Exercise 2.2

Exercise 2.2 of NCERT Class 9 Mathematics (Chapter: Polynomials) focuses on finding the value of polynomials at different points, understanding and calculating the zeros of polynomials, and verifying them with step-by-step solutions. All textbook questions and examples have been solved clearly to help students strengthen their fundamentals. This post is especially useful for Class 9 students preparing for exams who often face difficulty in solving polynomial problems. At the end of the post, we have provided important links for further study materials, detailed notes on the basics of polynomials, and MCQs quizzes that allow students to practice in an exam-like environment for better preparation.

Continue Reading →
Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.2

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

Continue Reading →
Exercise
Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

Continue Reading →
Exercise
September 14, 2025  |  By Academia Aeternum

Exercise 2.2

Maths - Exercise

1. Find the value of the polynomial \(5x – 4x^2 + 3\) at

Solution:
  1. \(x = 0\)
    $$\begin{aligned}p\left( x\right) &=5x-4x^2+3\\ p\left( 0\right) &=5-4\cdot({0})^2+3\\ &=5\cdot 0\cdot - 4\cdot 0+3\\ &=0+0+3\\ &=3\end{aligned}$$
  2. \(x = -1\)
    $$\begin{aligned}p\left( -1\right) &=5\times \left( -1\right)-4\times \left( {-1}^2\right) +3\\ &=-5\cdot 1-4\cdot 1+3\\ &=-5-4+3\\ &=-9+3\\ &=-6\end{aligned}$$
  3. \(x = 2\)
    $$\begin{aligned}p\left( 2\right) &=5\cdot 2-4\cdot (2)^2+3\\ &=5\cdot 2 -4\cdot 4 + 3\\ &=10-16+3\\ &=13-16\\ &=-3 \end{aligned}$$

2. Find \(p(0),~ p(1) ~and~ p(2)\) for each of the following polynomials:

Solution:
  1. \(p(y) = y^2 – y + 1\)
    • $$\begin{aligned} p\left( 0\right) &=0^{2}-0+1\\ &=1\end{aligned}$$
    • $$\begin{aligned}p\left( 1\right) &=1^{2}-1+1\\ &=1-1+1\\ &=1\end{aligned}$$
    • $$\begin{aligned}p\left( 2\right) &=2^{2}-2+1\\ &=4-2+1\\ &=5-2\\ &=3\end{aligned}$$
  2. \(p\left( t\right)=2+t+2t^{2}-t^{3}\)
    • $$\begin{aligned} p\left( 0\right) &=2+0+2\cdot 0^{2}+0^{3}\\ &=2+0+0+0\\ &=2\end{aligned}$$
    • $$\begin{aligned}P\left( 1\right) &=2+1+2\cdot 1^{2}-1^{3}\\ &=2+1+2\cdot 1-1\\ &=2+1+2-1\\ &=5-1\\ &=4\end{aligned}$$
    • $$\begin{aligned}p\left( 2\right) &=2+2+2\cdot 2^{2}-2^{3}\\ &=2+2+2\cdot 4-8\\ &=2+2+8-8\\ &=4\end{aligned}$$
  3. \(p\left( x\right) =x^{3}\)
    • $$\begin{aligned} p\left( 0\right) &=0^{3}\\ &=0\end{aligned}$$
    • $$\begin{aligned}P\left( 1\right) &=1^{3}\\ &=1\end{aligned}$$
    • $$\begin{aligned}P\left( 2\right) &=2^{3}\\ &=8\end{aligned}$$
  4. \(p\left( x\right) =\left( x-1\right) \times \left( x+1\right)\)
    • $$\begin{aligned}p\left( 0\right) &=\left( 0-1\right) \times \left( 0+1\right) \\ &=-1\times 1\\ &=-1\end{aligned}$$
    • $$\begin{aligned}p\left( 1\right) &=\left( 1-1\right) \times \left( 1+1\right) \\ &=0\times 2\\ &=0\end{aligned}$$
    • $$\begin{aligned}p\left( 2\right) &=\left( 2-1\right) \times \left( 2+1\right) \\ &=1\cdot 3\\ &=3\end{aligned}$$

