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Searching for NCERT Class 9 Maths Chapter 2 Exercise 2.2 solutions? You’re in the right place! This section provides clear and step-by-step solutions for all the questions in Exercise 2.2 of Chapter 2 – Polynomials. This exercise introduces students to the fundamental concepts of polynomials, including their definitions, types (like monomials, binomials and trinomials) and the degree of a polynomial. It helps learners identify and classify polynomials based on the number of terms and powers of variables. These solutions are designed to build a strong foundation in algebraic expressions, making it easier to understand more complex topics later. Ideal for exam preparation and conceptual clarity, this guide simplifies polynomial basics in a structured and student-friendly way—boosting confidence and encouraging a deeper understanding of algebra.
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Exercise 2.2
1.
(i) Find the value of the polynomial \( 5 x-4 x^{2}+3 \) at \(
\mathrm{x}=0\)
Answer
\(P(x)=5 x-4 x^{2}+3\)Now \(\mathrm{p}(a) \) will mean replacing the variable with the value that is put in the bracket.
\(\mathrm{p}(0)=5(0)-4(0)^{2}+3\)
\(\mathrm{p}(0)=0-0+3 \mathrm{p}(0)=3\)
(ii) Find the value of the polynomial \( 5 x-4 x^{2}+3 \) at \(
x=-1 \)
Answer
\( P(x)=5 x-4 x^{2}+3 \)Now \( \mathrm{p}(a) \) will mean replacing the variable with the value that is put in bracket.
\(\mathrm{p}(-1)=5(-1)-4(-1)^{2}+3\)
\(\mathrm{p}(-1)=-5-4+3\)
\(\mathrm{p}(-1)=-9+3\mathrm{p}(-1)=-6\)
(iii) Find the value of the polynomial \( 5 x-4 x^{2}+3 \) at \(
x=2 \)
Answer
\( \mathrm{p}(\mathrm{x})=5 \mathrm{x}-4 x^{2}+3 \)Now \( \mathrm{p} (a)\) will mean replacing the variable with the value that is put in the bracket.
\(\mathrm{p}(2)=5(2)-4(2)^{2}+3 \)
\(\mathrm{p}(2)=10-16+3\)
\(\mathrm{p}(2)=-6+3 \mathrm{p}(2)=-3\)
2.
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(i) Find \( p(0), p(1) \) and \( p(2) \) for each of the
following polynomials: \(\mathrm{p}(\mathrm{y})=y^{2}-y+1 \)
Answer
\( \mathrm{p}(\mathrm{y})=y^{2}-y+1\)\(\mathrm{P}(0)=(0)^{2}-(0)+1=1\)
\(\mathrm{P}(1)=(1)^{2}-(1)+1=1\)
\(\mathrm{P}(2)=(2)^{2}-(2)+1=3\)
(ii) Find \( p(0), p(1) \) and \( p(2) \) for each of the
following polynomials: \(\mathrm{p}(\mathrm{t})=2+\mathrm{t}+2
t^{2}-t^{3} \)
Answer
\( p(t)=2+t+2 t^{2}-t^{3}\)\(\mathrm{P}(0)=2+0+2(0)^{2}-(0)^{3}=2\)
\(\mathrm{P}(1)=2+(1)+2(1)^{2}-(1)^{3}\)
\(=2+1+2-1=4\)
\(\mathrm{P}(2)=2+2+2(2)^{2}-(2)^{3}\)
\(=2+2+8-8=4\)
(iii) Find \( p(0), p(1) \) and \( p(2) \) for each of the
following polynomials: \(\mathrm{p}(\mathrm{x})=x^{3} \)
Answer
\( \mathrm{p}(\mathrm{x})=x^{3} \)\(\mathrm{p}(0)=(0)^{3}=0\)
\(\mathrm{p}(2)=(2)^{3}=8\)
(iv) Find \( p(0), p(1) \) and \( p(2) \) for each of the
followingpolynomials:
\(\mathrm{p}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}+1) \)
Answer
\( \mathrm{p}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}+1) \) \(\mathrm{P}(0)=(0-1)(0+1)=(-1)(1)=-1\)
\(\mathrm{P}(1)=(1-1)(1+1)=0(2)=0\)
\(\mathrm{P}(2)=(2-1)(2+1)=1(3)=3\)
3.
