NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Exercise 2.2

1. Find the value of the polynomial $\text{5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$  at 

i) $\text{x = 0}$

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 5}\left( \text{0} \right)\text{ - 4}\left( {{\text{0}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 3}$

Hence, the value of the polynomial at $\text{x = 0}$ is $\text{3}$.

ii) $\text{x = -1}$ 

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = 5}\left( \text{-1} \right)\text{ - 4}\left[ {{\left( \text{-1} \right)}^{\text{2}}} \right]\text{ + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 5 - 4 + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 6}$

Hence, the value of the polynomial at $\text{x = -1}$ is $\text{-6}$.

iii) $\text{x = 2}$ 

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 5}\left( \text{2} \right)\text{ - 4}\left( {{\text{2}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 10 - 16 + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = - 3}$

Hence, the value of the polynomial at $\text{x = 2}$ is $\text{-3}$.

2. Find $\text{p}\left( \text{0} \right)$, $\text{p}\left( \text{1} \right)$  and $\text{p}\left( \text{2} \right)$  for each of the following polynomials:

i) $\text{p}\left( \text{y} \right)\text{ = }{{\text{y}}^{\text{2}}}\text{ - y + 1}$ 

Ans:

• $\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{2}}}\text{ - 0 + 1}$.

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 1}$.

• $\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{2}}}\text{ - 1 + 1 = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

• $\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{2}}}\text{ - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

ii) $\text{p}\left( \text{t} \right)\text{ = 2 + t + 2}{{\text{t}}^{\text{2}}}\text{ - }{{\text{t}}^{\text{3}}}$

Ans:

• $\text{p}\left( \text{0} \right)\text{ = 2 + 0 + 2}\left( {{\text{0}}^{\text{2}}} \right)\text{ - }\left( {{\text{0}}^{\text{3}}} \right)\text{ = 2}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 2}$.

• $\text{p}\left( \text{1} \right)\text{ = 2 + 1 + 2}\left( {{\text{1}}^{\text{2}}} \right)\text{ - }\left( {{\text{1}}^{\text{3}}} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 4}$.

• $\text{p}\left( \text{2} \right)\text{ = 2 + 2 + 2}\left( {{\text{2}}^{\text{2}}} \right)\text{ - }\left( {{\text{2}}^{\text{3}}} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 + 8 - 8}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 4}$.

iii) $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

Ans:

• $\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{3}}}\text{ = 0}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 0}$.

• $\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{3}}}\text{ = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

• $\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{3}}}\text{ = 8}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 8}$.

iv) $\text{p}\left( \text{x} \right)\text{ =}\left( \text{x - 1} \right)\left( \text{x + 1} \right)$

Ans:

• $\text{p}\left( \text{0} \right)\text{ =}\left( \text{0 - 1} \right)\left( \text{0 + 1} \right)\text{ = -1}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = -1}$.

• $\text{p}\left( \text{1} \right)\text{ =}\left( \text{1 - 1} \right)\left( \text{1 + 1} \right)\text{ = 0}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 0}$.

• $\text{p}\left( \text{2} \right)\text{ =}\left( \text{2 - 1} \right)\left( \text{2 + 1} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = }\left( \text{1} \right)\left( \text{3} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

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

i) $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$,$\text{x = -}\frac{\text{1}}{\text{3}}$

Ans: We are given: $\text{x = -}\frac{\text{1}}{\text{3}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$, then $\text{p}\left( \text{-}\frac{\text{1}}{\text{3}} \right)$ should be  $\text{0}$.

$\text{p}\left(\text{-}\frac{\text{1}}{\text{3}}\right)\text{=3}\left(\text{-}\frac{\text{1}}{\text{3}} \right)\text{ + 1 = -1 + 1 = 0}$

Hence, we can say that $\text{x = -}\frac{\text{1}}{\text{3}}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{= 5x - }\!\!\pi\!\!\text{ }$ ,$\text{x = }\frac{\text{4}}{\text{5}}$

Ans: We are given: $\text{x = }\frac{\text{4}}{\text{5}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 5x -  }\text{ }\!\!\pi\!\!\text{ }$, then $\text{p}\left( \frac{\text{4}}{\text{5}} \right)$ should be  $\text{0}$.

$\text{p}\left( \frac{\text{4}}{\text{5}} \right)\text{ = 5}\left( \frac{\text{4}}{\text{5}} \right)\text{ - 3}\text{.14 = 4 - 3}\text{.14 }\ne \text{ 0}$

Hence, we can say that $\text{x = }\frac{\text{4}}{\text{5}}$ is not a zero of the given polynomial.

iii) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1, x = 1, -1}$

Ans: We are given: $\text{x = 1}$  and  $\text{x = -1}$ . 

If they are zeros of polynomial  $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1}$ , then $\text{p}\left( \text{1} \right)$ and $\text{p}\left( \text{-1} \right)$ should both be $\text{0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\left( \text{1} \right)}^{\text{2}}}\text{-1 = 0}$

$\text{p}\left( \text{-1} \right)\text{ = }{{\left( \text{-1} \right)}^{\text{2}}}\text{-1 = 0}$

Hence, we can say that $\text{x = 1}$ and $\text{x = -1}$  are zeroes of the given polynomial.

iv) $\text{p(x) = (x+1)(x-2), x = -1, 2}$

Ans: We are given: $\text{x = -1}$ and $\text{x = 2}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = }\left( \text{x+1} \right)\left( \text{x-2} \right)$ , then $\text{p}\left( \text{-1} \right)$ and $\text{p}\left( \text{2} \right)$ should be $\text{0}$.

