NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

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Written by Team Trustudies
Updated at 2021-05-07


NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.1

Q1 ) Which of the following expression are polynomials in one variable and which are not?
State reason 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)\(x^{10} + y^3 + t^{50}\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) \(4x^2 - 3x + 7\) is a polynomail in one variable. As it is having only non-negative integral powers of x.


ii)\(y^2 + \sqrt{2}\) is also a polynomial in one variable. As it is having only non-negative integral power of y.


iii)\(3\sqrt{t} + t\sqrt{2}\) is not a polynomial. As the integral power of the variable is not a whole number.


iv)\(y + \frac{2}{y} \) is not a polynomial. As it having a negative integral power of y.


v) \(x^{10} + y^3 + t^{50}\) is a polynomial with three variables.

Q2 ) Write the coefficients of \(x^2\) in each of the following :
i)\(2 + x^2 + x\)
ii)\(2 - x^2 + x^3\)
iii)\(\frac{\pi{x^2}}{2} + x\)
iv)\(\sqrt{2}x - 1\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) The coefficient of \(x^2\) in \(2 + x^2 + x\) is \(1\).


ii) The coefficient of \(x^2\) in \(2 - x^2 + x^3\) is \(-1\).


iii) The coefficient of \(x^2\) in \(\frac{\pi{x^2}}{2} + x\) is \(\frac{\pi }{2} \).


iv) The coefficient of \(x^2\) in \(\sqrt{2}x - 1\) is \(0\).

Q3 ) Give one example each of a binomial of degree 35, and of a monomial of degree 100.



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

A binomial of degree 35 = \(4y^{35} + 3\sqrt{2}\)


and a monomial of degree 100 = \(9y^{100}\)

Q4 ) Write the degree of each of the following polynomials :
i)\(5x^3 + 4x^2 + 7x\)
ii)\(4 - y^2\)
iii)\(5t - \sqrt{7}\)
iv)\(3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\(5x^3 + 4x^2 + 7x\) is a polynomial with highest power of variable 3.
Hence, the degree of given polynomial is 3.


ii)\(4 - y^2\) is a polynomial with highest power of variable 2.
Hence, the degree of given polynomial is 2.


iii)\(5t - \sqrt{7}\) is a polynomial with highest power of variable 1.
Hence, the degree of given polynomial is 1.


iv)\(3\) is a polynomial with highest power of variable 0.
Hence, the degree of given polynomial is 0.

Q5 ) Classify 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)\(7x^3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) The degree of polynomial \(x^2 + x\) is 2. Hence, it is a quadratic polynomial.


ii) The degree of polynomial \(x - x^3\) is 3. Hence, it is a cubic polynomial.


iii) The degree of polynomial \(y + y^2 + 4\) is 2. Hence, it is a quadratic polynomial.


iv) The degree of polynomial \(1 + x\) is 1. Hence, it is a linear polynomial.


v) The degree of polynomial \(3t\) is 1. Hence, it is a linear polynomial.


vi) The degree of polynomial \(r^2\) is 2. Hence, it is a quadratic polynomial.


vii) The degree of polynomial \(7x^3\) is 3. Hence, it is a cubic polynomial.

NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2

Q1 ) Find the value of the polynomial \(5x - 4x^2 + 3\) at :
i) x = 0
ii) x = -1
iii) x = 2



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

Let f(x) = \(5x - 4x^2 + 3\)


i)The value of f(x) at x = 0 is :

\( = 5(0) - 4(0)^2 + 3\)
\( = 3 \)

\(\therefore \) f(0) = 3.


ii)The value of f(x) at x = -1 is :

\(= 5(-1) - 4(-1)^2 + 3\)
\( = -5 - 4 + 3 \)
\( = -6 \)

\(\therefore \) f(-1) = -6.


iii)The value of f(x) at x = 2 is :

\( = 5(2) - 4(2)^2 + 3\)
\(= 10 - 16 + 3 \)
\(= -3 \)

\(\therefore \) f(2) = -3.

Q2 ) Find 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)\(p(x) = (x - 1)(x + 1)\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

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

Firstly, let us put x = 0,
\(\therefore p(0) = 0^2 - 0 +1\)
\(\Rightarrow p(0) = 1 \)

Now, let us put x = 1,
\(\therefore p(1) = 1^2 - 1 +1\)
\(\Rightarrow p(1) = 1 \)

Now, let us put x = 2,
\(\therefore p(2) = 2^2 - 2 +1\)
\(\Rightarrow p(2) = 4 - 2 +1\)
\(\Rightarrow p(2) = 3 \)


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

Firstly, let us put x = 0,
\(\therefore p(0) = 2 + 0 + 2{0}^2 - {0}^3\)
\(\Rightarrow p(0) = 2 \)

Now, let us put x = 1,
\(\therefore p(1) = 2 + 1 + 2{1}^2 - {1}^3 \)
\(\Rightarrow p(1) = 3 + 2 - 1\)
\(\Rightarrow p(1) = 4 \)

Now, let us put x = 2,
\(\therefore p(2) = 2 + 2 + 2{2}^2 - {2}^3\)
\(\Rightarrow p(2) = 4 + 8 - 8\)
\(\Rightarrow p(2) = 4 \)


