August 9, 2024

# Relationship between Zeros and coefficients of a Polynomial

Relationship between Zeros and coefficients of a Polynomial

zeros of a polynomial : A real number $\alpha$ is called a zero of the polynomial $p(x)$

, if $p(\alpha&space;)=0$.

If “$\alpha$” is a zero of a polynomial  $p(x)$, then by factor theorem  $(x-\alpha&space;)$ is  a factor of a given polynomial. The relation between the zeros and the coefficients of a polynomial is given below.

Linear Polynomial:

The linear polynomial is an expression , in which the degree of the polynomial is 1 . The general form of a linear polynomial is  $ax+b$. Here, $x$ is a variable, “a” and “bare constant.

Let      be a linear polynomial,

then   means    .

So     =$-\frac{constant&space;\,&space;\,&space;\,&space;term}{coefficient&space;\,&space;\,&space;of\,&space;\,&space;x}$.

Quadratic polynomial:A polynomial of degree 2 is called a quadratic polynomial. A quadratic polynomial in one variable will have at most tree terms.  Any quadratic polynomial in $x$ will be of the form  $\large&space;ax^{2}+bx+c\,&space;\,&space;where&space;\,&space;\,&space;a\neq&space;0\,&space;\,&space;and&space;\,&space;\,&space;a,b,c\,&space;\,&space;are\,&space;\,&space;constants.$

Let $\alpha$ and $\beta$ be the zeros of the quadratic polynomial $p(x)=ax^{2}+bx+c,&space;\,&space;\,&space;a\neq&space;0$.

Then $(x-\alpha&space;)$ and $(x-\beta&space;)$ are factors of $p(x)$.$\therefore&space;\,&space;\,&space;\,&space;ax^{2}+bx+c=k(x-\alpha)(x-\beta&space;)$, where $k$ is a constant.

$\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;=k\left&space;[&space;x^{2}-(\alpha&space;+\beta&space;)x+\alpha&space;\beta&space;\right&space;]$

On comparing  coefficients of like powers of $x$ on both sides, we get

$k=a,&space;\,&space;\,&space;-k(\alpha&space;+\beta&space;)=b,\,&space;\,&space;k(\alpha&space;\beta)&space;=c$

$\Rightarrow&space;-a(\alpha&space;+\beta&space;)=b&space;\,&space;\,&space;and&space;\,&space;\,&space;\,&space;a(\alpha&space;\beta&space;)=c$    $(\because&space;k=a)$

$\Rightarrow&space;\alpha&space;+\beta&space;=-\frac{b}{a}&space;\,&space;\,&space;\,&space;and\,&space;\&space;\alpha\beta&space;=\frac{c}{a}$

$\therefore$   sum of zeros =    – $\frac{(cofficient\,&space;\,of\,&space;\,&space;x)}{(coefficient&space;\,&space;\,&space;of&space;\,&space;\,&space;x^{2})},$

product of zeros = $\frac{constant\,&space;\,term&space;}{coefficient\,&space;\,of&space;\,&space;\,x^{2}}$

Cubic polynomial :   A polynomial of  degree  3 is called  cubic polynomials. Any  cubic  polynomial can have at  most 4 terms.  Cubic polynomial can be written in the form  $ax^{3}+bx^{2}+cx+d,&space;a\neq&space;0$  and $a,\,&space;\,&space;b,\,&space;\,&space;c$ and $d$ are constants.

Let $\alpha$,  $\beta$  and $\gamma$ be the zeros of the cubic  polynomial $p(x)=ax^{3}+bx^{2}+cx+d,&space;\,&space;\,&space;a\neq&space;0$.

Then $(x-\alpha&space;)$$(x-\beta&space;)$  and $(x-\gamma&space;)$ are factors of $p(x)$.

$\therefore&space;\,&space;\,&space;\,&space;ax^{3}+bx^{2}+cx+d=k(x-\alpha)(x-\beta&space;)(x-\gamma&space;)$, for some constant $k$.

$\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;=k\left&space;[&space;x^{3}-(\alpha&space;+\beta+\gamma&space;)x^{2}+(\alpha&space;\beta&space;+\beta&space;\gamma&space;+\alpha&space;\gamma&space;)x-\alpha&space;\beta&space;\gamma&space;\right&space;]$

=$kx^{3}-k(\alpha&space;+\beta&space;+\gamma&space;)x^{2}+k(\alpha&space;\beta&space;+\beta&space;\gamma&space;+\gamma&space;\alpha&space;)x-k\alpha&space;\beta&space;\gamma$

On comparing  coefficients of like powers of $x$ on both sides, we get

$k=a,&space;\,&space;\,&space;-k(\alpha&space;+\beta+\gamma&space;)=b,\,&space;\,&space;k(\alpha&space;\beta+\gamma&space;+\gamma&space;\alpha&space;)&space;=c,\,&space;\,&space;-k(\alpha&space;\beta&space;\gamma&space;)=d$

$\Rightarrow&space;-a(\alpha&space;+\beta+\gamma&space;)=b,&space;\,&space;\,&space;a(\alpha&space;\beta+\beta&space;\gamma&space;+\gamma&space;\alpha&space;)=c,\,&space;\,&space;-a(\alpha&space;\beta&space;\gamma&space;)&space;=d$   $(\because&space;k=a)$

$\Rightarrow&space;\alpha&space;+\beta&space;+\gamma&space;=-\frac{b}{a},&space;\,&space;\,&space;\alpha\beta+\beta&space;\gamma+\gamma&space;\alpha&space;=\frac{c}{a},&space;\,&space;\,\alpha&space;\beta&space;\gamma&space;=-\frac{d}{a}$

$\therefore$ If $\alpha$,  $\beta$  and $\gamma$ be the zeros of the cubic  polynomial $p(x)=ax^{3}+bx^{2}+cx+d,&space;\,&space;\,&space;a\neq&space;0$, then

(i) $\alpha&space;+\beta&space;+\gamma&space;=\frac{b}{a}$          (ii) $\alpha&space;\beta&space;+\beta&space;\gamma&space;+\gamma&space;\alpha&space;=\frac{c}{a}$         (iii) $\alpha&space;\beta&space;\gamma&space;=-\frac{d}{a}$

Similarly, If α , β, γ, δ are roots of the  equation $ax^{4}+&space;bx^{3}&space;+&space;cx^{2}&space;+&space;dx&space;+e=0,&space;a\neq&space;0$, then

$\alpha&space;+\beta&space;+\gamma&space;+\delta&space;=-\frac{b}{a}$

$\alpha&space;\beta&space;+&space;\beta&space;\gamma+\gamma&space;\delta&space;+\delta&space;\alpha&space;+\delta&space;\beta&space;+\gamma&space;\alpha&space;=\frac{c}{a}$

$\alpha&space;\beta&space;\gamma&space;+\alpha&space;\gamma&space;\delta&space;+\alpha&space;\beta&space;\delta&space;+\beta&space;\gamma&space;\delta=-\frac{d}{a}$

$\alpha&space;\beta&space;\gamma&space;\delta&space;=\frac{e}{a}$.

Some practice questions based on polynomial are given in the following worksheet.

#### Bina singh

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