# ZEROS OF A POLYNOMIAL

## About "Zeros of a polynomial"

Zeros of a polynomial :

Here we are going to see how to find zero of a polynomials.

If the value of a polynomial is zero for some value of the variable then that value is known as zero of the polynomial.

Definition :

Let p(x) be a polynomial in x. If p(x) = 0, then we say that a is a zero of the polynomial p(x).

Let us consider the example given below.

## Zeros of linear polynomial

Example 1 :

Find the zeros of the following linear polynomial

p(x)  =  2x + 3

Solution :

p(x)  =  2x + 3

Now we have to think about the value of x, for which the given function will become zero.

For that let us factor out 2

p(x)  =  2 (x + 3/2)

Instead of "x" , if we apply -3/2 p(x) will become zero.

Hence -3/2 is the zero of the given linear polynomial.

Example 2 :

Find the zeros of the following linear polynomial

p(x)  =  4x - 1

Solution :

p(x)  =  4x - 1

Now we have to think about the value of x, for which the given function will become zero.

For that let us factor out 4

p(x)  =  4 (x - 1/4)

By applying the value 1/4 instead of x, the function p(x) will become zero.

Hence 1/4 is the zero of the given linear polynomial.

For a quadratic equation, there will be two zeros. In order to find those zeros, we may use the method called factoring.

Example 3 :

Find the zeros of the quadratic equation x² + 17 x + 60 by factoring.

Solution :

p(x)  =  x² + 17 x + 60

p(x)  =  x² + 12x + 5x + 60

p(x)  =  x (x + 12) + 5 (x + 12)

p(x)  =  (x + 5) (x + 12)

If x =  -5

p(x)  =  (-5 + 5) (-5 + 12)  =  0

If x =  -12

p(x)  =  (-12 + 5) (-12 + 12)  =  0

Hence the zeros are -5 and -12.

## Zeros of cubic polynomial

For a cubic equation, there will be three zeros. In order to find those zeros, we may use the methods

(i)  Factor theorem

(ii)  Synthetic division

Example 4 :

Find the zeros of the following polynomial 4 x³ - 7 x + 3

Solution :

Let p (x) = 4 x³ - 7 x + 3

x = 1

p (1) = 4 (1)³ -7 (1) + 3

= 4 - 7 + 3

= 7 - 7

= 0

So we can decide (x - 1) is a factor. To find other two factors, we have to use synthetic division. So, the factors are (x - 1) and (4 x² - 4 x - 3). By factoring this quadratic equation we get  (2 x + 3) (2 x - 1)

Hence the required three factors are (x - 1) (2 x + 3) (2 x - 1)

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