$\text{p}\left( \text{-1} \right)\text{ = }\left( \text{-1+1} \right)\left( \text{-1-2} \right)\text{ = }\left( \text{0} \right)\left( \text{-3} \right)\text{ = 0}$

$\text{p}\left( \text{2} \right)\text{ = }\left( \text{2+1} \right)\left( \text{2-2} \right)\text{ = }\left( \text{3} \right)\left( \text{0} \right)\text{ = 0}$

Hence, we can say that $\text{x = -1}$ and $\text{x = 2}$  are zeroes of the given polynomial.

v) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{, x = 0}$

Ans: We are given $\text{x = 0}$. 

If it is a zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}$  , then $\text{p}\left( \text{0} \right)$ should be $\text{0}$.

$\text{ p}\left( \text{0} \right)\text{ = }{{\left( \text{0} \right)}^{\text{2}}}\text{ = 0}$

Hence, we can say that $\text{x = 0}$ is a zero of the given polynomial.

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

Ans: We are given: $\text{x = -}\frac{\text{m}}{\text{l}}$. 

If it is a zero of the polynomial  $\text{p}\left( \text{x} \right)\text{ = lx + m}$ , then $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)$ should be  $\text{0}$ .

Here, $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ = l}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ + m = -m + m = 0}$

Hence, we can say that $\text{x = -}\frac{\text{m}}{\text{l}}$  is a zero of the given polynomial.

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

Ans: We are given:$\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$  and  $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3}{{\text{x}}^{\text{2}}}\text{-1}$, then $\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)$and $\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$   should be $\text{0}$ .

$\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{1}}{\text{3}} \right)\text{-1 = 1-1 = 0}$

$\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{4}}{\text{3}} \right)\text{-1 = 4-1 = 3}$

Hence, we can say that $\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$ is a zero of the given polynomial.

However, the value of $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$ is not a zero of the given polynomial.

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

Ans: We are given: $\text{x = }\frac{\text{1}}{\text{2}}$. 

If it is a zero of polynomial $\text{p}\left( \text{x} \right)\text{ = 2x+1}$ , then $\text{p}\left( \frac{\text{1}}{\text{2}} \right)$ should be $\text{0}$

Here, $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\text{ = 2}\left( \frac{\text{1}}{\text{2}} \right)\text{+1 = 1+1 = 2}$.

So, we get the value $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\ne \text{0}$  .

Hence, we can say that $\text{x = }\frac{\text{1}}{\text{2}}$ is not a zero of the given polynomial.

4. Find the zero of the polynomial in each of the following cases:

i) $\text{p}\left( \text{x} \right)\text{ = x+5}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x+5 = 0}$

$\Rightarrow \text{x = -5}$

Therefore, $\text{x = -5}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{ = x-5}$ 

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x-5 = 0}$

$\Rightarrow \text{x = 5}$

Therefore, $\text{x = 5}$ is a zero of the given polynomial.

iii) $\text{p}\left( \text{x} \right)\text{ = 2x+5}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{2x+5 = 0}$

$\Rightarrow \text{x = -}\frac{\text{5}}{\text{2}}$

Therefore, $\text{x = -}\frac{\text{5}}{\text{2}}$ is a zero of the given polynomial.

iv) $\text{p}\left( \text{x} \right)\text{ = 3x-2}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{3x-2 = 0}$

$\Rightarrow \text{x = }\frac{\text{2}}{\text{3}}$

Therefore, $\text{x = }\frac{\text{2}}{\text{3}}$  is a zero of the given polynomial.

v) $\text{p}\left( \text{x} \right)\text{ = 3x}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{3x = 0}$

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p}\left( \text{x} \right)\text{ = ax, a}\ne \text{0}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{ax = 0}$

It is also given that $\text{a}$ is non-zero. 

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vii) $\text{p}\left( \text{x} \right)\text{ = cx+d, c}\ne \text{0, c, d}$ are real numbers.

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{cx+d = 0}$

It is also given that $\text{c}$ is non-zero.

$\Rightarrow \text{x = -}\frac{\text{d}}{\text{c}}$

Therefore, $\text{x = -}\frac{\text{d}}{\text{c}}$  is a zero of the given polynomial.

Conclusion

The NCERT Solutions for Class 9th Maths Chapter 2 Exercise 2.2 primarily focuses on enhancing the understanding of polynomial expressions, their evaluation, zeros, and factorization. Ex 2.2 Class 9 is crucial for building a solid foundation in algebraic techniques, which are essential for higher mathematical learning. Students should pay particular attention to how to evaluate polynomials for given values, determine and verify the zeros of polynomials, and apply algebraic identities in the factorization process. The solutions provided, such as those by Vedantu, are designed to guide students step-by-step through each problem, ensuring a thorough grasp of concepts and methods.

Class 9 Maths Chapter 2: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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