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

Firstly, let us put x = 0,
\(\therefore, p(0) = {0}^3\)
\(\Rightarrow p(0) = 0 \)

Now, let us put x = 1,
\(\therefore p(1) = {1}^3\)
\(\Rightarrow p(1) = 1 \)

Now, let us put x = 2,
\(\therefore p(2) = {2}^3\)
\(\Rightarrow p(2) = 8\)


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

Firstly, let us put x = 0,
\(\therefore p(0) = (0 - 1)(0 + 1)\)
\(\Rightarrow p(0) = -1 \)

Now, let us put x = 1,
\(\therefore p(1) = (1 - 1)(1 + 1)\)
\(\Rightarrow p(1) = 0 \)

Now, let us put x = 2,
\(\therefore, p(2) = (2 - 1)(2 + 1)\)
\(\Rightarrow p(2) = 1 × 3\)
\(\Rightarrow p(2) = 3 \)

Q3 ) Verify 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)\(p(x) = 2x + 1, x = \frac{1}{2} \)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) \(p(x) = 3x + 1\)

To check, if \( x = \frac{-1}{3} \) is zero of given polynomial then \( p(\frac{-1}{3}) \) should be equal to 0.

\(\therefore p(\frac{-1}{3}) \)
\( \Rightarrow p(\frac{-1}{3} ) = 3(\frac{-1}{3} ) + 1\)
\( \Rightarrow p(\frac{-1}{3})= -1 +1 = 0 \)

\(\therefore \) \( x = \frac{-1}{3} \) is zero of given polynomial.


ii) \(p(x) = 5x - \pi\)

To check, if \( \frac{4}{5} \)is zero of given polynomial then \( p(\frac{4}{5} ) \) should be equal to 0.

\(\therefore p(\frac{4}{5} ) = 5(\frac{4}{5} ) - \pi \)
\( \Rightarrow p(\frac{4}{5} ) = 4 - \pi\)

\(\therefore x = \frac{4}{5} \) is not a zero of given polynomial.


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

To check, if x = 1 is zero of given polynomial then p(1) should be equal to 0.

\(\therefore p(1) = 1^2 - 1\)
\(\Rightarrow p(1) = 1 - 1 = 0 \)

\(\therefore x = 1 \) is a zero of given polynomial.

To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.

\(\therefore p(-1) = {-1}^2 - 1\)
\( \Rightarrow p(-1) = 1 - 1 = 0 \)

\(\therefore, x = 1 \) is too, a zero of given polynomial.


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

To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.

\(\therefore p(-1) = (-1 + 1)(-1 - 2) \)
\(\Rightarrow p(-1) = 0 × -3 = 0 \)

\(\therefore x = -1 \) is zero of given polynomial.

To check, if x = 2 is zero of given polynomial then p(2) should be equal to 0.

\(\therefore p(2) = (2 + 1)(2 - 2) \)
\( \Rightarrow p(2) = 3 × 0 = 0 \)

Therefore, x = 2 is too, a zero of given polynomial.


v) \(p(x) = x^2\)

To check, if x = 0 is zero of given polynomial then p(0) should be equal to 0.

\(\therefore p(0) = (0^2) = 0\)

Therefore, x = 0 is zero of given polynomial.


vi) \(p(x) = lx + m\)

To check, if \( x = \frac{-m}{l} \) is zero of given polynomial then \( p(\frac{-m}{l} ) \) should be equal to 0.

\(\therefore p(\frac{-m}{l} ) = l(\frac{-m}{l} ) + m \)
\(\Rightarrow p(\frac{-m}{l} )= -m + m = 0 \)
\(\therefore, x = \frac{-m}{l} \) is zero of given polynomial.


vii)\(p(x) = 3x^2 - 1\)

To check, if\( x = \frac{-1}{\sqrt{3}} \) is zero of given polynomial then p(\(\frac{-1}{\sqrt{3}} \)) should be equal to 0.

\( \therefore p(\frac{-1}{\sqrt{3}} ) = 3(\frac{-1}{\sqrt{3}} )^2 - 1\)
\(\Rightarrow p(\frac{-1}{\sqrt{3}} ) = 1 - 1 = 0 \)

\(\therefore x = \frac{-1}{\sqrt{3}} \) is zero of given polynomial.

To check, if \(x = \frac{2}{\sqrt{3}} \) is zero of given polynomial then \( p(\frac{2}{\sqrt{3}} \)) should be equal to 0.

\(\therefore p(\frac{2}{\sqrt{3}} ) = 3(\frac{2}{\sqrt{3}} )^2 - 1\)
\( \Rightarrow p(\frac{2}{\sqrt{3}} ) = 3 \)

\(\therefore, x = \frac{2}{\sqrt{3}} \) is not a zero of given polynomial.


viii)\(p(x) = 2x + 1\)

To check, if \( x = \frac{1}{2} \) is zero of given polynomial then\( p(\frac{1}{2} ) \) should be equal to 0.

\(\therefore p(\frac{1}{2} ) = 2(\frac{1}{2} ) + 1 \)
\( \Rightarrow p(\frac{1}{2} ) = 1 + 1 = 2 \)

\(\therefore x = \frac{1}{2} \) is not a zero of given polynomial.

