## Chapter 02 Polynomials

### 2.1 Introduction

In Class IX, you have studied polynomials in one variable and their degrees. Recall that if $p(x)$ is a polynomial in $x$, the highest power of $x$ in $p(x)$ is called the degree of the polynomial $p(x)$. For example, $4 x+2$ is a polynomial in the variable $x$ of degree $1,2 y^{2}-3 y+4$ is a polynomial in the variable $y$ of degree $2,5 x^{3}-4 x^{2}+x-\sqrt{2}$

is a polynomial in the variable $x$ of degree 3 and $7 u^{6}-\dfrac{3}{2} u^{4}+4 u^{2}+u-8$ is a polynomial in the variable $u$ of degree 6 . Expressions like $\dfrac{1}{x-1}, \sqrt{x}+2, \dfrac{1}{x^{2}+2 x+3}$ etc., are not polynomials.

A polynomial of degree 1 is called a linear polynomial. For example, $2 x-3$, $\sqrt{3} x+5, y+\sqrt{2}, x-\dfrac{2}{11}, 3 z+4, \dfrac{2}{3} u+1$, etc., are all linear polynomials. Polynomials such as $2 x+5-x^{2}, x^{3}+1$, etc., are not linear polynomials.

A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’. $2 x^{2}+3 x-\dfrac{2}{5}$, $y^{2}-2,2-x^{2}+\sqrt{3} x, \dfrac{u}{3}-2 u^{2}+5, \sqrt{5} v^{2}-\dfrac{2}{3} v, 4 z^{2}+\dfrac{1}{7}$ are some examples of quadratic polynomials (whose coefficients are real numbers). More generally, any quadratic polynomial in $x$ is of the form $a x^{2}+b x+c$, where $a, b, c$ are real numbers and $a \neq 0$. A polynomial of degree 3 is called a cubic polynomial. Some examples of a cubic polynomial are $2-x^{3}, x^{3}, \sqrt{2} x^{3}, 3-x^{2}+x^{3}, 3 x^{3}-2 x^{2}+x-1$. In fact, the most general form of a cubic polynomial is

$ a x^{3}+b x^{2}+c x+d, $

where, $a, b, c, d$ are real numbers and $a \neq 0$.

Now consider the polynomial $p(x)=x^{2}-3 x-4$. Then, putting $x=2$ in the polynomial, we get $p(2)=2^{2}-3 \times 2-4=-6$. The value ’ -6 ‘, obtained by replacing $x$ by 2 in $x^{2}-3 x-4$, is the value of $x^{2}-3 x-4$ at $x=2$. Similarly, $p(0)$ is the value of $p(x)$ at $x=0$, which is -4 .

If $p(x)$ is a polynomial in $x$, and if $k$ is any real number, then the value obtained by replacing $x$ by $k$ in $p(x)$, is called the value of $\boldsymbol{{}p}(\boldsymbol{{}x})$ at $\boldsymbol{{}x}=\boldsymbol{{}k}$, and is denoted by $p(k)$.

What is the value of $p(x)=x^{2}-3 x-4$ at $x=-1$ ? We have :

$ p(-1)=(-1)^{2}-{3 \times(-1)}-4=0 $

Also, note that $p(4)=4^{2}-(3 \times 4)-4=0$.

As $p(-1)=0$ and $p(4)=0,-1$ and 4 are called the zeroes of the quadratic polynomial $x^{2}-3 x-4$. More generally, a real number $k$ is said to be a zero of a polynomial $\boldsymbol{{}p}(\boldsymbol{{}x})$, if $p(k)=0$.

You have already studied in Class IX, how to find the zeroes of a linear polynomial. For example, if $k$ is a zero of $p(x)=2 x+3$, then $p(k)=0$ gives us $2 k+3=0$, i.e., $k=-\dfrac{3}{2}$.

In general, if $k$ is a zero of $p(x)=a x+b$, then $p(k)=a k+b=0$, i.e., $k=\dfrac{-b}{a}$. So, the zero of the linear polynomial $a x+b$ is $\dfrac{-b}{a}=\dfrac{-(\text{ Constant term })}{\text{ Coefficient of } x}$.

Thus, the zero of a linear polynomial is related to its coefficients. Does this happen in the case of other polynomials too? For example, are the zeroes of a quadratic polynomial also related to its coefficients?

In this chapter, we will try to answer these questions. We will also study the division algorithm for polynomials.

### 2.2 Geometrical Meaning of the Zeroes of a Polynomial

You know that a real number $k$ is a zero of the polynomial $p(x)$ if $p(k)=0$. But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.

