NCERT Solutions For Class 10 Maths: Polynomials

EXERCISE 2.1

Question 1: The graphs of $y = p(x)$ are given in Fig. 2.10 below, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$, in each case.

(Note: Since the figure 2.10 itself is not available, the solution relies on the geometrical meaning of zeroes described in the source, where the number of zeroes is equal to the number of points where the graph intersects the x-axis. We assume the standard six graphs typically shown in this exercise.)

Solution-

The zeroes of a polynomial $p(x)$ are precisely the x-coordinates of the points, where the graph of $y = p(x)$ intersects the x-axis.

(i) The graph is a straight line, parallel to the x-axis, which does not intersect it.
The number of zeroes is $\mathbf{0}$.

(ii) The graph intersects the x-axis at one point only.
The number of zeroes is $\mathbf{1}$.

(iii) The graph intersects the x-axis at three points.
The number of zeroes is $\mathbf{3}$.

(iv) The graph intersects the x-axis at one point only.
The number of zeroes is $\mathbf{1}$.

(v) The graph intersects the x-axis at one point only.
The number of zeroes is $\mathbf{1}$.

(vi) The graph intersects the x-axis at four points.
The number of zeroes is $\mathbf{4}$.

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EXERCISE 2.2

Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 – 2x – 8$
(ii) $4s^2 – 4s + 1$
(iii) $6x^2 – 3 – 7x$
(iv) $4u^2 + 8u$
(v) $t^2 – 15$
(vi) $3x^2 – x – 4$

Solution-

(i) $x^2 – 2x – 8$

Factorisation by splitting the middle term:
$x^2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4)$
$= (x + 2)(x – 4)$
The zeroes are $x = -2$ and $x = 4$.

Verification:
The polynomial is of the form $ax^2 + bx + c$, where $a=1, b=-2, c=-8$. Let $\alpha = -2$ and $\beta = 4$.

Sum of zeroes ($\alpha + \beta$):
$-2 + 4 = 2$
Using coefficients: $-\frac{b}{a} = - \frac{(-2)}{1} = 2$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$(-2) \times 4 = -8$
Using coefficients: $\frac{c}{a} = \frac{(-8)}{1} = -8$.
Verification: Product of zeroes $= \frac{c}{a}$.

(ii) $4s^2 – 4s + 1$

Factorisation by splitting the middle term:
$4s^2 – 2s – 2s + 1 = 2s(2s – 1) – 1(2s – 1)$
$= (2s – 1)(2s – 1)$
The zeroes are $s = \frac{1}{2}$ and $s = \frac{1}{2}$.

Verification:
The polynomial is of the form $as^2 + bs + c$, where $a=4, b=-4, c=1$. Let $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$.

Sum of zeroes ($\alpha + \beta$):
$\frac{1}{2} + \frac{1}{2} = 1$
Using coefficients: $-\frac{b}{a} = - \frac{(-4)}{4} = 1$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Using coefficients: $\frac{c}{a} = \frac{1}{4}$.
Verification: Product of zeroes $= \frac{c}{a}$.

(iii) $6x^2 – 3 – 7x$

Rewrite in standard form: $6x^2 – 7x – 3$.
Factorisation by splitting the middle term ($6 \times -3 = -18$; split -7x into -9x and 2x):
$6x^2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3)$
$= (3x + 1)(2x – 3)$
The zeroes are $x = -\frac{1}{3}$ and $x = \frac{3}{2}$.

Verification:
$a=6, b=-7, c=-3$. Let $\alpha = -\frac{1}{3}$ and $\beta = \frac{3}{2}$.

Sum of zeroes ($\alpha + \beta$):
$-\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6}$
Using coefficients: $-\frac{b}{a} = - \frac{(-7)}{6} = \frac{7}{6}$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$-\frac{1}{3} \times \frac{3}{2} = - \frac{3}{6} = - \frac{1}{2}$
Using coefficients: $\frac{c}{a} = \frac{-3}{6} = - \frac{1}{2}$.
Verification: Product of zeroes $= \frac{c}{a}$.

(iv) $4u^2 + 8u$

Factorisation:
$4u^2 + 8u = 4u(u + 2)$
The zeroes are $u = 0$ (from $4u=0$) and $u = -2$ (from $u+2=0$).

Verification:
The polynomial is $4u^2 + 8u + 0$, so $a=4, b=8, c=0$. Let $\alpha = 0$ and $\beta = -2$.

