Relationship of Zeros and Coefficients of Polynomials

Chapter Contents

Introduction
Sum and product of zeros of quadratic polynomial
Sum and product of zeros of cubic polynomial
Sum and product of zeros of biquadratic polynomial

Chapter
Contents

Introduction

The detailed discussion about zeros of polynomials and how to represent a polynomial on a graph can be found in the chapter Zeros of Polynomial and its Geometrical Meaning. Also, we learnt how to find the value of zeros of polynomial graphically for a linear polynomial, a cubic polynomial and a biquadratic polynomial.

This chapter is about the relationships between zeros and coefficients of a quadratic polynomial, zeros and coefficients of cubic polynomial along with zeros and coefficients of a biquadratic polynomial. The zeros and coefficients of a polynomial is related to each other in finding sum and product of roots of the polynomial.

Sum and product of zeros of quadratic polynomial

Consider a quadratic polynomial ax² + bx + c. Let α and β are two zeros of the polynomial. Then, we can find the sum of zeros and product of zeros from coefficients of x and x² with the following formulas for sum of zeros and product of zeros.

Formula

$Sum of zeros = - \frac{coefficient of x}{{coefficient of x}^{2}}$
$α + β = \frac{- b}{a}$
$Product of zeros = \frac{constant of term}{{constant of term}^{2}}$
$α β = \frac{c}{a}$

Example

Consider a polynomial x² + 5x + 6
Zeros of x² + 5x + 6 are -2 and -3
Here, coefficient of x² = 1
coefficient of x = 5
constant term = 6
Therefore, a = 1, b = 5, c = 6
Here, α = -2 and β = -3
$Sum of zeros = - \frac{coefficient of x}{{coefficient of x}^{2}}$
$α + β = \frac{- b}{a}$
$(-2) + (-3) = \frac{- 5}{1}$
-5 = -5
$Product of zeros = \frac{constant of term}{{constant of term}^{2}}$
$α β = \frac{c}{a}$
$(-2)(-3) = \frac{6}{1}$
6 = 6

Sum and product of zeros of cubic polynomial

Now, consider a cubic polynomial p(x)=ax³ + bx² + cx + d. Let α, β and γ are three zeros of polynomial. Sum of zeros and product of zeros can be found from the coefficients of x² and x³ with the following formulas for sum of zeros and product of zeros.

Formula

$Sum of zeros = - \frac{{coefficient of x}^{2}}{{coefficient of x}^{3}}$
$α + β + γ = \frac{- b}{a}$
$Sum of product of zeros taken two at a time = \frac{coefficient of x}{{coefficient of x}^{9}}$
$α β+β γ+α γ = \frac{c}{a}$
$Product of zeros = - \frac{constant of term}{{coefficient of x}^{3}}$
$α β γ = - \frac{d}{a}$

Example

Consider a cubic polynomial x³ + 2x² – 5x – 6
Zeros of x³ + 2x² – 5x – 6 are -1, 2 and -3
Here, coefficient of x³ = 1
coefficient of x² = 2
coefficient of x = -5
constant term = -6
Therefore, a = 1, b = 2, c = -5, d = -6
Here, α = -1, β = 2 and γ=-3
$Sum of zeros = - \frac{{coefficient of x}^{2}}{{coefficient of x}^{3}}$
$α + β + γ = \frac{- b}{a}$
$(-1) + (2) + (-3) = - \frac{2}{1}$
-2 = -2
$Sum of product of zeros taken two at a time = \frac{coefficient of x}{{coefficient of x}^{9}}$
$α β+β γ+α γ = \frac{c}{a}$
$(-1)(2) + (2)(-3) + (-3)(-1) = \frac{(-5)}{1}$
-2 -6 + 3 = -5
-5 = -5
$Product of zeros = - \frac{constant of term}{{coefficient of x}^{3}}$
$α β γ = - \frac{d}{a}$
$(-1)(2)(-3) = - \frac{(-6)}{1}$
6 = 6

Sum and product of zeros of biquadratic polynomial

Polynomial p(x)=ax⁴ + bx³ + cx² + dx + e, where a ≠ = 0 is a biquadratic polynomial. The graph of y = ax⁴ + bx³ + cx² + dx + e intersects the x-axis. These coordinates are the only zeros of the biquadratic polynomial.
Consider a polynomial ax⁴ + bx³ + cx² + dx + e. Let α, β, γ and δ are four zeros of the polynomial.
Sum of zeros and product of zeros can be found from the coefficients of x³ and x⁴ with the following formulas for sum of zeros and product of zeros.

Formula

$Sum of zeros = - \frac{{coefficient of x}^{3}}{{coefficient of x}^{4}}$
$α + β + γ + δ = - \frac{b}{a}$
$Sum of product of zeros taken two at a time = \frac{{coefficient of x}^{2}}{{coefficient of x}^{4}}$
$α β + β γ + δ γ + α δ + δ β + γα = \frac{c}{a}$
$Sum of product of zeros taken three at a time = - \frac{coefficient of x}{{coefficient of x}^{4}}$
$αβγ + βγδ + αβδ + αγδ = - \frac{d}{a}$
$Product of zeros = \frac{constant of term}{{coefficient of x}^{3}}$
$α β γ δ = \frac{e}{a}$

Example

Let x⁴ – 6x³ – 4x² + 54x – 45 be a biquadratic polynomial
Zeros of x⁴ – 6x³ – 4x² + 54x – 45 are 1, 3, 5 and -3
Here, coefficient of x⁴ = 1
coefficient of x³ = -6
coefficient of x² = -4
coefficient of x = 54
constant term = -45
∴ a = 1, b = -6, c = -4, d = 54, e = -45
Let zeros be, α = 1, β = 3, γ=5 and δ=-3
∴ $Sum of zeros = - \frac{{coefficient of x}^{3}}{{coefficient of x}^{4}}$
$α + β + γ + δ = \frac{- b}{a}$
$(1) + (3) + (5) + (-3) = - \frac{(-6)}{1}$
6 = 6
$Sum of product of zeros taken two at a time = \frac{{coefficient of x}^{2}}{{coefficient of x}^{4}}$
$α β + β γ + δ γ + α δ + δ β + γα = \frac{c}{a}$
$(1)(3) + (3)(5) + (5)(-3) + (-3)(1) + (-3)(3) + (5)(1) = \frac{-4}{1}$
3 + 15 – 15 – 3 – 9 + 5 = -4
-4 = -4
$Sum of product of zeros taken three at a time = - \frac{coefficient of x}{{coefficient of x}^{4}}$
$αβγ + βγδ + αβδ + αγδ = - \frac{d}{a}$
$(1)(3)(5)+(3)(5)(-3)+(1)(3)(-3)+(1)(5)(-3) = - \frac{54}{1}$
$15-45-9-15 = - \frac{54}{1}$
-54 = -54
$Product of zeros = \frac{constant of term}{{coefficient of x}^{3}}$
$α β γ δ = \frac{e}{a}$
$(1)(3)(5)(-3) = - \frac{45}{1}$
-45 = -45

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Last updated on: 03-05-2024