Polynomial function problem?

Question

Find the polynomial function of least degree that passes through (1,−8) and has the

following roots: −3,−1 (multiplicity 2), 0 (multiplicity 3), 2.

please show complete solution... thank you!

Answer 1

First of all, let's find the degree of our polynomial. It is said that it has 7 roots: -3, -1, -1, 0, 0, 0, 2, so the polynomial is at least of 7th degree (every polynomial has at most n real roots, where n is the polynomial's degree).

Let the polynomial be P(x). We know that P(-3)=0, P(-1)=0, P(0)=0 and P(2)=0. If a number p is a root of a polynomial P(x), this means that if you factorize the polynomial, there will be a factor (x-p). If p has multiplicity k, then there are k factors (x-p), or (x-p)^k is a factor of P(x). In this problem we know all the roots of the polynomial so we know how it is factorized:

P(x) = (x- (-3))*(x - (-1))*(x - (-1))*(x-0)*(x-0)*(x-0)*(x -2)*c =

= (x+3)*(x+1)^2*x^3*(x-2)*c

where c is a constant. c may be equal to 1 or not. To find the value of c we will use that P(x) passes through (1; -8), which means that P(1)= -8

P(1) = (1+3)*(1+1)^2*1^3*(1-2)*c = -8

4*4*1*( -1)*c= -8

-16c = -8

so c = 0.5

So, P(x) = 0.5*(x+3)*(x+1)^2*x^3*(x-2)

When you remove the brackets you will get the polynomial in its traditional order.

Hope this helps :)

Answer 2

Given the roots, the function will be :

y = x^3(x - 2)(x + 3)(x + 1)^2

Plugging in x = 1, gives y = 1^3(1 - 2))(1 + 3)(1 + 1)^2 = -16.

So to get y = -8, it's necessary to just divide by 2.

Therefore, the final function is :

y = x^3(x - 2)(x + 3)(x + 1)^2 / 2