 # Zeros of Polynomial

## Introduction

Let’s first understand what is zeros 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 zeros 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 zeros of polynomial. We can learn it by example, how we can find out the zeros 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 zeros 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 zeros of polynomial p(x).

## Geometrical meaning of zeros 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 zeros 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 zeros, 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 ∪ and another is open downwards or in shape of downwards ∩. These curves are also called as parabolas. Let’s find zeros of a quadratic polynomial in an example as below.

Example

Consider a quadratic polynomial, $$y = x^2 + 2x – 3$$
To find zeros 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 zeros 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 zeros of polynomial $$x^2 + 2x – 3$$ as graph $$y = x^2 + 2x – 3$$ intersects x-axis at -3 and 1. Therefore, for quadratic polynomial $$ax^2+bx+c$$, $$a \neq 0$$, zeros 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 zeros

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 zeros of quadratic polynomial $$ax^2+bx+c$$.

Graph of a quadratic polynomial with distinct zeros #### Graph of a quadratic polynomial with coincident zeros

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 zeros

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

### 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 zeros of a cubic polynomial in an example as below.

Example

Consider a cubic polynomial, $$y=x^3-x$$
To find zeros 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 zeros 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 zeros 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 zeros 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 zeros 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 zeros 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 zeros of biquadratic polynomial as these are points where the graph intersects x-axis.

## Relationship between zeros and coefficients of quadratic polynomial

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

Formula

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

Example

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

## Relationship between zeros 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 zeros of polynomial.

Formula

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

Example

Consider a cubic polynomial $$x^3+2x^2-5x-6$$
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, $$\alpha$$=-1, $$\beta$$=2 and $$\gamma$$=-3
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
$$\alpha +\beta+\gamma$$ = $$-\frac{b}{a}$$
(-1) + (2) + (-3)= $$-\frac{2}{1}$$
-2 = -2
Sum of product of zeros 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 zeros = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma = -\frac{d}{a}$$
(-1)(2)(-3) = -($$\frac{-6}{1}$$)
6 = 6

## Relationship between zeros 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 zeros of the polynomial.

Formula

Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^3}{coefficient \; \; of \; \; x^4}$$
$$\alpha + \beta + \gamma+ \delta$$ = $$-\frac{b}{a}$$
Sum of product of zeros 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 zeros 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 zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
$$\alpha \beta \gamma\delta = \frac{e}{a}$$

Example

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

### Q) What is zeros of polynomial?

The value which makes the value of polynomial equal to zero is called zeros of polynomial. i.e. k is said to be zeros of polynomial, when a polynomial p(x) becomes equal to zero for the value of k. i.e. when we put x = k in p(x) and p(k) = 0.

### Q) What is shape of graph of a linear polynomial?

The graph of a linear polynomial is always a straight line.

### Q) What is shape of graph of a quadratic polynomial?

The graph of a quadratic polynomial ax2 + bx + c is a parabola. It opens upward ∪ if a > 0 and opens downward ∩ if a < 0.

### Q) What is the number of zeros of polynomial of degree n?

The polynomial with degree n can have number of zeros equal to n or less than n. i.e. number of zeros ≤ n.

### Q) What is relationship between coefficient of polynomial and its zero?

Sum of zeros = $$-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}$$
Product of zeros = $$\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}$$

## Solved Examples

### 1) Find the zeros of x2 - 9 and verify its relationship between coefficient of polynomial and its zeros.

Solution

Here p(x) = x2 - 9
(x)2 -(3)2   using identity (a)2 - (b)2 = (a+b)(a-b)
(x+3)(x-3)
p(x) = 0
(x+3)(x-3) = 0
x+3 = 0 or x-3 = 0
x = -3 or x = 3
zeros of p(x) are -3, 3
Relationship between coefficients of polynomial and its zeros
p(x) = x2 - 9
compare it with ax2 + bx + c
p(x) = x2 + 0x - 9
Here a = 1, b = 0, c = -9
Sum of zeros = $$-\frac{b}{a}$$
-3 + 3 = 0
= $$-\frac{0}{1}$$
= $$-\frac{b}{a}$$
Product of zeros = $$\frac{c}{a}$$
= (-3)(3) = -9
= $$-\frac{9}{1}$$
= $$\frac{c}{a}$$

### 2) Form a quadratic polynomial whose sum of zeros is 5 and product of zeros is 6.

Solution

Sum of zeros = 5
Product of zeros = 6
As we know, quadratic polynomial is the form of x2 - (sum of zeros)x + product of zeros
By putting the above values, it becomes x2 - 5x + 6
Hence, x2 - 5x + 6 is a quadratic polynomial.

### 3) Verify relationship between coefficient of polynomial and its zeros if x3 - 9x2 - 12x + 20 has zeros -2, 1 and 10.

Solution

Here, compare x3 - 9x2 - 12x + 20 with ax3 + bx2 + cx + d
a = 1, b = -9, c = -12, d = 20
Zeros are -2, 1 and 10 (given)
α = -2
β = 1
γ = 10
Sum of zeros = $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
∴ α + β + γ = -2 + 1 + 10
= 9
= $$-\frac{(-9)}{-1}$$
= $$-\frac{(-b)}{a}$$
= $$-\frac{coefficient \; \; of \; \; x^2}{coefficient \; \; of \; \; x^3}$$
Sum of product of zeros taken two at a time = $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
αβ + βγ + γα = (-2)(1) + (1)(10) + (10)(-2)
= -2 + 10 -20
= -12
= $$\frac{(-12)}{1}$$
= $$\frac{(c)}{a}$$
= $$\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^3}$$
Product of zeros = $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$
αβγ = (-2)(1)(10)
= -20
= $$\frac{-20}{1}$$
= $$\frac{-(d)}{a}$$
= $$-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}$$

## Multiple Choice Questions

### 1) A quadratic polynomial has

1. 1 zero
2. 2 zeros
3. No zero
4. atmost 2 zeros

### 2) In quadratic polynomial ax2 + bx + c, c ≠ 0, if zeros are of equal sign, then

1. c and a have opposite sign
2. c and b have same sign
3. c and a have same sign
4. c and b have opposite sign

### 3) In quadratic polynomial ax2 + bx + c, c ≠ 0, if both zeros are negative, then

1. a, b and c all have same sign
2. a and b have same sign
3. a and c have opposite sign
4. b and c have same sign

### 4) Find the number of zeros from the following graph of polynomial. 1. 2
2. 3
3. 4
4. 1

1. a>0
2. a≤0
3. a≥0
4. a<0

1. a>0
2. a<0
3. a≥0
4. a≤0

1. 1 zero
2. 2 zeros
3. 3 zeros
4. No zeros

1. 1
2. 2
3. 3
4. 0

1. 5 and 5
2. -5 and 5
3. 2 and -5
4. -5 and -5

1. 3
2. 1
3. -3
4. -1