Consider first a linear polynomial $a x+b, a \neq 0$. You have studied in Class IX that the graph of $y=a x+b$ is a straight line. For example, the graph of $y=2 x+3$ is a straight line passing through the points $(-2,-1)$ and $(2,7)$.

$ \begin{array}{|l|c|c|} \hline x & -2 & 2 \\ \hline y=2 x+3 & -1 & 7 \\ \hline \end{array} $

From Fig. 2.1, you can see that the graph of $y=2 x+3$ intersects the $x$-axis mid-way between $x=-1$ and $x=-2$, that is, at the point $(-\dfrac{3}{2}, 0)$. You also know that the zero of $2 x+3$ is $-\dfrac{3}{2}$. Thus, the zero of the polynomial $2 x+3$ is the $x$-coordinate of the point where the graph of $y=2 x+3$ intersects the $x$-axis.

Fig. 2.1

In general, for a linear polynomial $a x+b, a \neq 0$, the graph of $y=a x+b$ is a straight line which intersects the $x$-axis at exactly one point, namely, $(\dfrac{-b}{a}, 0)$. Therefore, the linear polynomial $a x+b, a \neq 0$, has exactly one zero, namely, the $x$-coordinate of the point where the graph of $y=a x+b$ intersects the $x$-axis.

Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. Consider the quadratic polynomial $x^{2}-3 x-4$. Let us see what the graph[^0]of $y=x^{2}-3 x-4$ looks like. Let us list a few values of $y=x^{2}-3 x-4$ corresponding to a few values for $x$ as given in Table 2.1.

Table 2.1

$x$ | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|---|---|---|

$y=x^{2}-3 x-4$ | 6 | 0 | -4 | -6 | -6 | -4 | 0 | 6 |

If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.

In fact, for any quadratic polynomial $a x^{2}+b x+c, a \neq 0$, the graph of the corresponding equation $y=a x^{2}+b x+c$ has one of the two shapes either open upwards like $\bigcup$ or open downwards like $\bigcap$ depending on whether $a>0$ or $a<0$. (These curves are called parabolas.)

You can see from Table 2.1 that -1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that -1 and 4 are the $x$-coordinates of the points where the graph of $y=x^{2}-3 x-4$ intersects the $x$-axis. Thus, the zeroes of the quadratic polynomial $x^{2}-3 x-4$ are $x$-coordinates of the points where the graph of $y=x^{2}-3 x-4$ intersects the $x$-axis.

Fig. 2.2

This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial $a x^{2}+b x+c, a \neq 0$, are precisely the $x$-coordinates of the points where the parabola representing $y=a x^{2}+b x+c$ intersects the $x$-axis.

From our observation earlier about the shape of the graph of $y=a x^{2}+b x+c$, the following three cases can happen:

Case (i) : Here, the graph cuts $x$-axis at two distinct points A and $A^{\prime}$.

The $x$-coordinates of $A$ and $A^{\prime}$ are the two zeroes of the quadratic polynomial $a x^{2}+b x+c$ in this case (see Fig. 2.3).

Fig. 2.3

Case (ii) : Here, the graph cuts the $x$-axis at exactly one point, i.e., at two coincident points. So, the two points A and $A^{\prime}$ of Case (i) coincide here to become one point A (see Fig. 2.4).

Fig. 2.4

The $x$-coordinate of A is the only zero for the quadratic polynomial $a x^{2}+b x+c$ in this case.

Case (iii) : Here, the graph is either completely above the $x$-axis or completely below the $x$-axis. So, it does not cut the $x$-axis at any point (see Fig. 2.5).

Fig. 2.5

So, the quadratic polynomial $a x^{2}+b x+c$ has no zero in this case.

So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree 2 has at most two zeroes.

Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be? Let us find out. Consider the cubic polynomial $x^{3}-4 x$. To see what the graph of $y=x^{3}-4 x$ looks like, let us list a few values of $y$ corresponding to a few values for $x$ as shown in Table 2.2.

Table 2.2

$x$ | -2 | -1 | 0 | 1 | 2 |
---|---|---|---|---|---|

$y=x^{3}-4 x$ | 0 | 3 | 0 | -3 | 0 |

Locating the points of the table on a graph paper and drawing the graph, we see that the graph of $y=x^{3}-4 x$ actually looks like the one given in Fig. 2.6.

We see from the table above that $-2,0$ and 2 are zeroes of the cubic polynomial $x^{3}-4 x$. Observe that $-2,0$ and 2 are, in fact, the $x$-coordinates of the only points where the graph of $y=x^{3}-4 x$ intersects the $x$-axis. Since the curve meets the $x$-axis in only these 3 points, their $x$-coordinates are the only zeroes of the polynomial.