Q4 ) Find 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)\(p(x) = ax, a \ne 0\)
vii)\(p(x) = cx + d ,c \ne 0\), c,d, are real numbers.



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) We have, \(p(x) = x + 5\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow x + 5 = 0 \)
\(\Rightarrow x = -5 \)

Hence, -5 is the zero of given ploynomial.


ii)We have, \(p(x) = x - 5\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow x - 5 = 0\)
\(\Rightarrow x = 5 \)

Hence, 5 is the zero of given ploynomial.


iii)We have, \(p(x) = 2x + 5\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore, p(x) = 0 \)
\(\Rightarrow 2x + 5 = 0\)
\(\Rightarrow x = \frac{-5}{2} \)

Hence,\(\frac{-5}{2} \) is the zero of given ploynomial.


iv) We have, \(p(x) = 3x - 2\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow 3x - 2 = 0\)
\(\Rightarrow x = \frac{2}{3} \)

Hence, \(\frac{2}{3} \) is the zero of given ploynomial.


v) We have, \(p(x) = 3x\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow 3x = 0\)
\(\Rightarrow x = 0 \)

Hence, 0 is the zero of given ploynomial.


vi) We have, \(p(x) = ax\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow ax = 0\)
\(\Rightarrow x = 0, a \ne 0\)

Hence, 0 is the zero of given ploynomial.


vii) We have, \(p(x) = cx + d\)

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

\(\therefore p(x) = 0 \)
\(\Rightarrow cx + d = 0\)
\(\Rightarrow x = \frac{-d}{c} , c \ne 0\)

Hence, \(\frac{-d}{c} \) is the zero of given ploynomial.

NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.3

Q1 ) Find the remainder when \(x^3 + 3x^2 + 3x + 1\) is divided by
i)\(x + 1\)
ii)\(x - \frac{1}{2} \)
iii)\(x\)
iv)\(x + \pi\)
v)\(5 + 2x\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

Let \(p(x) = x^3 + 3x^2 + 3x + 1\)


i)The zero of \(x + 1\) is -1
[\(\because x+1= 0 \Rightarrow x = -1 \) ]

\(\therefore p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1\)
\(\Rightarrow p(-1) = -1 + 3 - 3 +1\)
\(\Rightarrow p(-1) = 0\)

Hence, By remainder theorem, required remainder = 0.


ii)The zero of \(x - \frac{1}{2} \) is x = \(\frac{1}{2} \)
[\(\because (x - \frac{1}{2} = 0 )\)]

\(\therefore p(\frac{1}{2} ) = (\frac{1}{2} )^3 + 3(\frac{1}{2} )^2 + 3(\frac{1}{2} )+ 1\)
\( \Rightarrow p(\frac{1}{2}) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} +1\)
\(\Rightarrow p(\frac{1}{2} ) = \frac{27}{8} \)

Hence, By remainder theorem, required remainder = \( \frac{27}{8} \) .


iii)The zero of \(x\) is x = 0 [\(\because (x = 0) \) ]

\(\therefore p(0) = (0)^3 + 3(0)^2 + 3(0) + 1\)
\(\Rightarrow p(-1) = 1\)

Hence, By remainder theorem, required remainder = 1.


iv) The zero of \(x + \pi\) is \(x = -\pi\) [\(\because (x + \pi = 0)\)]

\(\therefore p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1\)

\(\Rightarrow p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1\)

Hence, By remainder theorem, required remainder\( = -\pi^3 + 3\pi^2 - 3\pi + 1\).


v) The zero of \(5 + 2x\) is x = \(\frac{-5}{2} \) [\(\because (5 + 2x = 0) \)]

\(\therefore p(\frac{-5}{2} ) = (\frac{-5}{2})^3 + 3(\frac{-5}{2})^2 + 3(\frac{-5}{2} )+ 1\)
\(\Rightarrow p(\frac{-5}{2} ) = \frac{-125}{8} + \frac{75}{4} - \frac{15}{2} + 1\)
\(\Rightarrow p(\frac{-5}{2} ) = \frac{-27}{8} \)

Hence, By remainder theorem, required remainder = \(\frac{-27}{8} \).

Q2 ) Find the remainder when \(x^3 - ax^2 + 6x - a\) is divided by x - a.



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

Let \(p(x) = x^3 - ax^2 + 6x - a\)
The zero of \(x - a\) is x = a
[\(\because (x - a = 0) \) ]

\(\therefore p(a) = (a)^3 - a(a)^2 + 6(a) - a\)
\(\Rightarrow p(a) = a^3 - a^3 + 6a - a\)
\(\Rightarrow p(a) = 5a\)

Hence, By remainder theorem, required remainder = 5a.

Q3 ) Check whether 7 + 3x is a factor of \(3x^3 + 7x\).



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

Let \(p(x) = 3x^3 + 7x\)

To check, whether 7 + 3x is a factor of \(3x^3 + 7x\).
By factor theorem, \(7 + 3x = 0\)
\(\therefore x = \frac{-7}{3} \)

\(\therefore p(\frac{-7}{3} ) = 3(\frac{-7}{3} )^3 + 7(\frac{-7}{3} )\)
\(\Rightarrow p(\frac{-7}{3} ) = - 3×(\frac{343}{27} ) - \frac{49}{3} \)
\( \Rightarrow p(\frac{-7}{3} ) = \frac{-1470}{27} \)
\(\Rightarrow p(\frac{-7}{3} ) = \frac{-490}{9} \)

Hence, 7 + 3x is not the factor of \(3x^3 + 7x\) since the remainder is not zero.

NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.4

Q1 ) Determine which of the following polynomials has 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)\(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

The zero of x + 1 is –1.


i)Let \(p(x) = x^3 + x^2 + x + 1\)
\(\therefore \) To check, whether x + 1 is a factor of \(x^3 + x^2 + x + 1\).

By factor theorem,
\(\therefore \) \(p(–1) = {–1}^3 + {–1}^2 + {-1} + 1\)
\(\Rightarrow \) \(p(–1) = –1 + 1 -1 + 1\)
\(\Rightarrow \) \(p(-1) = 0\)

Hence, x + 1 is the factor of \(x^3 + x^2 + x + 1\)


ii)Let \(p(x) = x^4 + x^3 + x^2 + x + 1\)
\(\therefore \) To check, whether x + 1 is a factor of \(x^4 + x^3 + x^2 + x + 1\).

By factor theorem,
\(\therefore p(–1) = (-1)^4 + (–1)^3 + (–1)^2 + (-1) + 1\)
\(\Rightarrow \) \(p(–1) = 1 – 1 + 1 - 1 + 1\)
\(\Rightarrow \) \(p(-1) = 1\)

Hence, x + 1 is not the factor of \(x^4 + x^3 + x^2 + x + 1\)


iii) Let \(p(x) = x^4 + 3x^3 + 3x^2 + x + 1\)
\(\therefore \) To check, whether x + 1 is a factor of \(x^4 + 3x^3 + 3x^2 + x + 1\).

By factor theorem,
\(\therefore p(–1) = {-1}^4 + 3{–1}^3 + 3{–1}^2 + {-1} + 1\)
\(\Rightarrow p(–1) = 1 – 3 + 3 - 1 + 1\)
\(\Rightarrow p(-1) = 1\)

Hence, x + 1 is not the factor of \(x^4 + 3x^3 + 3x^2 + x + 1\)


iv) Let \(p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)
\(\therefore \) To check, whether x + 1 is a factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\).

By factor theorem,
\(\therefore p(–1) = {–1}^3 - {–1}^2 - (2 + \sqrt{2}){-1} + \sqrt{2}\)
\(\Rightarrow p(–1) = – 1 - 1 + 2 + \sqrt{2} + \sqrt{2}\)
\(\Rightarrow p(-1) = 2\sqrt{2}\)

Hence, x + 1 is not the factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)

Q2 ) Use 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)\(p(x) = x^3 - 4x^2 + x + 6\), \(g(x) = x - 3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)The zero of g(x) = x + 1 is –1.

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

so, To check, whether x + 1 is a factor of \(2x^3 + x^2 - 2x - 1\).

By factor theorem,
\(\therefore p(–1) = 2{–1}^3 + {–1}^2 - 2{-1} - 1\)
\(\Rightarrow p(–1) = –2 + 1 + 2 - 1\)
\(\Rightarrow p(-1) = 0\)

Hence, g(x) = x + 1 is the factor of \(2x^3 + x^2 - 2x - 1\)


ii)The zero of \(g(x) = x + 2\) is –2.

Let \(p(x) = x^3 + 3x^2 + 3x + 1\)

So, To check, whether x + 2 is a factor of \(x^3 + 3x^2 + 3x + 1\).

By factor theorem,
\(\therefore p(–2) = (–2)^3 + 3(–2)^2 + 3(–2) + 1\)
\(\Rightarrow p(–2) = –8 + 12 - 6 + 1\)
\(\Rightarrow p(-2) = -1\)

Hence, \(g(x) = x + 2\) is not the factor of \(x^3 + 3x^2 + 3x + 1\)


iii)The zero of \(g(x) = x - 3\) is 3.

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

So, To check, whether x - 3 is a factor of \(p(x) = x^3 - 4x^2 + x + 6\).

By factor theorem,

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

Hence, \(g(x) = x - 3\) is the factor of \(p(x) = x^3 - 4x^2 + x + 6\)

Q3 ) Find 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)\(p(x) = kx^2 - 3x + k\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

The zero of x - 1 is 1.


i)So, as x - 1 is a factor of \(p(x) = x^2 + x + k\),
\(\therefore p(1) = 0\)
\(\Rightarrow 0 = (1)^2 + 1 + k\)
\(\Rightarrow 0 = 1 + 1 + k\)
\(\Rightarrow 0 = 2 + k\)
\(\Rightarrow k = -2\)


ii)So, as x - 1 is a factor of \(p(x) = 2x^2 + kx + \sqrt{2}\),
\(\therefore p(1) = 0\)
\(\Rightarrow 0 = 2(1)^2 + k(1) + \sqrt{2}\)
\(\Rightarrow 0 = 2 + k + \sqrt{2}\)
\(\Rightarrow k = -2 - \sqrt{2}\)


iii)So, as x - 1 is a factor of \(p(x) = kx^2 - \sqrt{2}x + 1\),
\(\therefore p(1) = 0\)
\(\Rightarrow 0 = k(1)^2 - ( \sqrt{2})1 + 1\)
\(\Rightarrow 0 = k - \sqrt{2} + 1\)
\( \Rightarrow k = \sqrt{2} - 1\)


iv)So, as x - 1 is a factor of \(p(x) = kx^2 - 3x + k\),
\(\therefore p(1) = 0\)
\(\Rightarrow 0 = k(1)^2 - 3(1) + k\)
\(\Rightarrow 0 = k - 3 + k\)
\( \Rightarrow 3 = 2k\)
\(\Rightarrow k = \frac{3}{2} \)

