Question

Find a polynomial f(x) of degree 3 that has the following zeros: -8 (multiplicity of 2), 6.Leave answer in factor form

Answer 1

Polynomials

A polynomial of degree 3 can be written in factor form as follows:

$p(x)=a(x-x_1)(x-x_2)(x-x_3)$

Where x1, x2, and x3 are the zeros of the polynomial, and a is the leading coefficient.

We are given the three zeros: x1= -8, x2 = -8, x3 = 6

Note we have repeated the root -8 because it has a multiplicity of 2.

Substituting these values, we have:

$p(x)=a(x-(-8))(x-(-8))(x-6)$

Operating:

$p(x)=a(x+8)(x+8)(x-6)$

We can set the value of a to any non-zero real number. If a=1, our polynomial is:

$p(x)=(x+8)(x+8)(x-6)$

For a=-2, we have:

$p(x)=-2(x+8)(x+8)(x-6)$

We have given two different solutions, and you can give more by changing the value of a.