Let us take a few more examples. Consider the cubic polynomials $x^{3}$ and $x^{3}-x^{2}$. We draw the graphs of $y=x^{3}$ and $y=x^{3}-x^{2}$ in Fig. 2.7 and Fig. 2.8 respectively.

Fig. 2.6

Fig. 2.7

Fig. 2.8

Note that 0 is the only zero of the polynomial $x^{3}$. Also, from Fig. 2.7, you can see that 0 is the $x$-coordinate of the only point where the graph of $y=x^{3}$ intersects the $x$-axis. Similarly, since $x^{3}-x^{2}=x^{2}(x-1), 0$ and 1 are the only zeroes of the polynomial $x^{3}-x^{2}$. Also, from Fig. 2.8, these values are the $x$-coordinates of the only points where the graph of $y=x^{3}-x^{2}$ intersects the $x$-axis.

From the examples above, we see that there are at most 3 zeroes for any cubic polynomial. In other words, any polynomial of degree 3 can have at most three zeroes.

**Remark :** In general, given a polynomial $p(x)$ of degree $n$, the graph of $y=p(x)$ intersects the $x$-axis at atmost $n$ points. Therefore, a polynomial $p(x)$ of degree $n$ has at most $n$ zeroes.

**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 $p(x)$.

Fig. 2.9

**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?)

### EXERCISE 2.1

**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.

Fig. 2.10

## Show Answer

**Solution**

(i) The number of zeroes is 0 as the graph does not cut the $x$-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the $x$-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the $x$-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the $x$-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the $x$-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the $x$-axis at 3 points.

### 2.3 Relationship between Zeroes and Coefficients of a Polynomial

You have already seen that zero of a linear polynomial $a x+b$ is $-\dfrac{b}{a}$. We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, say $p(x)=2 x^{2}-8 x+6$. In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term. So, here we need to split the middle term ’ $-8 x^{\text{’ }}$ as a sum of two terms, whose product is $6 \times 2 x^{2}=12 x^{2}$. So, we write

$ \begin{aligned} 2 x^{2}-8 x+6 & =2 x^{2}-6 x-2 x+6=2 x(x-3)-2(x-3) \\ & =(2 x-2)(x-3)=2(x-1)(x-3) \end{aligned} $

So, the value of $p(x)=2 x^{2}-8 x+6$ is zero when $x-1=0$ or $x-3=0$, i.e., when $x=1$ or $x=3$. So, the zeroes of $2 x^{2}-8 x+6$ are 1 and 3 . Observe that :

$ \begin{aligned} & \text{ Sum of its zeroes }=1+3=4=\dfrac{-(-8)}{2}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}} \\ & \text{ Product of its zeroes }=1 \times 3=3=\dfrac{6}{2}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \end{aligned} $

Let us take one more quadratic polynomial, say, $p(x)=3 x^{2}+5 x-2$. By the method of splitting the middle term,

$ \begin{aligned} 3 x^{2}+5 x-2 & =3 x^{2}+6 x-x-2=3 x(x+2)-1(x+2) \\ & =(3 x-1)(x+2) \end{aligned} $

Hence, the value of $3 x^{2}+5 x-2$ is zero when either $3 x-1=0$ or $x+2=0$, i.e., when $x=\dfrac{1}{3}$ or $x=-2$. So, the zeroes of $3 x^{2}+5 x-2$ are $\dfrac{1}{3}$ and -2 . Observe that :

$ \begin{aligned} & \text{ Sum of its zeroes }=\dfrac{1}{3}+(-2)=\dfrac{-5}{3}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}} \\ & \text{ Product of its zeroes }=\dfrac{1}{3} \times(-2)=\dfrac{-2}{3}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \end{aligned} $

In general, if $\alpha$[^1] and $\beta$[^1] are the zeroes of the quadratic polynomial $p(x)=a x^{2}+b x+c$, $a \neq 0$, then you know that $x-\alpha$ and $x-\beta$ are the factors of $p(x)$. Therefore,

$ \begin{aligned} a x^{2}+b x+c & =k(x-\alpha)(x-\beta), \text{ where } k \text{ is a constant } \\ & =k[x^{2}-(\alpha+\beta) x+\alpha \beta] \\ & =k x^{2}-k(\alpha+\beta) x+k \alpha \beta \end{aligned} $

\missing

Comparing the coefficients of $x^{2}, x$ and constant terms on both the sides, we get

$ \begin{aligned} a=k, b & =-k(\alpha+\beta) \text{ and } c=k \alpha \beta . \\ \text {This gives}\qquad \boldsymbol{{}\alpha}+\boldsymbol{{}\beta} & =\dfrac{-\boldsymbol{{}b}}{\boldsymbol{{}a}}, \\ \boldsymbol{{}\alpha} \boldsymbol{{}\beta} & =\dfrac{\boldsymbol{{}c}}{\boldsymbol{{}a}} \end{aligned} $

$ \begin{aligned} & \text{ i.e., } \\ & \text{ sum of zeroes }=\alpha+\beta=-\dfrac{b}{a}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}, \\ & \text{ product of zeroes }=\alpha \beta=\dfrac{c}{a}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} . \end{aligned} $

Let us consider some examples.