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(i) Verify whether the following are zeroes of the polynomial,
indicated against them. \( p(x)=3 x+1,x=-\frac{ 1 }{ 3 }\)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
Here \( p(x)=3 x+1 \), for \( \mathrm{x}=-\frac{ 1 }{ 3 } \)
\(p\left(\frac{-1}{3}\right)=3 \times \frac{-1}{3}+1\)
\(p\left(\frac{-1}{3}\right)=-1+1\)
\(p\left(\frac{-1}{3}\right)=0\)
Therefore, \( x=-\frac{ 1 }{ 3 } \) is a zero of polynomial \( p(x)=3 x+1 \)
(ii) Verify whether the following are zeroes of the polynomial,
indicated against them. \(\mathrm{p}(\mathrm{x})=5 \mathrm{x}-\pi,
\mathrm{x}=4 / 5 \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
here \( p(x)=5 x-\pi \), for \(x=\frac{ 4 }{ 5 } \),
\(p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi=4-\pi\)
Since, \( p\left(\frac{4}{5}\right) \neq 0 \)
Therefore, \( \mathrm{x}=\frac{4}{5} \) is not a zero of polynomial \( \mathrm{p}(\mathrm{x})=5\mathrm{x}-\pi \)
(iii) Verify whether the following are zeroes of the polynomial,
indicated against them. \(\mathrm{p}(\mathrm{x})=x^{2}-1,
\mathrm{x}=1,-1 \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( x=1 \) and \( x=-1 \) are zeros of polynomial \( p(x)=x^{2}-1 \) then \( p (1)\) and \( p(-1) \) should be 0
At, \( p(1)=(1)^{2}-1=0 \) and,
At, \( p(-1)=(-1)^{2}-1=0 \)
Hence, \( x=1 \) and \(-1\) are zeros of polynomial \( p(x)=x^{2}-1 \)
(iv) Verify whether the following are zeroes of the polynomial,
indicated against them.
\(\mathrm{p}(\mathrm{x})=(\mathrm{x}+1)(\mathrm{x}-2),
\mathrm{x}=-1,2 \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( x=-1 \) and \( x=2 \) are zeros of polynomial \( p(x)=(x+1)(x-2) \) then \( \mathrm{p}(-1) \) and \( \mathrm{p}(2) \) should be 0
At, \( p(-1)=(-1+1)(-1-2)=0(-3)=0 \) and,
At, \( p(2)=(2+1)(2-2)=3(0)=0 \)
Hence, \( x=-1 \) and \( x=2 \) are zeros of polynomial \( p(x)=(x+1)(x-2) \)
(v) Verify whether the following are zeroes of the polynomial,
indicated against them. \(\mathrm{p}(\mathrm{x})=x^{2},
\mathrm{x}=0 \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( x=0 \) is a zero of polynomial \( p(x)=x^{2} \)
Then, \( p(0) \) should be zero
Here, \( p(0)=(0)^{2}=0 \)
Hence, \( x=0 \) is the zero of the polynomial \( p(x)=x^{2} \)
(vi) Verify whether the following are zeroes of the polynomial,
indicated against them. \(\mathrm{p}(\mathrm{x})=l
x+\mathrm{m},=-\frac{m}{l} \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( \mathrm{x}=\frac{-m}{l} \) is a zero of the polynomial \(\mathrm{p}(\mathrm{x})=l \mathrm{x}+\mathrm{m} \) then \( \mathrm{p}\left(\frac{-m}{l}\right) \)should be \(0\)
\(\mathrm{p}\left(\frac{-m}{l}\right) =l\left(\frac{-m}{l}\right)+\mathrm{m}\)
\(=-m+m\)
\(=0\)
Therefore, \( x=\frac{-m}{l} \) is the zero of the polynomial \( p(x)=1 \mathrm{x}+\mathrm{m}\).
(vii) Verify whether the following are zeroes of the polynomial,
indicated against them. \(\mathrm{p}(\mathrm{x})=3 x^{2}-1,
\mathrm{x}=\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( \mathrm{x}=\frac{-1}{\sqrt{3}} \) and \(\mathrm{x}=\frac{2}{\sqrt{3}} \) are zeros of the polynomial \( \mathrm{p}(\mathrm{x})=3 x^{2}-1\), then \( \mathrm{p}\left(\frac{-1}{\sqrt{3}}\right) \) and \(\mathrm{p}\left(\frac{2}{\sqrt{3}}\right) \) should be \(0\)
At, \(p\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{-1}{\sqrt{3}}\right)^{2}-1\)
\(=3\left(\frac{1}{3}\right)-1\)
\(=1-1=0 \text { and, }\)
\(\text { At, } p\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-1\)
\(=3\left(\frac{4}{3}\right)-1\)
\(=4-1=3\)
Therefore \( \mathrm{x}=\left(\frac{-1}{\sqrt{3}}\right) \) is a zero of the polynomial \(\mathrm{p}(\mathrm{x})=3 x^{2}+1 \)But, \( x=\frac{2}{\sqrt{3}} \) is not a zero of the polynomial.