Sum of zeroes ($\alpha + \beta$):
$0 + (-2) = -2$
Using coefficients: $-\frac{b}{a} = - \frac{8}{4} = -2$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$0 \times (-2) = 0$
Using coefficients: $\frac{c}{a} = \frac{0}{4} = 0$.
Verification: Product of zeroes $= \frac{c}{a}$.

(v) $t^2 – 15$

Using the identity $a^2 - b^2 = (a-b)(a+b)$:
$t^2 – 15 = t^2 – (\sqrt{15})^2 = (t - \sqrt{15})(t + \sqrt{15})$
The zeroes are $t = \sqrt{15}$ and $t = -\sqrt{15}$.

Verification:
The polynomial is $t^2 + 0t – 15$, so $a=1, b=0, c=-15$. Let $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

Sum of zeroes ($\alpha + \beta$):
$\sqrt{15} + (-\sqrt{15}) = 0$
Using coefficients: $-\frac{b}{a} = - \frac{0}{1} = 0$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$(\sqrt{15}) \times (-\sqrt{15}) = -15$
Using coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$.
Verification: Product of zeroes $= \frac{c}{a}$.

(vi) $3x^2 – x – 4$

Factorisation by splitting the middle term ($3 \times -4 = -12$; split -x into -4x and 3x):
$3x^2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4)$
$= (x + 1)(3x – 4)$
The zeroes are $x = -1$ and $x = \frac{4}{3}$.

Verification:
$a=3, b=-1, c=-4$. Let $\alpha = -1$ and $\beta = \frac{4}{3}$.

Sum of zeroes ($\alpha + \beta$):
$-1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}$
Using coefficients: $-\frac{b}{a} = - \frac{(-1)}{3} = \frac{1}{3}$.
Verification: Sum of zeroes $= - \frac{b}{a}$.

Product of zeroes ($\alpha\beta$):
$(-1) \times \frac{4}{3} = - \frac{4}{3}$
Using coefficients: $\frac{c}{a} = \frac{-4}{3}$.
Verification: Product of zeroes $= \frac{c}{a}$.

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Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4}, -1$
(ii) $\sqrt{2}, \frac{1}{3}$
(iii) $0, \sqrt{5}$
(iv) $1, 1$
(v) $-\frac{1}{4}, \frac{1}{4}$
(vi) $4, 1$

Solution-

Let the quadratic polynomial be $ax^2 + bx + c$, and its zeroes be $\alpha$ and $\beta$.
We know that $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
A general form of the required polynomial is $k[x^2 - (\alpha + \beta)x + \alpha\beta]$, where $k$ is a real constant.

(i) Sum of zeroes: $\frac{1}{4}$, Product of zeroes: $-1$

$k[x^2 - (\frac{1}{4})x + (-1)]$
$k[x^2 - \frac{1}{4}x - 1]$
If we choose $k=4$ to clear the fraction, the polynomial is:
$4x^2 - x - 4$.

(ii) Sum of zeroes: $\sqrt{2}$, Product of zeroes: $\frac{1}{3}$

$k[x^2 - (\sqrt{2})x + \frac{1}{3}]$
$k[x^2 - \sqrt{2}x + \frac{1}{3}]$
If we choose $k=3$ to clear the fraction, the polynomial is:
$3x^2 - 3\sqrt{2}x + 1$.

(iii) Sum of zeroes: $0$, Product of zeroes: $\sqrt{5}$

$k[x^2 - (0)x + \sqrt{5}]$
$k[x^2 + \sqrt{5}]$
If we choose $k=1$, the polynomial is:
$x^2 + \sqrt{5}$.

(iv) Sum of zeroes: $1$, Product of zeroes: $1$

$k[x^2 - (1)x + 1]$
If we choose $k=1$, the polynomial is:
$x^2 - x + 1$.

(v) Sum of zeroes: $-\frac{1}{4}$, Product of zeroes: $\frac{1}{4}$

$k[x^2 - (-\frac{1}{4})x + \frac{1}{4}]$
$k[x^2 + \frac{1}{4}x + \frac{1}{4}]$
If we choose $k=4$ to clear the fraction, the polynomial is:
$4x^2 + x + 1$.

(vi) Sum of zeroes: $4$, Product of zeroes: $1$

$k[x^2 - (4)x + 1]$
If we choose $k=1$, the polynomial is:
$x^2 - 4x + 1$.