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.

**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).

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.

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.

Graph of quadratic polynomial

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.

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

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

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.

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.

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

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

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

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.

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

The graph of a quadratic polynomial ax^{2} + bx + c is a parabola. It opens upward ∪ if a > 0
and opens downward ∩ if a < 0.

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

Sum of zeros = \(-\frac{coefficient \; \; of \; \; x}{coefficient \; \; of \; \; x^2}\)

Product of zeros = \(\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^2}\)

**Solution**

**Here p(x) = x ^{2} - 9**

(x)

(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

p(x) = x

compare it with ax

p(x) = x

Here a = 1, b = 0, c = -9

-3 + 3 = 0

= \(-\frac{0}{1}\)

= \(-\frac{b}{a}\)

= (-3)(3) = -9

= \(-\frac{9}{1}\)

= \(\frac{c}{a}\)

**Solution**

Sum of zeros = 5

Product of zeros = 6

As we know, quadratic polynomial is the form of x^{2} - (sum of zeros)x + product of zeros

By putting the above values, it becomes x^{2} - 5x + 6

Hence, x^{2} - 5x + 6 is a quadratic polynomial.

**Solution**

Here, compare x^{3} - 9x^{2} - 12x + 20 with ax^{3} + bx^{2} + 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

= \(\frac{-20}{1}\)

= \(\frac{-(d)}{a}\)

= \(-\frac{constant \; \; of \; \; term}{coefficient \; \; of \; \; x^3}\)

- c and a have opposite sign
- c and b have same sign
- c and a have same sign
- c and b have opposite sign

- a, b and c all have same sign
- a and b have same sign
- a and c have opposite sign
- b and c have same sign

- a>0
- a≤0
- a≥0
- a<0

- a>0
- a<0
- a≥0
- a≤0

- 1 zero
- 2 zeros
- 3 zeros
- No zeros

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