}

5Factorise:
(i)

4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz

(ii)

2x^{2} + y^{2} + 8z^{2} - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz

Show solution

Identity used:

(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca

(i) $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$ :
Write as

(2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)

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

\boxed{(2x + 3y - 4z)^2}

(ii) $2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$ :
Write as

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

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

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

6Write the following cubes in expanded form:
(i)

(2x + 1)^{3}

(ii)

(2a - 3b)^{3}

(iii)

\left[\frac{3}{2}x + 1\right]^{3}

(iv)

\left[x - \frac{2}{3}y\right]^{3}

Show solution

Identities used:

(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 = a^3 + 3a^2b + 3ab^2 + b^3

(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

(i) $(2x+1)^3$ :

a=2x, b=1

= (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + 1^3

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

(ii) $(2a-3b)^3$ :

a=2a, b=3b

= (2a)^3 - 3(2a)^2(3b) + 3(2a)(3b)^2 - (3b)^3

= 8a^3 - 3(4a^2)(3b) + 3(2a)(9b^2) - 27b^3

= \mathbf{8a^3 - 36a^2b + 54ab^2 - 27b^3}

(iii) $\left(\dfrac{3}{2}x + 1\right)^3$ :

a=\dfrac{3x}{2}, b=1

= \left(\frac{3x}{2}\right)^3 + 3\left(\frac{3x}{2}\right)^2(1) + 3\left(\frac{3x}{2}\right)(1)^2 + 1

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

= \mathbf{\dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1}

(iv) $\left(x - \dfrac{2}{3}y\right)^3$ :

a=x, b=\dfrac{2y}{3}

= x^3 - 3x^2\cdot\frac{2y}{3} + 3x\cdot\frac{4y^2}{9} - \frac{8y^3}{27}

= \mathbf{x^3 - 2x^2y + \dfrac{4xy^2}{3} - \dfrac{8y^3}{27}}

7Evaluate the following using suitable identities:
(i)

(99)^{3}

(ii)

(102)^{3}

(iii)

(998)^{3}

Show solution

Identity used:

(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3

(i) $(99)^3 = (100 - 1)^3$ :

= 100^3 - 3(100)^2(1) + 3(100)(1)^2 - 1^3

= 1000000 - 30000 + 300 - 1

= \mathbf{970299}

(ii) $(102)^3 = (100 + 2)^3$ :

= 100^3 + 3(100)^2(2) + 3(100)(4) + 8

= 1000000 + 60000 + 1200 + 8

= \mathbf{1061208}

(iii) $(998)^3 = (1000 - 2)^3$ :

= 1000^3 - 3(1000)^2(2) + 3(1000)(4) - 8

= 1000000000 - 6000000 + 12000 - 8

= \mathbf{994011992}

8Factorise each of the following:
(i)

8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}

(ii)

8a^{3} - b^{3} - 12a^{2}b + 6ab^{2}

(iii)

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

(iv)

64a^{3} - 27b^{3} - 144a^{2}b + 108ab^{2}

(v)

27p^{3} - \dfrac{1}{216} - \dfrac{9}{2}p^{2} + \dfrac{1}{4}p

Show solution

Identities used:

(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2

(a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2

(i) $8a^3 + b^3 + 12a^2b + 6ab^2$ :

= (2a)^3 + b^3 + 3(2a)^2(b) + 3(2a)(b)^2 = (2a + b)^3

\boxed{(2a + b)^3}

(ii) $8a^3 - b^3 - 12a^2b + 6ab^2$ :

= (2a)^3 - b^3 - 3(2a)^2(b) + 3(2a)(b)^2 = (2a - b)^3

\boxed{(2a - b)^3}

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

= 3^3 - (5a)^3 - 3(3)^2(5a) + 3(3)(5a)^2 = (3 - 5a)^3

\boxed{(3 - 5a)^3}

(iv) $64a^3 - 27b^3 - 144a^2b + 108ab^2$ :

= (4a)^3 - (3b)^3 - 3(4a)^2(3b) + 3(4a)(3b)^2 = (4a - 3b)^3

\boxed{(4a - 3b)^3}

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

= (3p)^3 - \left(\frac{1}{6}\right)^3 - 3(3p)^2\left(\frac{1}{6}\right) + 3(3p)\left(\frac{1}{6}\right)^2

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

\boxed{\left(3p - \dfrac{1}{6}\right)^3}

9Verify:
(i)

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

(ii)

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

Show solution

(i) Verify $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ :

Expand the RHS:

(x+y)(x^2 - xy + y^2)

= x(x^2 - xy + y^2) + y(x^2 - xy + y^2)

= x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3

= x^3 + y^3 \quad \checkmark

Hence verified.

(ii) Verify $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ :

Expand the RHS:

(x-y)(x^2 + xy + y^2)

= x(x^2 + xy + y^2) - y(x^2 + xy + y^2)

= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3

= x^3 - y^3 \quad \checkmark

Hence verified.

