MATHS QUERY

# Zeroes of Polynomial

## Introduction

Let’s first understand what is zeroes of polynomial. We can represent a polynomial on graph with a curve which intersects the axis at specific points. The values of the intersecting points on x-axis are the values that make the polynomial equal to zero. These points on x-axis which make polynomial equal to zero are called zeroes of a polynomial. Below is an example of polynomial which is drawn as a curve on graph.

Graph of polynomial

In this graph, the curve of polynomial y = x + 4 intersects the x-axis at x = – 4 and y-axis at y = 4 or the intersecting points on the x-axis for x = -4 is (-4,0) and and on the y-axis for y = 4 is (0,4).
As said above in the introduction, the values of x that lie on x-axis i.e. x = -4 is the zero of polynomial.
How?
We can check it by putting the value of x = – 4 in polynomial y = x + 4, if y comes equal to 0 for x = -4, then value of x is zero of polynomial.
Substitute value x = – 4 in polynomial y = x + 4.
y = (-4) + 4
y = -4 + 4
y = 0
$$\therefore$$, x=-4 is zero of polynomial.

## Definition

A real number k is the zero of polynomial p(x) if p(k)= 0.
Let’s see, what it says. If p(x) is a polynomial in x and k is any real number, then value obtained by replacing x by k in p(x) is called the value of p(x) at x = k and is denoted by p(k).
If value of k in polynomial can make the value of polynomial to zero, then k is called zeroes of polynomial. We can learn it by example, how we can find out the zeroes of a polynomial.
Zeros of a polynomial are determined by putting different value of x, which can make value of polynomial to 0.

Example

Find zeroes of polynomial $$p(x) = x^2 – 5x + 6$$
$$p(x) = x^2 – 5x + 6$$
Put value x = 2
$$p(2) = (2)^2 – 5(2) + 6$$
$$= 4 – 10 + 6$$
$$= 10 – 10$$
$$= 0$$
$$p(2) = 0$$
$$\therefore$$ 2 is zero of polynomial p(x)
Put value x = 3
$$p(3) = (3)^2 -5(3) + 6$$
$$= 9 – 15 + 6$$
$$= 15 – 15$$
= 0
$$\therefore$$ 3 is also zero of polynomial p(x)
So, 2 and 3 are two zeroes of polynomial p(x).

## Geometrical meaning of zeroes of polynomial

Any polynomial can be drawn on a graph with a specific curve. We have already introduced above in Introduction section how the polynomial curves are drawn on a graph. The shapes of the curves of polynomial varies with polynomial degree. Linear polynomial has curves of unique shape and are different from quadratic polynomial and even biquadratic has its own unique shape. Let’s learn what are the shapes and how to find zeroes of polynomials for linear, quadratic, cubic and biquadratic polynomials.

### Linear polynomial

The general form of a linear polynomial is $$ax + b$$, where $$a \neq 0$$. The graph of linear polynomial is a straight line and it intersects the x-axis at exactly one point. So, the linear polynomial has only one zero.

Example

Consider a linear polynomial, y = 2x + 6
To find the zeroes, we put y = 0
2x + 6 = 0
x = -3
$$\therefore$$ -3 is zero of y = 2x + 6 linear polynomial

Graph of linear polynomial

The general form of a quadratic polynomial is $$ax^2+bx+c$$, where $$a \neq 0$$. The graph of quadratic polynomial has two shapes, one is known as open upwards or in shape of U and another is open downwards or in shape of downwards U. These curves are also called as parabolas. Let’s find zeroes of a quadratic polynomial in an example as below.

Example

Consider a quadratic polynomial, $$y = x^2 + 2x – 3$$
To find zeroes of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 0
$$y = (0)^2 + 2(0) – 3$$
y = 0 + 0 – 3
y = 0 – 3
y = -3
Put x = -3
$$y = (-3)^2 + 2(-3) – 3$$
y = 9 – 6 – 3
y = 9 – 9
y = 0
Put x = 1
$$y = (1)^2 + 2(1) – 3$$
y = 1 + 2 – 3
y = 3 – 3
y = 0
The zeroes of quadratic polynomial $$x^2 + 2x – 3$$ will be the x coordinates of points where the graph $$y=x^2 + 2x – 3$$ intersects the x-axis.
Therefore, -3 and 1 are zeroes of polynomial $$x^2 + 2x – 3$$ as graph $$y = x^2 + 2x – 3$$ intersects x-axis at -3 and 1.

