# CBSE 10th Mathematics | Relation Between The Zeros And The Coefficients Of A Polynomial

## Relation Between The Zeros And The Coefficients Of A Polynomial

### 1.    Quadratic polynomial: $\displaystyle a{{x}^{2}}+bx+c=0$;     a ¹ 0.

Let a, b are two zeros of the given polynomial.

Sum of zeros (a + b) = $\displaystyle -\frac{\text{coefficient}\,\text{of}\,\,x}{\text{coefficient}\,\text{of}\,\,{{x}^{2}}}=-\frac{b}{a}$

Product of zeros (ab) = $\displaystyle \frac{constant}{coefficient\,of\,\,{{x}^{2}}}=\frac{c}{a}$

### 2.    Cubic polynomial: $\displaystyle a{{x}^{3}}+b{{x}^{2}}+cx+d=0$; a ¹ 0

Let a, b and g are three zeros of the given polynomial.

(i) Sum of zeros $\displaystyle (\alpha +\beta +\gamma )=-\frac{\text{coefficient of }{{x}^{\text{2}}}}{coefficient\,of\,\,{{x}^{3}}}=-\frac{b}{a}$

(ii) Product of zeros taken two at a time

(ab + bg + ga) = $\displaystyle \frac{\text{coefficient of }x}{\text{coefficient of }{{x}^{\text{3}}}}=\frac{c}{a}$

(iii) Product of zeros (abg) = $\displaystyle -\frac{\text{constant}}{\text{coefficient of }{{x}^{\text{3}}}}=-\frac{d}{a}$

### 3.    Formation of Quadratic Polynomial:

Let a, b are the zeros, then required polynomial is

k[x2 – (sum of roots)x + (product of roots)] or k[(x–a)(x–b)]

where k is a non-zero constant

### 4.    Formation of Cubic Polynomial

Let a, b, g are the zeros then required polynomial is

$\displaystyle k[{{x}^{3}}-(\alpha +\beta +\gamma ){{x}^{2}}+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma ]$     or $\displaystyle k[(x-\alpha )(x-\beta )(x-\gamma )]$

where k is a non-zero constant

### Question:

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients.

(i) $\displaystyle {{x}^{2}}-2x-8$    (ii) $\displaystyle 6{{x}^{2}}-3$        (iii) $\displaystyle 4{{u}^{2}}+8u$

### Solution:

(i)    Given quadratic polynomial is $\displaystyle y={{x}^{2}}-2x-8={{x}^{2}}-4x+2x-8$

= $\displaystyle x(x-4)+2(x-4)=(x+2)(x-4)$

= $\displaystyle (x+2)(x-4)$

y = 0 gives x = –2, 4. These are two zeros such that a = –2, b = 4.

\    Sum of zeros (a + b) = –2 + 4 = 2 = $\displaystyle -\frac{(-2)}{1}=-\frac{b}{a}$

Product of zeros (ab) = – 2 × 4 = – 8 = $\displaystyle -\frac{8}{1}$ = $\displaystyle \frac{c}{a}$

(ii)    Given quadratic polynomial is $\displaystyle 6{{x}^{2}}-3$

y = $\displaystyle 6{{x}^{2}}-3=3(2{{x}^{2}}-1)=3[{{(\sqrt{2}x)}^{2}}-{{(1)}^{2}}]=3(\sqrt{2}x+1)(\sqrt{2}x-1)$

y = 0 gives$\displaystyle x=-\frac{1}{\sqrt{2}},\,\,\frac{1}{\sqrt{2}}$. These are two zeros such that $\displaystyle \alpha =-\frac{1}{\sqrt{2}},\,\,\beta =\frac{1}{\sqrt{2}}$

Sum of zeros (a + b) = $\displaystyle -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=0=-\frac{0}{6}=-\frac{b}{a}$

Product of zeros (ab) = $\displaystyle -\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=-\frac{1}{2}=\frac{-3}{6}$ = $\displaystyle \frac{c}{a}$

(iii)    $\displaystyle y=4{{u}^{2}}+8u=4u(u+2)$

y = 0 gives $\displaystyle 4u=0$ and $\displaystyle u+2=0$; i.e. u = 0, –2.

a =0, b = –2

Sum of zeros (a + b) = 0 – 2 = –2 = $\displaystyle -\frac{8}{4}$ = $\displaystyle -\frac{b}{a}$

Product of zeros (ab) = 0 × (–2) = $\displaystyle \frac{0}{4}=\frac{c}{a}$.

### Question:

Find a quadratic polynomial, the sum and product of whose zeros are $\displaystyle \frac{1}{2}$ and –1 respectively.

### Solution:

Here $\displaystyle \alpha +\beta =\frac{1}{2}$, ab = –1

\    Required polynomial is

$\displaystyle f(x)=k[{{x}^{2}}-(\alpha +\beta )x+(\alpha \beta )]$

=$\displaystyle k\left( {{x}^{2}}-\frac{1}{2}x-1 \right)$ where k is a non-zero constant.

### Question:

Form a cubic polynomial with zeros a = 3, b = 2, g = –1 .

### Solution:

a = 3, b = 2, g = –1

Required polynomial     = $\displaystyle k[(x-\alpha )(x-\beta )(x-\gamma )]$

= $\displaystyle k[(x-3)(x-2)(x+1)]$

= $\displaystyle k[({{x}^{2}}-5x+6)(x+1)]$

= $\displaystyle k[{{x}^{3}}-5{{x}^{2}}+6x+{{x}^{2}}-5x+6]$

= $\displaystyle k[{{x}^{3}}-4{{x}^{2}}+x+6]$

where k is a non-zero constant.