**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)=\dfrac{-(7)}{1}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}, \\ \text{ product of zeroes } & =(-2) \times(-5)=10=\dfrac{10}{1}=\dfrac{\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=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}, \\ \text{ product of zeroes } & =(\sqrt{3})(-\sqrt{3})=-3=\dfrac{-3}{1}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}} \text{. } \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=\dfrac{-b}{a} \text{, } $

$ \text{and}\qquad \alpha \beta=2=\dfrac{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 $k(x^{2}+3 x+2)$, 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 $x=4,-2, \dfrac{1}{2}$. Since $p(x)$ can have atmost three zeroes, these are the zeores of $2 x^{3}-5 x^{2}-14 x+8$. Now,

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

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 \dfrac{1}{2}\right\}+\left\{\dfrac{1}{2} \times 4\right\} \\ &=-8-1+2=-7=\dfrac{-14}{2}=\dfrac{ \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 & =\dfrac{-b}{a}, \\ \alpha \beta+\beta \gamma+\gamma \alpha & =\dfrac{c}{a}, \\ \alpha \beta \gamma & =\dfrac{-d}{a} . \end{aligned} $

Let us consider an example.

**Example 5** : Verify that $3,-1,-\dfrac{1}{3}$ 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}-(5 \times 3^{2})-(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(-\dfrac{1}{3})=3 \times(-\dfrac{1}{3})^{3}-5 \times(-\dfrac{1}{3})^{2}-11 \times(-\dfrac{1}{3})-3, \\ & \quad=-\dfrac{1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3=-\dfrac{2}{3}+\dfrac{2}{3}=0 \end{aligned} $

Therefore, $3,-1$ and $-\dfrac{1}{3}$ are the zeroes of $3 x^{3}-5 x^{2}-11 x-3$.

So, we take $\alpha=3, \beta=-1$ and $\gamma=-\dfrac{1}{3}$.

Now,

$ \begin{aligned} & \alpha+\beta+\gamma=3+(-1)+(-\dfrac{1}{3})=2-\dfrac{1}{3}=\dfrac{5}{3}=\dfrac{-(-5)}{3}=\dfrac{-b}{a}, \\ & \alpha \beta+\beta \gamma+\gamma \alpha=3 \times(-1)+(-1) \times(-\dfrac{1}{3})+(-\dfrac{1}{3}) \times 3=-3+\dfrac{1}{3}-1=\dfrac{-11}{3}=\dfrac{c}{a}, \\ & \alpha \beta \gamma=3 \times(-1) \times(-\dfrac{1}{3})=1=\dfrac{-(-3)}{3}=\dfrac{-d}{a} . \end{aligned} $

### EXERCISE 2.2

**1.** Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^{2}-2 x-8$

(ii) $4 s^{2}-4 s+1$

(iii) $6 x^{2}-3-7 x$

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

(v) $t^{2}-15$

(vi) $3 x^{2}-x-4$

## Show Answer

**Solution**

$$ \begin{equation*} x^{2}-2 x-8=(x-4)(x+2) \tag{i} \end{equation*} $$

The value of $x^{2}-2 x-8$ is zero when $x-4=0$ or $x+2=0$, i.e., when $x=4$ or $x=-2$

Therefore, the zeroes of $x^{2}-2 x-8$ are 4 and -2 .

Sum of zeroes $=4-2=2=\dfrac{-(-2)}{1}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}$

Product of zeroes $=4 \times(-2)=-8=\dfrac{(-8)}{1}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}}$

(ii) $4 s^{2}-4 s+1=(2 s-1)^{2}$

The value of $4 s^{2}-4 s+1$ is zero when $2 s-1=0$, i.e., $\quad s=\dfrac{1}{2}$

Therefore, the zeroes of $4 s^{2}-4 s+1$ are $\dfrac{1}{2}$ and $\dfrac{1}{2}$.