Therefore, for quadratic polynomial $$ax^2+bx+c$$, $$a \neq 0$$, zeroes of polynomial are x-coordinates of points where the parabola $$y=ax^2+bx+c$$ intersects x-axis.
Further, for the graph of $$y=ax^2+bx+c$$, there are three cases that arises with different shapes of graph. Let’s have a look at it one by one.

#### Graph of a quadratic polynomial with distinct zeroes

First case of graph is when the graph cuts x-axis at two distinct points A and B. The x-coordinate of A and B are two zeroes of quadratic polynomial $$ax^2+bx+c$$.

Graph of a quadratic polynomial with distinct zeroes

#### Graph of a quadratic polynomial with coincident zeroes

When graph of the polynomial $$ax^2+bx+c$$ cuts x-axis at exactly one point i.e. at two coincident points. So the two points A and B coincide here to become one point A. The x-coordinates of A is only one zero for quadratic polynomial $$ax^2+bx+c$$.

Graph of a quadratic polynomial with coincident roots

#### Graph of a quadratic polynomial with no zero zeroes

Here graph is completely above x-axis and completely below x-axis. It does not cut the x-axis at any point. So, quadratic polynomial has no zero in this case.

Graph of a quadratic polynomial with no zero zeroes

Therefore, we can summarize from the above three cases, geometrically, a quadratic polynomial can have either two distinct zeroes or two equal zeroes or no zero. In other words, a quadratic polynomial can have at most two zeroes.

### Cubic polynomial

The general form of a cubic polynomial is $$ax^3+bx^2+cx+d$$, where $$a \neq 0$$. The graph of cubic polynomial intersects the x-axis at points, the coordinates of these points are the only zeros of the cubic polynomial.
Let’s find zeroes of a cubic polynomial in an example as below.

Example

Consider a cubic polynomial, $$y=x^3-x$$
To find zeroes of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 1
$$y=(1)^3-(1)$$
y=1-1
y=0
Put x = 0
$$y=(0)^3-(0)$$
y=0-0
y=0
Put x = -1
$$y=(-1)^3-(-1)$$
y=-1+1
y=0
Put x = -2
$$y=(-2)^3-(-2)$$
y=-(-8)+2
y=-6
The zeroes of cubic polynomial $$y=x^3-x$$ will be the x coordinates of points where the graph $$y=x^3-x$$ intersects the x-axis.
Here 1, 0 and 1 are zeroes of cubic polynomial as these are points where the graph intersects x-axis.

Graph of cubic polynomial

The general form of a biquadratic polynomial is $$ax^4+bx^3+cx^2+dx+c$$, where $$a \neq 0$$. The graph of biquadratic polynomial intersects the x-axis at points, the coordinates of these points are the only zeros of the biquadratic polynomial.
Let’s find zeroes of a biquadratic polynomial in an example as below.

Example

Consider a biquadratic polynomial, $$y=x^4-x^3-38x^2+36x+72$$
To find zeroes of polynomial, let’s start putting value of x and find the corresponding y.
Put x = 0
y=(0)4-(0)3-38(0)2+36(0)+72
y = (0) – (0) – (0) + 36(0) + 72
y = 72
Put x = 2
y=(2)4-(2)3-38(2)2+36(2)+72
y = 16 – 8 – 152 + 72 + 72
y = 0
Put x = 6
y=(6)4-(6)3-38(6)2+36(6)+72
y = 1296 – 216 – 1368 + 216 + 72
y=0
Put x = -1
y=(-1)4-(1)3-38(1)2+36(1)+72
y = 1 + 1 – 38 – 36 + 72
y=0
Put x = -6
y=(-6)4-(-6)3-38(-6)2+36(-6)+72
y = 1296 + 216 – 1368 – 216 + 72
y=0
The zeroes of cubic polynomial $$y=x^4-x^3-38x^2+36x+72$$ will be the x coordinates of points where the graph $$y=x^4-x^3-38x^2+36x+72$$ intersects the x-axis.
Here, -6, -1, 2 and 6 are zeroes of biquadratic polynomial as these are points where the graph intersects x-axis.