Q4 ) Factorize :
i)\(12x^2 - 7x + 1\)
ii)\(2x^2 + 7x + 3\)
iii)\(6x^2 + 5x - 6\)
iv)\(3x^2 - x - 4\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\(12x^2 - 7x + 1 \)
\( = 12x^2 - 4x -3x + 1\)
(by splitting middle term)

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


ii) \(2x^2 + 7x + 3 \)
\(= 2x^2 + 6x + x + 3\)...
(by splitting middle term)

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


iii) \(6x^2 + 5x - 6 \)
\( = 6x^2 + 9x - 4x - 6\)
(by splitting middle term)

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


iv) \(3x^2 - x - 4 \)
\( = 3x^2 - 4x + 3x - 4\)
(by splitting middle term)

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

Q5 ) Factorize :
i)\(x^3 - 2x^2 - x + 2\)
ii)\(x^3 - 3x^2 -9x - 5\)
iii)\(x^3 + 13x^2 + 32x + 20\)
iv)\(2y^3 + y^2 - 2y - 1\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)Let \(p(x) = x^3 - 2x^2 - x + 2\)

Here, constant term is 2. So, finding the factors of +2, we get, ±1 and ±2.

Now, \(p(1) = (1)^3 - 2(1)^2 - (1) + 2\)
\(= 1 - 2 - 1 + 2\)
= 0

Therefore, by trial method, we find that,
p(1) = 0.

Hence, x - 1 is a factor of p(x).

\(\therefore x^3 - 2x^2 - x + 2 \)
\( = x^3 - x^2 - x^2 + x - 2x + 2\)
\(= x^2(x - 1) -x(x - 1) - 2(x - 1)\)
\(= (x - 1)(x^2 - x - 2)\)
now, splitting the middle term,
\(= (x - 1)(x^2 - 2x + x - 2)\)
\(= (x - 1)[x(x - 2) + 1(x - 2)]\)
\(= (x - 1)(x - 2)(x + 1)\)


ii)Let \(p(x) = x^3 - 3x^2 -9x - 5\)
Here, constant term is -5. So, finding the factors of -5, we get, 1 and -5 or vice versa.

Now, \(p(5) = (5)^3 - 3(5)^2 - 9(5) - 5\)
\(= 125 - 75 - 45 - 5\)
= 0

Therefore, by trial method, we find that, p(5) = 0.

Hence, x - 5 is a factor of p(x).

\(\therefore x^3 - 3x^2 -9x - 5 \)
\( = x^3 - 5x^2 + 2x^2 - 10x + x - 5\)
\(= x^2(x - 5) + 2x(x - 5) + 1(x - 5)\)
\(= (x - 5)(x^2 + 2x + 1)\)
now, splitting the middle term,
\(= (x - 5)(x^2 + x + x + 1)\)
\(= (x - 5)[x(x + 1) + 1(x + 1)]\)
\(= (x - 5)(x + 1)^2\)


iii)Let \(p(x) = x^3 + 13x^2 + 32x + 20\)

Here, constant term is +20. So, finding the factors of 20, we get, ±1, ±2, ±4 and ±5 or vice versa.

Now, \(p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20\)
\(= -1 + 13 - 32 + 20\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

\(\therefore x^3 + 13x^2 + 32x + 20 \)
\( = x^3 + 2x^2 + 11x^2 + 22x + 10x + 20\)
\(= x^2(x + 2) + 11x(x + 2) + 10(x + 2)\)
\(= (x + 2)(x^2 + 11x + 10)\)
now, splitting the middle term,
\(= (x + 2)(x^2 + 10x + x + 10)\)
\(= (x + 2)[x(x + 10) + 1(x + 10)]\)
\(= (x + 2)(x + 1)(x + 10)\)


iv)Let \(p(x) = 2y^3 + y^2 - 2y - 1\)

Here, constant term is -1. So, finding the factors of -1, we get, -1.

