# Relationship between Zeroes and Coefficients of a Polynomial

Complete above steps to enable this content

You have already seen that zero of a linear polynomial ax + b is

So, we write

So, the value of p(x) =

**Sum of its zeroes = 1 + 3 = = **

**Product of its zeroes = 1 x 3 = = **

Complete above steps to enable this content

In general, α and β are the zeroes of the quadratic polynomial p(x) =

= k[

= k

Comparing the coefficients of

a = k, b = – k(α + β ) and c = kαβ

This gives α + β =

αβ =

Complete above steps to enable this content

Consider the Equation

You can factorize using the methods you have learnt or look at the graph above and we can see that the zeros of the polynomial are

So (x-1) and (x-3) are factors for the above equation.

This shows that the sum of the zeros of a quadratic polynomial is equal to the negative of the coefficient of the linear term (b) divided by the coefficient of the quadratic term (a), and the product of the zeros is equal to the constant term (c) divided by the coefficient of the quadratic term (a).

Let us now look at cubic polynomials.

Let us consider p(x) =

You can check that p(x) = 0 for x =

Since p(x) can have atmost three zeroes, these are the zeores of

Sum of zeros = 4 + (-2) +

Product of zeros =

For cubic polynomials we can also get one more identity. Mix the zeros two at a time and we get the following:

In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial

Complete above steps to enable this content

**Example 2**

**Find the zeroes of the quadratic polynomial **

Complete above steps to enable this content

**Example 3**

Find the zeroes of the polynomial

Complete above steps to enable this content

**Example 4**

Verify that 3, –1,

- Comparing the given polynomial with
+ax 3 + cx + d. we get a =bx 2 , b = , c = , d = - Further: p(3) = 3 x
- (5 x3 3 ) - (11 x 3) -3 =3 2 – – – (substitute the value of '3' in given eq) - Hence we get the answer is
- p(-1)= 3 ×
– 5 ×− 1 3 – 11 × (–1) – 3 = –− 1 2 – + – (substitute the value of '-1' in given eq) - Hence we get the answer is
- p(
− ) = 3 x1 3 - 5 x− 1 3 3 - 11 x− 1 3 2 - 3 =− 1 3 + + + (substitute the value of − in given eq)1 3 - Hence we get the answer is
− +2 3 =2 3 - Therefore, 3, –1 and
− are the zeroes of1 3 3 -x 3 5 – 11x – 3x 2 - So, we take α = 3, β= -1, γ =
− 1 3 - Now, α + β + γ = 3 + (-1) + (
− ) = 2 -1 3 =1 3 - αβ + βγ + γα = 3x(-1) + (-1)x(
− ) + (1 3 − )x3 = 3 +1 3 - 1 =1 3 - αβγ = 3x(-1)x(
− ) =1 3 - We found the required answers.