Chapter 2: Polynomials (Algebra)

Step-by-step NCERT solutions for Class 10 Maths Chapter 2, Polynomials (2026-27 reprint): Exercise 2.1 on counting zeroes from graphs, and Exercise 2.2 on finding zeroes of quadratic polynomials, verifying the relationship between zeroes and coefficients, and forming a quadratic polynomial from the sum and product of its zeroes. All 3 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-2-polynomials

For a quadratic $a x^{2} + b x + c$ with zeroes $α$ and $β$ : $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ . A quadratic with sum of zeroes $S$ and product $P$ is $k (x^{2} - S x + P)$ for any non-zero constant $k$ .

Exercise 2.1

The graphs of y = p(x) are given in Fig. 2.10 for some polynomials p(x). Find the number of zeroes of p(x) in each case.

Solution

The zeroes of $p (x)$ are the $x$ -coordinates of the points where the graph meets the $x$ -axis, so we count those points.

The graph is a line parallel to the $x$ -axis and never meets it: 0 zeroes.
It cuts the $x$ -axis once: 1 zero.
It cuts the $x$ -axis three times: 3 zeroes.
It cuts the $x$ -axis twice: 2 zeroes.
It cuts the $x$ -axis four times: 4 zeroes.
It meets the $x$ -axis at three points (it touches it at one of them): 3 zeroes.

(i) 0 (ii) 1 (iii) 3 (iv) 2 (v) 4 (vi) 3

Exercise 2.2

Find the zeroes of the quadratic polynomials and verify the relationship between the zeroes and the coefficients: (i) x² – 2x – 8 (ii) 4s² – 4s + 1 (iii) 6x² – 3 – 7x (iv) 4u² + 8u (v) t² – 15 (vi) 3x² – x – 4

Solution

(i) $x^{2} - 2 x - 8 = (x - 4) (x + 2)$ , so the zeroes are $4$ and $- 2$ .

Sum $= 4 + (- 2) = 2 = - \frac{- 2}{1}$ ✓
Product $= 4 \times (- 2) = - 8 = \frac{- 8}{1}$ ✓

(ii) $4 s^{2} - 4 s + 1 = (2 s - 1)^{2}$ , so the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$ .

Sum $= 1 = - \frac{- 4}{4}$ ✓
Product $= \frac{1}{4} = \frac{1}{4}$ ✓

(iii) $6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = (3 x + 1) (2 x - 3)$ , so the zeroes are $- \frac{1}{3}$ and $\frac{3}{2}$ .

Sum $= - \frac{1}{3} + \frac{3}{2} = \frac{7}{6} = - \frac{- 7}{6}$ ✓
Product $= - \frac{1}{3} \times \frac{3}{2} = - \frac{1}{2} = \frac{- 3}{6}$ ✓

(iv) $4 u^{2} + 8 u = 4 u (u + 2)$ , so the zeroes are $0$ and $- 2$ .

Sum $= - 2 = - \frac{8}{4}$ ✓
Product $= 0 = \frac{0}{4}$ ✓

(v) $t^{2} - 15 = (t - 15) (t + 15)$ , so the zeroes are $15$ and $- 15$ .

Sum $= 0 = - \frac{0}{1}$ ✓
Product $= - 15 = \frac{- 15}{1}$ ✓

(vi) $3 x^{2} - x - 4 = 3 x^{2} - 4 x + 3 x - 4 = (3 x - 4) (x + 1)$ , so the zeroes are $\frac{4}{3}$ and $- 1$ .

Sum $= \frac{4}{3} - 1 = \frac{1}{3} = - \frac{- 1}{3}$ ✓
Product $= - \frac{4}{3} = \frac{- 4}{3}$ ✓

(i) 4, –2 (ii) 1/2, 1/2 (iii) –1/3, 3/2 (iv) 0, –2 (v) √15, –√15 (vi) 4/3, –1; in each case sum = –b/a and product = c/a.

Find a quadratic polynomial with the given numbers as the sum and product of its zeroes respectively: (i) 1/4, –1 (ii) √2, 1/3 (iii) 0, √5 (iv) 1, 1 (v) –1/4, 1/4 (vi) 4, 1

Solution

Use $x^{2} - (sum) x + (product)$ , then multiply by a constant to clear fractions:

$x^{2} - \frac{1}{4} x - 1$ , i.e. $4 x^{2} - x - 4$
$x^{2} - 2 x + \frac{1}{3}$ , i.e. $3 x^{2} - 32 x + 1$
$x^{2} + 5$ (sum 0, so there is no $x$ term)
$x^{2} - x + 1$
$x^{2} + \frac{1}{4} x + \frac{1}{4}$ , i.e. $4 x^{2} + x + 1$
$x^{2} - 4x + 1$

(Any non-zero multiple of each answer is also correct.)

(i) 4x² – x – 4 (ii) 3x² – 3√2x + 1 (iii) x² + √5 (iv) x² – x + 1 (v) 4x² + x + 1 (vi) x² – 4x + 1

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