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

Solution:
  1. $$\begin{aligned}p\left( x\right) &=3x+1,x=-\frac{1}{3}\\ P\left( \frac{-1}{3}\right) &=3\cdot \left( -\frac{1}{3}\right) +1\\ &=-3\times \dfrac{1}{3}+1\\ &=-1+1\\ &=0\end{aligned}$$ hence, \(x=-\frac{1}{3}\) is zero of polynomial.
  2. $$\begin{aligned}p\left( x\right) &=5x-\pi ,x=4/5\\ P\left( 4/5\right) &=5\times \dfrac{4}{5}-\pi \\ &=4-\pi\\ &\neq 0\end{aligned}$$ hence, \(x=4/5\) is not a zero of the polynomial.
  3. $$\begin{aligned}p\left( x\right) &=x^{2}-1,x=1,-1\\ p\left( 1\right) &=1^{2}-1\\ &=0\\\\ p\left( -1\right)&=\left( -1\right) ^{2}-1\\ &=1-1\\ &=0\end{aligned}$$ hence, \(x=1~and ~x=-1\) are zero of polynomial
  4. $$\begin{aligned}p\left( x\right) &=\left( x+1\right) \times \left( x-2\right) ,x=-1,2\\ p\left( -1\right) &=\left( -1+1\right) \times \left( -1-2\right) \\ &=0\times \left( -3\right) \\ &=0\\\\ p\left( 2\right) &=\left( 2+1\right) \times \left( 2-2\right) \\ &=3\times 0\\ &=0\end{aligned}$$ hence, \(x=1~and ~x=2\) are zero of polynomial
  5. $$\begin{aligned}p\left( x\right) &=x^{2},x=0\\ p\left( 0\right) &=0^{2}\\ &=0\end{aligned}$$ hence, \(x=0\) is zero of polynomial
  6. $$\begin{aligned}p\left( x\right) =lx+m,\quad x&=-\dfrac{m}{l}\\ p\left( \frac{-m}{l }\right) &=l\left( \dfrac{-m}{l}\right) +m\\ &=-\dfrac{l\cdot m}{l}+m\\ &=-m+m\\ &=0\end{aligned}$$ hence, \(x=-\dfrac{m}{l}\) is zero of polynomial
  7. $$\begin{aligned}p\left( x\right) &=3x^{2}-1,\quad x=\dfrac{-1}{\sqrt{3}},\dfrac{2}{\sqrt{3}}\\ p\left( -\dfrac{1}{\sqrt{3}}\right) &=3\cdot \left( -\dfrac{1}{\sqrt{3}}\right) ^{2}-1\\ &=3\cdot \dfrac{1}{3}-1\\ &=1-1\\ &=0\\\\ p\left( \dfrac{2}{\sqrt{3}}\right) &=3\cdot \left( \dfrac{2}{\sqrt{3}}\right) ^{2}-1\\ &=3\cdot \dfrac{4}{3}-1\\ &=4-1\\ &=3\end{aligned}$$ hence, \(x=\dfrac{-1}{\sqrt{3}}\) is zero of polynomial but \(x=\dfrac{2}{\sqrt{3}}\) is not a zero of polynomial
  8. $$\begin{aligned}p\left( x\right) &=2x+1,\quad x=\frac{1}{2}\\ p\left(\frac{1}{2}\right) &=2\cdot \dfrac{1}{2}+1\\ &=1+1\\ &=2\end{aligned}$$ hence, \(x=\dfrac{1}{2}\) is not a zero of polynomial

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

Solution:
  1. \(p(x) = x + 5\)
    $$\begin{aligned}p\left( x\right) &=x+5\\ x+5&=0\\ x&=-5\end{aligned}$$
  2. \( p(x) = x – 5\\\) $$\begin{aligned}p\left( x\right) &=x-5\\ x-5&=0\\ x&=5\end{aligned}$$
  3. \(p(x) = 2x + 5\\\) $$\begin{aligned}p\left( x\right) &=2x+5\\ 2x+5&=0\\ 2x&=-5\\ x&=-\frac{5}{2}\end{aligned}$$
  4. \( p(x) = 3x – 2\\\) $$\begin{aligned}p\left( x\right) &=3x-2\\ 3x-2&=0\\ 3x&=2\\ x&=\frac{2}{3}\end{aligned}$$
  5. \( p(x) = 3x\\\) $$\begin{aligned}p\left( x\right) &=3x\\ 3x&=0\\ x&=0\end{aligned}$$
  6. \( p(x) = ax,\quad a ≠ 0\\\) $$\begin{aligned}p\left( x\right) &=ax,\quad a\neq 0\\ ax&=0\\ x&=0\end{aligned}$$
  7. \(p(x) = cx + d, \quad c ≠ 0, c, d are real numbers.\\\) $$\begin{aligned} cx+d&=0\\ cx&=-d\\ x&=-\dfrac{d}{c}\end{aligned}$$

Recent posts


    Important Links

    Leave Your Message & Comments