Now, \(p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1\)
\(= -2 + 1 + 2 - 1\)
= 0

Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

\(\therefore 2y^3 + y^2 - 2y - 1 \)
\( = 2y^3 + 2y^2 - y^2 - y - y -1\)
\(= 2y^2(y + 1) - y(x + 1) - 1(y + 1)\)
\(= (y + 1)(2y^2 - y - 1)\)
now, splitting the middle term,
\(= (y + 1)(2y^2 - 2y + y - 1)\)
\(= (y + 1)[2y(y - 1) + 1(y - 1)]\)
\(= (y - 1)(y + 1)(2y + 1)\)

NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.5

Q1 ) Use suitable identities to find the following products :
i)\((x + 4)(x + 10)\)
ii)\((x + 8)(x - 10)\)
iii)\((3x - 4)(3x - 5)\)
iv)\((y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )\)
v)\((3 - 2x)(3 + 2x)\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\((x + 4)(x + 10)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab] \)

We have,
\(=(x + 4)(x + 10) \)
\( = x^2 + (4 + 10)x + (4)(10)\)
\(= x^2 + 14x + 40\)


ii) \((x + 8)(x - 10)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab]\)
We have,
\(= (x + 8)(x - 10) \)
\( = x^2 + (8 - 10)x + (8)(-10)\)
\(= x^2 - 2x - 80\)

iii) \((3x + 4)(3x - 5)\)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab]\)
We have,
\((3x + 4)(3x - 5) \)
\( = (3x)^2 + (4 - 5)x + (4)(5)\)
\(= 9x^2 - x + 20\)


iv) \((y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )\)
\([\because (a^2 - b^2) = (a + b)(a - b)] \)
We have,
\((y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} ) \)
\( = (y^2)^2 - (\frac{3}{2} )^2\)
\(= y^4 - \frac{9}{4} \)


v) \((3 - 2x)(3 + 2x)\)
\([\because (a^2 - b^2) = (a + b)(a - b)]\)
We have,
\((3 - 2x)(3 + 2x) \)
\( = (3)^2 - (2x)^2)\)
\(= 9 - 4x^2\)

Q2 ) Evaluate the following products without multiplying directly
i)103 × 107
ii)95 × 96
iii)104 × 96



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)103 × 107 = (100 + 3)(100 + 7)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab]\)

We have,
\((100 + 3)(100 + 7) \)
\( = {100}^2 + (3 + 7)100 + (3)(7)\)
\(= 10000 + 1000 + 21\)
\(= 11021\)


ii)95 × 96 = (100 - 5)(100 - 4)
\([\because (x + a)(x + b) = x^2 + (a + b)x + ab] \)

We have,
\((100 - 5)(100 - 4) \)
\( = {100}^2 - (5 + 4)100 + (4)(5)\)
\(= 10000 - 900 + 20\)
\(= 9120\)


iii)104 × 96 = (100 + 4)(100 - 4)
\([\because (a^2 - b^2) = (a + b)(a - b)]\)

We have,
\((100 + 4)(100 - 4) \)
\( = {100}^2 - {4}^2\)
\(= 10000 - 16\)
\(= 9984\)

Q3 ) Factorize the following using appropriate identities :
i)\(9x^2 + 6xy + y^2\)
ii)\(4y^2 - 4y + 1\)
iii)\(x^2 - \frac{y^2}{100} \)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i) \(9x^2 + 6xy + y^2 \)
\( = {3x}^2 + 2(3x)(y) + {y}^2\)
\([\because (a + b)^2 = a^2 + 2ab + b^2\)
\( = (3x + y)^2 ] \)

Therefore, the factors are (3x + y)(3x + y).


ii) \(4y^2 - 4y + 1 \)
\( = {2y}^2 - 2(2y)(1) + {1}^2\)
\([\because (a + b)^2 = a^2 + 2ab + b^2\)
\( = (2y - 1)^2 ]\)

Therefore, the factors are (2y - 1)(2y - 1).


iii)\(x^2 - \frac{y^2}{100} \)
\( = (x +\frac{y}{10} )(x - \frac{y}{10} )\)
\([\because (a^2 - b^2) = (a + b)(a - b))] \)
\( =(x + \frac{y}{10} )(x - \frac{y}{10} ) \)

Therefore, the factors are \( (x + \frac{y}{10} )(x - \frac{y}{10} ). \)

Q4 ) Factorize the following using appropriate 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)\([\frac{1}{4} a - \frac{1}{2} b + 1]^2\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\((x + 2y + 4z)^2 \)
\( = (x)^2 + (2y)^2 +(4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz\)


ii)\((2x - y + z)^2 \)
\( = (2x)^2 + (-y)^2 + (z)^2 - 2(2x)(y) - 2(y)(z) + 2(2x)(z)\)
\([ \because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz\)


iii)\((-2x + 3y + 2z)^2 \)
\( = (-2x)^2 + (3y)^2 +(2z)^2 - 2(2x)(3y) + 2(3y)(2z) - 2(2x)(2z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz\)


iv)\((3a - 7b - c)^2 \)
\( = (3a)^2 + (-7b)^2 + (-c)^2 - 2(3a)(7b) + 2(-7b)(-c) - 2(3a)(c)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac\)


v)\((-2x + 5y - 3z)^2 \)
\( = (-2x)^2 + (5y)^2 +(-3z)^2 - 2(2x)(5y) - 2(5y)(3z) + 2(2x)(3z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz\)


vi)\([\frac{1}{4} a - \frac{1}{2} b + 1]^2 \)
\( = (\frac{1}{4} a )^2 + (\frac{-1}{2} b )^2 + (1)^2 - 2(\frac{1}{4} a )(\frac{1}{2} b ) - 2(\frac{-1}{2} b )(1) + 2(\frac{1}{4} a )(1)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= \frac{a^2}{16} + \frac{b^2}{4} + 1 - \frac{ab}{4} - b + \frac{a}{2} \)