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(viii) Verify whether the following are zeroes of the polynomial,
indicated against them. \( p(x)=2 x+1,x=\frac{1}{2} \)
Answer
For any value of \( x \), say a, to be a zero of polynomial
\(p(a)=0 \) Keeping that in mind,
If \( \mathrm{x}=\frac{1}{2} \) is a zero of polynomial \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}+1 \) then \( \mathrm{p}(\frac{ 1 }{ 2 }) \) should be zero
\(\text { At, } p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1\)
\(=1+1=2\)
Therefore, \( \mathrm{x}=\frac{1}{2} \) is not a zero of given polynomial \(\mathrm{p}(\mathrm{x})=2 \mathrm{x}+1 \)
4.
(i) Find the zero of the polynomial in each of the following
cases: \( p(x)=x+5 \)
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0\)
\(\mathrm{p}(\mathrm{x})=\mathrm{x}+5\)
\(\mathrm{p}(\mathrm{x})=0\)
\(\mathrm{x}+5=0\)
\(\mathrm{x}=-5\)
Therefore, \( \mathrm{x}=-5 \) is a zero of the polynomial \( \mathrm{p(x)=x}+5 \)
(ii) Find the zero of the polynomial in each of the following
cases: \( p(x)=x-5 \)
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= \) \(-2\), then \( f(x)=0 \)
\( p(x)=x-5 \)
\(p(x)=0\)
\(x-5=0\)
\(x=5\)
Therefore, \( x=5 \) is a zero of the polynomial \( p(x)=x-5 \)
(iii) Find the value of the polynomial \( p(x)=2 x+5\)
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0 \)
\( p(x)=2 x+5\)
\(p(x)=0\)
\(2 x+5=0\)
\(2 x=-5\)
\(x=\frac{-5}{2}\)
Therefore, \( x=\frac{-5}{2} \) is a zero of the polynomial \( p(x)=2 x+5\)
(iv) Find the value of the polynomial \( p(x)=3 x-2\)
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at
which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0 \)
\( \mathrm{p}(\mathrm{x})=3 \mathrm{x}-2\)
\(\mathrm{p}(\mathrm{x})=0\)
\(3 \mathrm{x}-2=0\)
\(\mathrm{x}=\frac{2}{3}\)
Therefore, \( x=\frac{2}{3} \) is a zero of the polynomial \( p(x)=3 x-2 \)
(v) Find the value of the polynomial \(\mathrm{p}(\mathrm{x})=3
\mathrm{x} \)
Answer
Zero of a polynomial means the value of the variable at which
thepolynomial becomes zero.
For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0 \)
\(p(x)=3 x\)
\(p(x)=0\)
\(3 x=0\)
\(x=0\)
Therefore, \( \mathrm{x}=0 \) is a zero of the polynomial \( \mathrm{p}(\mathrm{x})=3
\mathrm{x} \)
(vi) Find the value of the polynomial
\(\mathrm{p}(\mathrm{x})=\mathrm{ax}, \mathrm{a} \neq 0 \)
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0 \)
\(p(x)=a x\)
\(p(x)=0\)
\(a x=0\)
\(x=0\)
Therefore, \( x=0 \) is a zero of the polynomial \( p(x)=a x \)
(vii) Find the value of the polynomial
\(\mathrm{p}(\mathrm{x})=\mathrm{cx}+\mathrm{d}, \mathrm{c} \neq 0
\), \( \mathrm{c}, \mathrm{d}\) are real numbers.
Answer
Zero of a polynomial means the value of the variable at which the
polynomial becomes zero.For example for a polynomial \( f(x) \), zeroes of \( f(x) \) means the values of \( x \) at which \( f(x)=0 \)
Let \( f(x)=x+2 \), so it can clearly be seen that if you put \( x= -2\), then \( f(x)=0 \)
\( \mathrm{p}(\mathrm{x})=\mathrm{cx}+\mathrm{d}\)
\(\mathrm{p}(\mathrm{x})=0\)
\(\mathrm{cx}+\mathrm{d}=0\)
\(\mathrm{x}=-\frac{d}{c}\)
Therefore, \( \mathrm{x}=-\frac{d}{c} \) is a zero of polynomial \(\mathrm{p}(\mathrm{x})=\mathrm{cx}+\mathrm{d} \)