10Factorise each of the following:
(i)

27y^{3} + 125z^{3}

(ii)

64m^{3} - 343n^{3}

Show solution

Identity used:

x^3 + y^3 = (x+y)(x^2 - xy + y^2)

and

x^3 - y^3 = (x-y)(x^2+xy+y^2)

(i) $27y^3 + 125z^3 = (3y)^3 + (5z)^3$ :

= (3y + 5z)\left[(3y)^2 - (3y)(5z) + (5z)^2\right]

= \mathbf{(3y + 5z)(9y^2 - 15yz + 25z^2)}

(ii) $64m^3 - 343n^3 = (4m)^3 - (7n)^3$ :

= (4m - 7n)\left[(4m)^2 + (4m)(7n) + (7n)^2\right]

= \mathbf{(4m - 7n)(16m^2 + 28mn + 49n^2)}

11Factorise:

27x^{3} + y^{3} + z^{3} - 9xyz

Show solution

Given:

27x^3 + y^3 + z^3 - 9xyz

Write as:

(3x)^3 + y^3 + z^3 - 3(3x)(y)(z)

Identity used:

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Here

a = 3x,\ b = y,\ c = z

= (3x + y + z)\left[(3x)^2 + y^2 + z^2 - (3x)(y) - (y)(z) - (z)(3x)\right]

= \mathbf{(3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3xz)}

12Verify that

x^{3} + y^{3} + z^{3} - 3xyz = \dfrac{1}{2}(x + y + z)\left[(x - y)^{2} + (y - z)^{2} + (z - x)^{2}\right]

Show solution

We need to show LHS = RHS.

Expand the RHS:

\frac{1}{2}(x+y+z)\left[(x-y)^2 + (y-z)^2 + (z-x)^2\right]

First expand the bracket:

(x-y)^2 + (y-z)^2 + (z-x)^2

= x^2 - 2xy + y^2 + y^2 - 2yz + z^2 + z^2 - 2zx + x^2

= 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx

= 2(x^2 + y^2 + z^2 - xy - yz - zx)

So RHS becomes:

\frac{1}{2}(x+y+z) \cdot 2(x^2+y^2+z^2-xy-yz-zx)

= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

By Identity VIII:

(x+y+z)(x^2+y^2+z^2-xy-yz-zx) = x^3+y^3+z^3-3xyz

Hence RHS = LHS.

\quad\checkmark

13If

x + y + z = 0

, show that

x^{3} + y^{3} + z^{3} = 3xyz

.Show solution

Given:

x + y + z = 0

To prove:

x^3 + y^3 + z^3 = 3xyz

Proof:
We know the identity (Identity VIII):

x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

Since

x + y + z = 0

, substituting:

x^3 + y^3 + z^3 - 3xyz = 0 \times (x^2 + y^2 + z^2 - xy - yz - zx) = 0

Therefore:

x^3 + y^3 + z^3 = 3xyz \qquad \blacksquare

14Without actually calculating the cubes, find the value of each of the following:
(i)

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

(ii)

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

Show solution

Concept: If

x + y + z = 0

, then

x^3 + y^3 + z^3 = 3xyz

.

(i) $(-12)^3 + 7^3 + 5^3$ :
Let

x = -12,\ y = 7,\ z = 5

x + y + z = -12 + 7 + 5 = 0

Since the sum is 0:

x^3 + y^3 + z^3 = 3xyz = 3(-12)(7)(5) = 3 \times (-420) = \mathbf{-1260}

(ii) $(28)^3 + (-15)^3 + (-13)^3$ :
Let

x = 28,\ y = -15,\ z = -13

x + y + z = 28 - 15 - 13 = 0

Since the sum is 0:

x^3 + y^3 + z^3 = 3xyz = 3(28)(-15)(-13) = 3 \times 5460 = \mathbf{16380}

15Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area:

25a^{2} - 35a + 12

(ii) Area:

35y^{2} + 13y - 12

Show solution

Concept: Area of rectangle = length × breadth. We factorise the given expression.

(i) Area $= 25a^2 - 35a + 12$ :
Split middle term: product

= 25 \times 12 = 300

, sum

= -35

. Numbers:

-15

and

-20

25a^2 - 35a + 12 = 25a^2 - 15a - 20a + 12

= 5a(5a - 3) - 4(5a - 3) = (5a - 3)(5a - 4)

Possible expressions: Length $= (5a - 3)$ , Breadth $= (5a - 4)$

(ii) Area $= 35y^2 + 13y - 12$ :
Split middle term: product

= 35 \times (-12) = -420

, sum

= 13

. Numbers:

28

and

-15

35y^2 + 13y - 12 = 35y^2 + 28y - 15y - 12

= 7y(5y + 4) - 3(5y + 4) = (7y - 3)(5y + 4)

Possible expressions: Length $= (7y - 3)$ , Breadth $= (5y + 4)$

16What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume:

3x^{2} - 12x

(ii) Volume:

12ky^{2} + 8ky - 20k

Show solution

Concept: Volume of cuboid = l × b × h. We factorise the given expression.

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

= 3x(x - 4)

Possible dimensions:

\mathbf{3,\ x,\ (x-4)}

(ii) Volume $= 12ky^2 + 8ky - 20k$ :
First take out common factor

k

= k(12y^2 + 8y - 20) = 4k(3y^2 + 2y - 5)

Factorise

3y^2 + 2y - 5

: product

= -15

, sum

= 2

. Numbers:

5

and

-3

3y^2 + 2y - 5 = 3y^2 + 5y - 3y - 5 = y(3y+5) - 1(3y+5) = (y-1)(3y+5)

So:

12ky^2 + 8ky - 20k = 4k(y-1)(3y+5)

Possible dimensions:

\mathbf{4k,\ (y-1),\ (3y+5)}

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