Examples - Chapter 2 Polynomials class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples  (Revised) - Chapter 2 - Polynomials - Ncert Solutions class 10 Maths

Example 1 :

Look at the graphs in Fig. 2.9 given below. Each is the graph of $y=p(x)$, where $p(x)$ is a polynomial. For each of the graphs, find the number of zeroes of $p(x)$.

Solution :
(i) The number of zeroes is 1 as the graph intersects the $x$-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the $x$-axis at two points.
(iii) The number of zeroes is 3 . (Why?)

(iv) The number of zeroes is 1 . (Why?)
(v) The number of zeroes is 1 . (Why?)
(vi) The number of zeroes is 4 . (Why?)

Example 2 :

Find the zeroes of the quadratic polynomial $x^2+7 x+10$, and verify the relationship between the zeroes and the coefficients.
Solution :

We have
$
x^2+7 x+10=(x+2)(x+5)
$

So, the value of $x^2+7 x+10$ is zero when $x+2=0$ or $x+5=0$, i.e., when $x=-2$ or $x=-5$. Therefore, the zeroes of $x^2+7 x+10$ are -2 and -5 . Now,
$
\begin{aligned}
& \text { sum of zeroes }=-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^2}, \\
& \text { product of zeroes }=(-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2} .
\end{aligned}
$

Example 3 :

Find the zeroes of the polynomial $x^2-3$ and verify the relationship between the zeroes and the coefficients.
Solution :

Recall the identity $a^2-b^2=(a-b)(a+b)$. Using it, we can write:
$
x^2-3=(x-\sqrt{3})(x+\sqrt{3})
$

So, the value of $x^2-3$ is zero when $x=\sqrt{3}$ or $x=-\sqrt{3}$.
Therefore, the zeroes of $x^2-3$ are $\sqrt{3}$ and $-\sqrt{3}$.
Now,
$
\begin{aligned}
& \text { sum of zeroes }=\sqrt{3}-\sqrt{3}=0=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^2}, \\
& \text { product of zeroes }=(\sqrt{3})(-\sqrt{3})=-3=\frac{-3}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2} .
\end{aligned}
$

Example 4 :

Find a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 , respectively.
Solution :

Let the quadratic polynomial be $a x^2+b x+c$, and its zeroes be $\alpha$ and $\beta$. We have
$
\alpha+\beta=-3=\frac{-b}{a},
$
and
$
\alpha \beta=2=\frac{c}{a} .
$

If $a=1$, then $b=3$ and $c=2$.
So, one quadratic polynomial which fits the given conditions is $x^2+3 x+2$.
You can check that any other quadratic polynomial that fits these conditions will be of the form $k\left(x^2+3 x+2\right)$, where $k$ is real.

Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients?
Let us consider $p(x)=2 x^3-5 x^2-14 x+8$.
You can check that $p(x)=0$ for $x=4,-2, \frac{1}{2}$. Since $p(x)$ can have atmost three zeroes, these are the zeores of $2 x^3-5 x^2-14 x+8$. Now,

$
\begin{aligned}
& \text { sum of the zeroes }=4+(-2)+\frac{1}{2}=\frac{5}{2}=\frac{-(-5)}{2}=\frac{-\left(\text { Coefficient of } x^2\right)}{\text { Coefficient of } x^3}, \\
& \text { product of the zeroes }=4 \times(-2) \times \frac{1}{2}=-4=\frac{-8}{2}=\frac{- \text { Constant term }}{\text { Coefficient of } x^3} .
\end{aligned}
$

However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have
$
\begin{aligned}
\{4 \times(-2)\}+\left\{(-2) \times \frac{1}{2}\right\} & +\left\{\frac{1}{2} \times 4\right\} \\
& =-8-1+2=-7=\frac{-14}{2}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^3} .
\end{aligned}
$

In general, it can be proved that if $\alpha, \beta, \gamma$ are the zeroes of the cubic polynomial $a x^3+b x^2+c x+d$, then

$
\begin{aligned}
\alpha+\beta+\gamma & =\frac{-b}{a}, \\
\alpha \beta+\beta \gamma+\gamma \alpha & =\frac{c}{a}, \\
\alpha \beta \gamma & =\frac{-d}{a} .
\end{aligned}
$

Let us consider an example.

Example 5 :

Verify that $3,-1,-\frac{1}{3}$ are the zeroes of the cubic polynomial $p(x)=3 x^3-5 x^2-11 x-3$, and then verify the relationship between the zeroes and the coefficients.
Solution :

Comparing the given polynomial with $a x^3+b x^2+c x+d$, we get
$
\begin{aligned}
& a=3, b=-5, c=-11, d=-3 . \text { Further } \\
& p(3)=3 \times 3^3-\left(5 \times 3^2\right)-(11 \times 3)-3=81-45-33-3=0, \\
& p(-1)=3 \times(-1)^3-5 \times(-1)^2-11 \times(-1)-3=-3-5+11-3=0, \\
& p\left(-\frac{1}{3}\right)=3 \times\left(-\frac{1}{3}\right)^3-5 \times\left(-\frac{1}{3}\right)^2-11 \times\left(-\frac{1}{3}\right)-3, \\
& \quad=-\frac{1}{9}-\frac{5}{9}+\frac{11}{3}-3=-\frac{2}{3}+\frac{2}{3}=0
\end{aligned}
$

Therefore, $3,-1$ and $-\frac{1}{3}$ are the zeroes of $3 x^3-5 x^2-11 x-3$.
So, we take $\alpha=3, \beta=-1$ and $\gamma=-\frac{1}{3}$.
Now,
$
\begin{aligned}
& \alpha+\beta+\gamma=3+(-1)+\left(-\frac{1}{3}\right)=2-\frac{1}{3}=\frac{5}{3}=\frac{-(-5)}{3}=\frac{-b}{a} \\
& \alpha \beta+\beta \gamma+\gamma \alpha=3 \times(-1)+(-1) \times\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}\right) \times 3=-3+\frac{1}{3}-1=\frac{-11}{3}=\frac{c}{a} \\
& \alpha \beta \gamma=3 \times(-1) \times\left(-\frac{1}{3}\right)=1=\frac{-(-3)}{3}=\frac{-d}{a}
\end{aligned}
$