Polynomials
NCERT Solutions • Class 9 Maths • Chapter 2Exercise 2.2
1. Find the value of the polynomial $5x – 4x^2 + 3$ at:
(i) $x = 0$
$5(0) – 4(0)^2 + 3 = 0 – 0 + 3$ $\rightarrow$ Value: 3
$5(0) – 4(0)^2 + 3 = 0 – 0 + 3$ $\rightarrow$ Value: 3
(ii) $x = -1$
$5(-1) – 4(-1)^2 + 3 = -5 – 4(1) + 3 = -9 + 3$ $\rightarrow$ Value: -6
$5(-1) – 4(-1)^2 + 3 = -5 – 4(1) + 3 = -9 + 3$ $\rightarrow$ Value: -6
(iii) $x = 2$
$5(2) – 4(2)^2 + 3 = 10 – 4(4) + 3 = 10 – 16 + 3$ $\rightarrow$ Value: -3
$5(2) – 4(2)^2 + 3 = 10 – 4(4) + 3 = 10 – 16 + 3$ $\rightarrow$ Value: -3
2. Find $p(0), p(1)$ and $p(2)$ for each of the following polynomials:
(i) $p(y) = y^2 – y + 1$
$p(0) = 0 – 0 + 1 = \mathbf{1}$ | $p(1) = 1 – 1 + 1 = \mathbf{1}$ | $p(2) = 4 – 2 + 1 = \mathbf{3}$
$p(0) = 0 – 0 + 1 = \mathbf{1}$ | $p(1) = 1 – 1 + 1 = \mathbf{1}$ | $p(2) = 4 – 2 + 1 = \mathbf{3}$
(ii) $p(t) = 2 + t + 2t^2 – t^3$
$p(0) = 2$ | $p(1) = 2+1+2-1 = \mathbf{4}$ | $p(2) = 2+2+8-8 = \mathbf{4}$
$p(0) = 2$ | $p(1) = 2+1+2-1 = \mathbf{4}$ | $p(2) = 2+2+8-8 = \mathbf{4}$
(iii) $p(x) = x^3$
$p(0) = 0^3 = \mathbf{0}$ | $p(1) = 1^3 = \mathbf{1}$ | $p(2) = 2^3 = \mathbf{8}$
$p(0) = 0^3 = \mathbf{0}$ | $p(1) = 1^3 = \mathbf{1}$ | $p(2) = 2^3 = \mathbf{8}$
(iv) $p(x) = (x – 1)(x + 1)$
$p(0) = (-1)(1) = \mathbf{-1}$ | $p(1) = (0)(2) = \mathbf{0}$ | $p(2) = (1)(3) = \mathbf{3}$
$p(0) = (-1)(1) = \mathbf{-1}$ | $p(1) = (0)(2) = \mathbf{0}$ | $p(2) = (1)(3) = \mathbf{3}$
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p(x) = 3x + 1, x = -\frac{1}{3}$
$p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0$ $\rightarrow$ Yes
$p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0$ $\rightarrow$ Yes
(ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
$p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi = 4 – \pi \neq 0$ $\rightarrow$ No
$p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi = 4 – \pi \neq 0$ $\rightarrow$ No
(iii) $p(x) = x^2 – 1, x = 1, -1$
$p(1) = 1^2 – 1 = 0$ (Yes) | $p(-1) = (-1)^2 – 1 = 0$ (Yes)
$p(1) = 1^2 – 1 = 0$ (Yes) | $p(-1) = (-1)^2 – 1 = 0$ (Yes)
(iv) $p(x) = (x + 1)(x – 2), x = -1, 2$
$p(-1) = 0$ (Yes) | $p(2) = 0$ (Yes)
$p(-1) = 0$ (Yes) | $p(2) = 0$ (Yes)
(v) $p(x) = x^2, x = 0$
$p(0) = 0^2 = 0$ $\rightarrow$ Yes
$p(0) = 0^2 = 0$ $\rightarrow$ Yes
(vi) $p(x) = lx + m, x = -\frac{m}{l}$
$p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0$ $\rightarrow$ Yes
$p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0$ $\rightarrow$ Yes
(vii) $p(x) = 3x^2 – 1, x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
$p(-\frac{1}{\sqrt{3}}) = 0$ (Yes) | $p(\frac{2}{\sqrt{3}}) = 3(\frac{4}{3}) – 1 = 3 \neq 0$ (No)
$p(-\frac{1}{\sqrt{3}}) = 0$ (Yes) | $p(\frac{2}{\sqrt{3}}) = 3(\frac{4}{3}) – 1 = 3 \neq 0$ (No)
(viii) $p(x) = 2x + 1, x = \frac{1}{2}$
$p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 2 \neq 0$ $\rightarrow$ No
$p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 2 \neq 0$ $\rightarrow$ No
4. Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
$x + 5 = 0 \Rightarrow$ $x = -5$
$x + 5 = 0 \Rightarrow$ $x = -5$
(ii) $p(x) = x – 5$
$x – 5 = 0 \Rightarrow$ $x = 5$
$x – 5 = 0 \Rightarrow$ $x = 5$
(iii) $p(x) = 2x + 5$
$2x + 5 = 0 \Rightarrow$ $x = -\frac{5}{2}$
$2x + 5 = 0 \Rightarrow$ $x = -\frac{5}{2}$
(iv) $p(x) = 3x – 2$
$3x – 2 = 0 \Rightarrow$ $x = \frac{2}{3}$
$3x – 2 = 0 \Rightarrow$ $x = \frac{2}{3}$
(v) $p(x) = 3x$
$3x = 0 \Rightarrow$ $x = 0$
$3x = 0 \Rightarrow$ $x = 0$
(vi) $p(x) = ax, a \neq 0$
$ax = 0 \Rightarrow$ $x = 0$
$ax = 0 \Rightarrow$ $x = 0$
(vii) $p(x) = cx + d, c \neq 0$
$cx + d = 0 \Rightarrow$ $x = -\frac{d}{c}$
$cx + d = 0 \Rightarrow$ $x = -\frac{d}{c}$