Q5 ) Factorize :
i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz\)
ii)\(2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz \)
\(= (2x)^2 + (3y)^2 - (4z)^2 + 2(2x)(3y) - 2(3y)(4z) - 2(2x)(4z)\)
\(=(2x + 3y - 4z)^2\)



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

Q6 ) Write the following cubes in expanded form.
i)\((2x + 1)^3\)
ii)\((2a - 3b)^3\)
iii)\([\frac{3x}{2} + 1]^3\)
iv)\((x - \frac{2y}{3} )^3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\((2x + 1)^3 \)
\( = (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)\)
\([\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]\)
\(= 8x^3 + 1 + 6x(2x + 1))\)
\(= 8x^3 + 1 + 12x^2 + 6x\)
\(= 8x^3 + 12x^2 + 6x + 1\)


ii)\((2a - 3b)^3 \)
\( = (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]\)
\(= 8a^3 - 27b^3 - 18ab(2a - 3b))\)
\(= 8a^3 - 27b^3 - 36a^2b + 54ab^2)\)
\(= 8a^3 - 36a^2b + 54ab^2 - 27b^3\)


iii)\([\frac{3x}{2} + 1]^3 \)
\( = (\frac{3x}{2} )^3 + (1)^3 + 3(\frac{3x}{2} )(1)(\frac{3x}{2} + 1)\)
\([\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]\)
\(= \frac{27x^3}{8} + 1 + \frac{9x}{2} (\frac{3x}{2} + 1))\)
\(= \frac{27x^3}{8} + 1 + \frac{27x^2}{4} + \frac{9x}{2} \)
\(= \frac{27x^3}{8} + \frac{27x^2}{4} + \frac{9x}{2} + 1\)


iv)\((x - \frac{2y}{3} )^3 = (x)^3 - (\frac{2y}{3} )^3 - 3(x)(\frac{2y}{3} )(x - \frac{2y}{3} )\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)] \)
\(= x^3 - \frac{8y^3}{27} - 2xy(x - \frac{2y}{3} )\)
\(= x^3 - \frac{8y^3}{27} - 2x^2y + \frac{4xy^2}{3} \)
\(= x^3 - 2x^2y + \frac{4xy^2}{3} - \frac{8y^3}{27} \)

Q7 ) Evaluate the following using suitable identities.
i)\({99}^3\)
ii)\({102}^3\)
iii)\({998}^3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\({99}^3 \)
\( = (100 - 1)^3 \)
\( = (100)^3 - (1)^3 - 3(100)(1)(100 - 1)\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]\)
\(= 1000000 - 1 - 300(99)\)
\(= 1000000 - 1 - 2700)\)
\(= 970299\)


ii)\({102}^3 \)
\( = (100 + 2)^3 \)
\( = (100)^3 + (2)^3 + 3(100)(2)(100 + 2)\)
\([\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]\)
\(= 1000000 + 8 + 600(102)\)
\(= 1000000 + 8 + 61200)\)
\(= 1061208\)


iii)\({998}^3 \)
\( = (1000 - 2)^3 \)
\( = (1000)^3 - (2)^3 - 3(1000)(2)(1000 - 2)\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)] \)
\(= 1000000000 - 8 - 6000(998)\)
\(= 1000000000 - 1 - 5988000)\)
\(= 994011992\)

Q8 ) Factorize each of the following :
i)\(8a^3 + b^3 + 12a^2b + 6ab^2\)
ii)\(8a^3 - b^3 - 12a^2b + 6ab^2\)
iii)\(27 - 125a^3 - 135a + 225a^2\)
iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
v)\(27p^3 - \frac{1}{216} - \frac{9p^2}{2} + \frac{p}{4} \)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\(8a^3 + b^3 + 12a^2b + 6ab^2 \)
\(= (2a)^3 + b^3 +3(2a)(b)(2a + b))\)
\([\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]\)
\(= (2a)^3 + b^3\)


ii)\(8a^3 - b^3 - 12a^2b + 6ab^2 \)
\(= (2a)^3 - b^3 -3(2a)(b)(2a - b)\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)] \)
\(= (2a)^3 - b^3\)


iii)\(27 - 125a^3 - 135a + 225a^2 \)
\(= (3)^3 - (5a)^3 -3(3)(5a)(3 - 5a)\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]\)
\(= (3)^3 - 5a^3\)


iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
\( = (4a)^3 - (3b)^3 -3(4a)(4b)(4a - 3b))\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]\)
\(= (4a)^3 - (3b)^3\)


v)\((27p^3 - \frac{1}{216} - \frac{9p^2}{2} + \frac{p}{4} \)
\( = (3p)^3 - (\frac{1}{6} )^3 -3(3p)(\frac{1}{6})(3p - 1/6))\)
\([\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)] \)
\(= (3p)^3 - (\frac{1}{6})^3\)

