## 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 ax^{2} + 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 x^{2}
with the following formulas for sum of zeros and product of zeros.

$\text{Sum of zeros}=\u2013\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{2}}$

$\mathrm{\alpha \; +\; \beta}=\frac{\u2013b}{a}$

$\text{Product of zeros}=\frac{\text{constant of term}}{{\text{constant of term}}^{2}}$

$\mathrm{\alpha \; \beta}=\frac{c}{a}$

**Consider a polynomial x ^{2} + 5x + 6**

Zeros of x

^{2}+ 5x + 6 are -2 and -3

Here, coefficient of x

^{2}= 1

coefficient of x = 5

constant term = 6

Therefore, a = 1, b = 5, c = 6

Here, α = -2 and β = -3

$\text{Sum of zeros}=\u2013\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{2}}$

$\mathrm{\alpha \; +\; \beta}=\frac{\u2013b}{a}$

$\mathrm{(-2)\; +\; (-3)}=\frac{\u20135}{1}$

-5 = -5

$\text{Product of zeros}=\frac{\text{constant of term}}{{\text{constant of term}}^{2}}$

$\mathrm{\alpha \; \beta}=\frac{c}{a}$

$\mathrm{(-2)(-3)}=\frac{6}{1}$

6 = 6

## Sum and product of zeros of cubic polynomial

Now, consider a cubic polynomial p(x)=ax^{3} + bx^{2} + 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
x^{2} and x^{3} with the following formulas for sum of zeros and product of zeros.

$\text{Sum of zeros}=\u2013\frac{{\text{coefficient of x}}^{2}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; +\; \beta \; +\; \gamma}=\frac{\u2013b}{a}$

Sum of product of zeros taken two at a time
$=\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{9}}$

$\mathrm{\alpha \; \beta +\beta \; \gamma +\alpha \; \gamma}=\frac{c}{a}$

$\text{Product of zeros}=\u2013\frac{\text{constant of term}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; \beta \; \gamma}=\u2013\frac{d}{a}$

**Consider a cubic polynomial x ^{3} + 2x^{2} – 5x – 6**

Zeros of x

^{3}+ 2x

^{2}– 5x – 6 are -1, 2 and -3

Here, coefficient of x

^{3}= 1

coefficient of x

^{2}= 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

$\text{Sum of zeros}=\u2013\frac{{\text{coefficient of x}}^{2}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; +\; \beta \; +\; \gamma}=\frac{\u2013b}{a}$

$\mathrm{(-1)\; +\; (2)\; +\; (-3)}=\u2013\frac{2}{1}$

-2 = -2

Sum of product of zeros taken two at a time

**$=\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{9}}$**

$\mathrm{\alpha \; \beta +\beta \; \gamma +\alpha \; \gamma}=\frac{c}{a}$

$\mathrm{(-1)(2)\; +\; (2)(-3)\; +\; (-3)(-1)}=\frac{\mathrm{(-5)}}{1}$

-2 -6 + 3 = -5

-5 = -5

$\text{Product of zeros}=\u2013\frac{\text{constant of term}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; \beta \; \gamma}=\u2013\frac{d}{a}$

$\mathrm{(-1)(2)(-3)}=\u2013\frac{\mathrm{(-6)}}{1}$

6 = 6

$\mathrm{\alpha \; \beta +\beta \; \gamma +\alpha \; \gamma}=\frac{c}{a}$

$\mathrm{(-1)(2)\; +\; (2)(-3)\; +\; (-3)(-1)}=\frac{\mathrm{(-5)}}{1}$

-2 -6 + 3 = -5

-5 = -5

$\text{Product of zeros}=\u2013\frac{\text{constant of term}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; \beta \; \gamma}=\u2013\frac{d}{a}$

$\mathrm{(-1)(2)(-3)}=\u2013\frac{\mathrm{(-6)}}{1}$

6 = 6

## Sum and product of zeros of biquadratic polynomial

Polynomial p(x)=ax^{4} + bx^{3} + cx^{2} + dx + e, where a ≠ = 0 is a biquadratic
polynomial. The graph of
y = ax^{4} + bx^{3} + cx^{2} + dx + e intersects the x-axis.
These coordinates are the only zeros of the biquadratic polynomial.

Consider a polynomial ax^{4} + bx^{3} + cx^{2} + 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
x^{3} and x^{4} with the following formulas for sum of zeros and product of zeros.

$\text{Sum of zeros}=\u2013\frac{{\text{coefficient of x}}^{3}}{{\text{coefficient of x}}^{4}}$

$\mathrm{\alpha \; +\; \beta \; +\; \gamma \; +\; \delta}=\u2013\frac{b}{a}$

Sum of product of zeros taken two at a time
$=\frac{{\text{coefficient of x}}^{2}}{{\text{coefficient of x}}^{4}}$

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

Sum of product of zeros taken three at a time
$=\u2013\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{4}}$

$\mathrm{\alpha \beta \gamma \; +\; \beta \gamma \delta \; +\; \alpha \beta \delta \; +\; \alpha \gamma \delta}=\u2013\frac{d}{a}$

$\text{Product of zeros}=\frac{\text{constant of term}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; \beta \; \gamma \; \delta}=\frac{e}{a}$

**Let x ^{4} – 6x^{3} – 4x^{2} + 54x – 45 be a biquadratic polynomial**

Zeros of x

^{4}– 6x

^{3}– 4x

^{2}+ 54x – 45 are 1, 3, 5 and -3

Here, coefficient of x

^{4}= 1

coefficient of x

^{3}= -6

coefficient of x

^{2}= -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

∴ $\text{Sum of zeros}=\u2013\frac{{\text{coefficient of x}}^{3}}{{\text{coefficient of x}}^{4}}$

$\mathrm{\alpha \; +\; \beta \; +\; \gamma \; +\; \delta}=\frac{\u2013b}{a}$

$\mathrm{(1)\; +\; (3)\; +\; (5)\; +\; (-3)}=\u2013\frac{\mathrm{(-6)}}{1}$

6 = 6

Sum of product of zeros taken two at a time $=\frac{{\text{coefficient of x}}^{2}}{{\text{coefficient of x}}^{4}}$

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

(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 $=\u2013\frac{\text{coefficient of x}}{{\text{coefficient of x}}^{4}}$

$\mathrm{\alpha \beta \gamma \; +\; \beta \gamma \delta \; +\; \alpha \beta \delta \; +\; \alpha \gamma \delta}=\u2013\frac{d}{a}$

(1)(3)(5)+(3)(5)(-3)+(1)(3)(-3)+(1)(5)(-3) $=\u2013\frac{54}{1}$

$\mathrm{15-45-9-15}=\u2013\frac{54}{1}$

-54 = -54

$\text{Product of zeros}=\frac{\text{constant of term}}{{\text{coefficient of x}}^{3}}$

$\mathrm{\alpha \; \beta \; \gamma \; \delta}=\frac{e}{a}$

$\mathrm{(1)(3)(5)(-3)}=\u2013\frac{45}{1}$

-45 = -45