### Question:

Find a quadratic polynomial whose zeros are 2 and –3.

### Solution:

Required polynomial

$\displaystyle =k[(x-2)\,[x-(-3)]$

$\displaystyle =k[(x-2)\,(x+3)]$

$\displaystyle =k({{x}^{2}}+x-6)$

where k is a non-zero constant

### Question:

If a and b are the zeros of the polynomial $\displaystyle f(x)={{x}^{2}}-5x+k$ such that
ab = 1, find the value of k.

### Solution:

Since a and b are the zeros of the polynomial $\displaystyle f(x)={{x}^{2}}-5x+k.$

\    $\displaystyle \alpha +\beta =-\left( \frac{-5}{1} \right)=5$ and $\displaystyle \alpha \beta =\frac{k}{1}=k$

Now,    $\displaystyle \alpha -\beta =1$                    [Given]

Þ    $\displaystyle {{(\alpha -\beta )}^{2}}=1$

Þ    $\displaystyle {{(\alpha +\beta )}^{2}}-4\alpha \beta =1$    $\displaystyle [{{(\alpha -\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -4\alpha \beta ]$

Þ    25 – 4k = 1

Þ    24 = 4k

Þ    k = 6

Hence, the value of k is 6.

### Question:

Verify that the numbers given alongside of the cubic polynomials below are the zeros. Also verify the relationship between the zeros and co-efficients in each case.

$\displaystyle {{x}^{3}}-4{{x}^{2}}+5x-2$; 2, 1, 1

### Solution:

Let $\displaystyle P(x)={{x}^{3}}-4{{x}^{2}}+5x-2$

On comparing with $\displaystyle a{{x}^{3}}+b{{x}^{2}}+cx+d$

$\displaystyle a=1$, $\displaystyle b=-4$, c = 5, d = –2

Given zeros are 2, 1, 1.

$\displaystyle P(2)={{(2)}^{3}}-4{{(2)}^{2}}+5(2)-2=8-16+10-2=0$

$\displaystyle P(1)={{1}^{3}}-4{{(1)}^{2}}+5(1)-2=1-4+5-2=0$

\    2, 1, 1 are zeros of $\displaystyle P(x)$

a = 2, b = 1, g = 1

$\displaystyle \alpha +\beta +\gamma =2+1+1=4=-\frac{(-4)}{1}=-\frac{b}{a}$

$\displaystyle \alpha \beta \gamma =2\times 1\times 1=$ $\displaystyle -\frac{(-2)}{1}$ = $\displaystyle -\frac{d}{a}$

$\displaystyle \alpha \beta +\beta \gamma +\gamma \alpha =2\times 1+1\times 1+1\times 2$ = 2 + 1 +2 = $\displaystyle 5=\frac{5}{1}$ = $\displaystyle \frac{c}{a}$

Hence the result.

### Question:

Write a rational expression whose numerator is a quadratic polynomial with zeros 2 and –1 and denominator is a quadratic polynomial with zeros $\displaystyle \frac{1}{2}$ and 3.

### Solution:

Zeros of numerator are 2 and –1.

a = 2, b = –1, $\displaystyle \alpha +\beta =2-1=1$, $\displaystyle \alpha \beta =2\times (-1)=-2$

Numerator is $\displaystyle k[{{x}^{2}}-(1)x+(-2)]=k({{x}^{2}}-x-2)$

Zeros of denominator are $\displaystyle \frac{1}{2}$, 3.

$\displaystyle \alpha =\frac{1}{2}$, $\displaystyle \beta =3$, $\displaystyle \alpha +\beta =\frac{1}{2}+3=\frac{7}{2}$, $\displaystyle \alpha \beta =\frac{1}{2}\times 3=\frac{3}{2}$

Denominator is $\displaystyle k\prime \,\left[ {{x}^{2}}-\frac{7}{2}x+\frac{3}{2} \right]=\frac{k\prime }{2}\left[ 2{{x}^{2}}-7x+3 \right]$

= $\displaystyle k[2{{x}^{2}}-7x+3]$    where k = $\displaystyle \frac{k\prime }{2}$.

\    Rational expression is = $\displaystyle \frac{\text{Numerator}}{Denominator}=\frac{{{x}^{2}}-x-2}{2{{x}^{2}}-7x+3}$

### Question:

Find a quadratic polynomial whose zeros are reciprocals of the zeros of the polynomial $\displaystyle f(x)=a{{x}^{2}}+bx+c,\,\,a\ne 0,\,\,c\ne 0.$

### Solution:

Let a, b be the zeros of the polynomial $\displaystyle f(x)=a{{x}^{2}}+bx+c.$ Then,

$\displaystyle \alpha +\beta =-\frac{b}{a}$ and $\displaystyle \alpha \beta =\frac{c}{a}$

Let S and P denote respectively the sum and product of the zeros of a polynomial whose zeros are $\displaystyle \frac{1}{\alpha }$ and $\displaystyle \frac{1}{\beta }$. Then,

$\displaystyle S=\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c}$ and $\displaystyle P=\frac{1}{\alpha }\times \frac{1}{\beta }=\frac{1}{\alpha \beta }=\frac{1}{\frac{c}{a}}=\frac{a}{c}$

Hence, the required polynomial $\displaystyle g(x)$ is given by

$\displaystyle g(x)=k({{x}^{2}}-Sx+P)=k\left( {{x}^{2}}+\frac{bx}{c}+\frac{a}{c} \right),$ where k is any non-zero constant.