Q9 ) Verify :
i)\(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
ii)\(x^3 - y^3 = (x - y)(x^2 + xy + y^2) \)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)We know that,
\((x + y)^3 = x^3 + y^3 + 3xy(x + y)\)
\(\Rightarrow x^3 + y^3 = (x + y)^3 - 3xy(x + y)\)
\(= (x + y)[(x + y)^2 - 3xy]\)
\(= (x + y)[(x^2 + y^2 +2xy - 3xy]\)
\(= (x + y)[(x^2 + y^2 - xy]\)
\(= R.H.S\)
Hence, Proved.


ii)We know that,
\((x - y)^3 = x^3 - y^3 - 3xy(x - y)\)
\(\Rightarrow x^3 - y^3 = (x - y)^3 + 3xy(x - y)\)
\(= (x - y)[(x - y)^2 + 3xy]\)
\(= (x - y)[(x^2 + y^2 - 2xy + 3xy]\)
\(= (x - y)[(x^2 + y^2 + xy]\)
\(= R.H.S\)
Hence, Proved.

Q10 ) Factorize each of the following :
i)\(27y^3 + 125z^3\)
ii)\(64m^3 + 343n^3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)\(27y^3 + 125z^3 \)
\( = (3y)^3 + (5z)^3\)
\(= (3y + 5z)[((3y)^2 + (5y)^2 - (3y)(5z)]\)
\(= (3y + 5z)[(9y^2 - 15yz + 25z^2)]\)


ii)\(64m^3 + 343n^3 \)
\( = (4m)^3 + (7n)^3\)
\(= (4m + 7n)[((4m)^2 + (7n)^2 - (4m)(7n)]\)
\(= (4m + 7n)[(16m^2 - 48mn + 49n^2)]\)

Q11 ) Factorize \(27x^3 + y^3 + z^3 - 9xyz\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

\(27x^3 + y^3 + z^3 - 9xyz \)
\( = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)\)
\(= (3x + y +z)((3x)^2 + y^2 + z^2 - 3xy - yz - z(3x)\)
\([ \because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) ]\)
We get,
\(= (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)\)

Q12 ) Verify that \(x^3 + y^3 + z^3 - 3xyz = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

We know that, \(x^3 + y^3 + z^3 - 3xyz \)
\(= (x + y + z)[x^2 + y^2 + z^2 - xy - yz - zx]\)
\( = \frac{1}{2} (x + y + z)[2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx]\)
\( = \frac{1}{2} (x + y + z)[x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx]\)
\( = \frac{1}{2} (x + y + z)[x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx]\)
\( = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)

Q13 ) If show that x + y + z = 0, show that \(x^3 + y^3 + z^3 = 3xyz\).



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
As it is given that (x + y + z = 0)
\(x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)\)
\(\Rightarrow x^3 + y^3 + z^3 = 3xyz\)
Hence, proved.

Q14 ) Without actually calculating the cubes, find the value of each of the following :
i)\((-12)^3 + (5)^3 + (7))^3\)
ii)\((28)^3 + (-15)^3 + (-13)^3\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)

Also, if x + y + z = 0 then, \(x^3 + y^3 + z^3 = 3xyz\)


i) Here, -12 + 7 + 5 = 0

\(\therefore (-12)^3 + (5)^3 + (7))^3 \)
\( = 3(-12)(7)(5)\)
\(= -1260\)


ii)Here, 28 - 15 - 13 = 0
\(\therefore (28)^3 + (-15)^3 + (-13))^3 \)
\( = 3(-13)(28)(-15)\)
\(= 16380\)

Q15 ) Give 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 \(35y^2 + 13y - 12\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)We have area of rectangle \(= 25a^2 - 35a + 12\)
\(= 25a^2 - 20a - 15a + 12\)
\(= 5a(5a - 4) - 3(5a - 4)\)
\(= (5a - 4)(5a - 3)\)
One possible answers : Length = 5a - 3, Breadth = 5a - 4.

ii)We have area of rectangle \(= 35y^2 + 13y - 12\)
\(= 35y^2 - 15y + 28y - 12\)
\(= 5y(7y - 3) + 4(7y - 3)\)
\(= (7y - 3)(5y + 4)\)
One possible answers : Length = 7y - 3, Breadth = 5y + 4.

Q16 ) What are the possible expressions for the dimensions of the cuboids whose volumes are given below ?
i)Volume \(3x^2 - 12x\)
ii)Volume \(12ky^2 + 8ky - 20k\)



NCERT Solutions for Class 9 Maths Chapter 2 Polynomials


Answer :

i)We have Volume of cuboid \(3x^2 - 12x = 3x(x - 4)\)
One possible expressions for the dimensions of the cuboid is 3, x and x-4.

ii)We have Volume of cuboid \(12ky^2 + 8ky - 20k\)
\(= 12ky^2 + 20ky - 12ky - 20k\)
\(= 4ky(3y - 5) - 4k(3y - 5)\)
\(= (3y - 5)(4ky - 4k)\)
\(= (3y - 5)4k(y - 1)\)
One possible expressions for the dimensions of the cuboid is 4k, 3y - 5 and y - 1.



FAQs Related to NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
There are total 33 questions present in ncert solutions for class 9 maths chapter 2 polynomials
There are total 11 long question/answers in ncert solutions for class 9 maths chapter 2 polynomials
There are total 5 exercise present in ncert solutions for class 9 maths chapter 2 polynomials
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