## Relationship between zeroes and coefficients of quadratic polynomial

Consider a quadratic polynomial $$ax^2+bx+c$$. Let $$\alpha$$ and $$\beta$$ are two zeroes of polynomial.

Formula

Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
$$\alpha + \beta = \frac{-b}{a}$$
Product of zeroes = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$
$$\alpha \beta = \frac{c}{a}$$

Example

Consider a polynomial $$x^2+5x+6$$
Zeroes 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, $$\alpha$$=-2 and $$\beta$$=-3
Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
$$\alpha +\beta$$ = $$-\frac{b}{a}$$
(-2) + (-3) = $$-\frac{5}{1}$$
-5 = -5
Product of zeroes = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$
$$\alpha \beta = \frac{c}{a}$$
(-2)(-3) = $$\frac{6}{1}$$
6 = 6

## Relationship between zeroes and coefficients of cubic polynomial

Now, consider a cubic polynomial $$p(x)=ax^3+bx^2+cx+d$$. Let $$\alpha$$, $$\beta$$ and $$\gamma$$ are three zeroes of polynomial.

Formula

Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
$$\alpha + \beta + \gamma$$ = $$-\frac{b}{a}$$
Sum of product of zeroes taken two at a time= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta+\beta \gamma+\alpha \gamma= \frac{c}{a}$$
Product of zeroes = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma = -\frac{d}{a}$$

Example

Consider a cubic polynomial $$x^3+2x^2-5x-6$$
Zeroes 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, $$\alpha$$=-1, $$\beta$$=2 and $$\gamma$$=-3
Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
$$\alpha +\beta+\gamma$$ = $$-\frac{b}{a}$$
(-1) + (2) + (-3)= $$-\frac{2}{1}$$
-2 = -2
Sum of product of zeroes taken two at a time= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta+\beta \gamma+\alpha \gamma= \frac{c}{a}$$
(-1)(2) + (2)(-3) + (-3)(-1)= $$\frac{-5}{1}$$
-2 -6 + 3 = -5
-5 = -5
Product of zeroes = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma = -\frac{d}{a}$$
(-1)(2)(-3) = -($$\frac{-6}{1}$$)
6 = 6

## Relationship between zeroes and coefficients of biquadratic polynomial

Polynomial $$p(x)=ax^4+bx^3+cx^2+dx+e$$, where $$a \neq=0$$ 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 biquadratic polynomial.
Consider a polynomial $$ax^4+bx^3+cx^2+dx+e$$. Let $$\alpha, \beta, \gamma and \delta$$ are four zeroes of the polynomial.

Formula

Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x^3}{coefficient \; \; of \; \; x^4}$$
$$\alpha + \beta + \gamma+ \delta$$ = $$-\frac{b}{a}$$
Sum of product of zeroes taken two at a time= $$\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^4}$$
$$\alpha \beta+\beta \gamma+\delta \gamma+\alpha \delta+\delta\beta+\gamma\alpha= \frac{c}{a}$$
Sum of product of zeroes taken three at a time= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^4}$$
$$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta= -\frac{d}{a}$$
Product of zeroes = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma\delta = \frac{e}{a}$$

Example

Let $$x^4-6x^3-4x^2+54x-45$$ be a biquadratic polynomial
Zeroes 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
$$\therefore$$ a=1, b=-6, c=-4, d=54, e=-45
Let zeroes be, $$\alpha$$=1, $$\beta$$=3, $$\gamma$$=5 and $$\delta$$=-3
$$\therefore$$ Sum of zeroes = $$-\frac{coefficient \; \; of \; \; x^3}{coefficient \; \; of \; \; x^4}$$
$$\alpha +\beta+\gamma+\delta$$ = $$-\frac{b}{a}$$
(1) + (3) + (5) + (-3)= $$-\frac{-6}{1}$$
6 = 6
Sum of product of zeroes taken two at a time= $$\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^4}$$
$$\alpha \beta+\beta \gamma+\gamma\delta+\alpha \delta+\delta\beta+\gamma\alpha$$ = $$\frac{c}{a}$$
(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 zeroes taken three at a time= $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^4}$$
$$\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta= -\frac{d}{a}$$
(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 zeroes = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma\delta = \frac{e}{a}$$
(1)(3)(5)(-3)= $$\frac{-45}{1}$$
-45 = -45