Sum of zeroes $=\dfrac{1}{2}+\dfrac{1}{2}=1=\dfrac{-(-4)}{4}=\dfrac{-(\text{ Coefficient of } s)}{(\text{ Coefficient of } s^{2})}$

Product of zeroes $=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } s^{2}}$

$$ \begin{equation*} 6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3) \tag{iii} \end{equation*} $$

The value of $6 x^{2}-3-7 x$ is zero when $3 x+1=0$ or $2 x-3=0$, i.e., $x=\dfrac{-1}{3}$ or $x=\dfrac{3}{2}$

Therefore, the zeroes of $6 x^{2}-3-7 x$ are $\dfrac{-1}{3}$ and $\dfrac{3}{2}$.

Sum of zeroes $=\dfrac{-1}{3}+\dfrac{3}{2}=\dfrac{7}{6}=\dfrac{-(-7)}{6}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}$

Product of zeroes $=\dfrac{-1}{3} \times \dfrac{3}{2}=\dfrac{-1}{2}=\dfrac{-3}{6}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}}$

(iv) $4 u^{2}+8 u=4 u^{2}+8 u+0$

$=4 u(u+2)$

The value of $4 u^{2}+8 u$ is zero when $4 u=0$ or $u+2=0$, i.e., $u=0$ or $u=-2$

Therefore, the zeroes of $4 u^{2}+8 u$ are 0 and -2 .

Sum of zeroes $=0+(-2)=-2=\dfrac{-(8)}{4}=\dfrac{-(\text{ Coefficient of } u)}{\text{ Coefficient of } u^{2}}$

Product of zeroes $=0 \times(-2)=0=\dfrac{0}{4}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } u^{2}}$

(v) $t^{2}-15$

$ \begin{aligned} & =t^{2}-0 t-15 \\ & =(t-\sqrt{15})(t+\sqrt{15}) \end{aligned} $

The value of $t^{2}-15$ is zero when $t-\sqrt{15}=0$ or $t+\sqrt{15}=0$, i.e., when

**2.** Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) $\dfrac{1}{4},-1$

(ii) $\sqrt{2}, \dfrac{1}{3}$

(iii) $0, \sqrt{5}$

(iv) 1,1

(v) $-\dfrac{1}{4}, \dfrac{1}{4}$

(vi) 4,1

## Show Answer

**Solution**

(i)

$ \dfrac{1}{4},-1 $

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$. $\alpha+\beta=\dfrac{1}{4}=\dfrac{-b}{a}$

$\alpha \beta=-1=\dfrac{-4}{4}=\dfrac{c}{a}$

If $a=4$, then $b=-1, c=-4$

Therefore, the quadratic polynomial is $4 x^{2}-x$ - 4 .

(ii) $\sqrt{2}, \dfrac{1}{3}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=\sqrt{2}=\dfrac{3 \sqrt{2}}{3}=\dfrac{-b}{a}$

$\alpha \beta=\dfrac{1}{3}=\dfrac{c}{a}$

If $a=3$, then $b=-3 \sqrt{2}, c=1$

Therefore, the quadratic polynomial is $3 x^{2}-3 \sqrt{2} x+1$.

(iii) $0, \sqrt{5}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=0=\dfrac{0}{1}=\dfrac{-b}{a}$

$\alpha \times \beta=\sqrt{5}=\dfrac{\sqrt{5}}{1}=\dfrac{c}{a}$

If $a=1$, then $b=0, c=\sqrt{5}$

Therefore, the quadratic polynomial is $x^{2}+\sqrt{5}$.

(iv) 1,1

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=1=\dfrac{1}{1}=\dfrac{-b}{a}$

$\alpha \times \beta=1=\dfrac{1}{1}=\dfrac{c}{a}$

If $a=1$, then $b=-1, c=1$

Therefore, the quadratic polynomial is $x^{2}-x+1$.

(v) $-\dfrac{1}{4}, \dfrac{1}{4}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and

### 2.4 Summary

In this chapter, you have studied the following points:

**1.** Polynomials of degrees 1,2 and 3 are called linear, quadratic and cubic polynomials respectively.

**2.** A quadratic polynomial in $x$ with real coefficients is of the form $a x^{2}+b x+c$, where $a, b, c$ are real numbers with $a \neq 0$.

**3.** 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.

**4.** A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.

**5.** If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $a x^{2}+b x+c$, then

$$ \alpha+\beta=-\dfrac{b}{a}, \quad \alpha \beta=\dfrac{c}{a} . $$

**6.** 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=\dfrac{-b}{a}, \\ & \alpha \beta+\beta \gamma+\gamma \alpha=\dfrac{c}{a}, \\ & \text{ and } \quad \alpha \beta \gamma=\dfrac{-d}{a} . \end{aligned} $