1. ## I need help checking my answer!

The problem:

find all zeros of: x^4 - 4x^3 - 5x^2 +36x - 36 = 0 such that 2 is a zero of multiplicity 2

I found 2 and 3 are zeros thus; (x - 2)^2 (x - 3) (x^2 + x - 6)
I then applied the quadratic formula to (x^2 + x - 6) resulting in (-1 +- 5) / 2

I then put x^4 - 4x^3 - 5x^2 +36x - 36 into y= on the calculator and only found 2 and 3 to be the only zeros

If you don't mind,can you check this problem out? I want to be sure I'm doing it right, got myself an exam tomorrow and this part is where I don't feel quite confident...

2. ## Re: I need help checking my answer!

Hey nojnotnow.

You are given a 4th degree polynomial and a root of multiplicity two. This means your polynomial has the form (x-2)^2*(x^2 + bx + c) = 0.

You have factorized the polynomial wrong because you have (x-2)^2 * (x-3) * a quadratic which will give you a fifth degree polynomial.

If 2 and 3 are the only two zeroes then x^4 - 4x^3 - 5x^2 + 36x - 36 = (x-2)^2*(x-3)^2 and an expansion should verify this if it is the case.

Also try dividing your polynomial by (x-2)^2 or x^2 - 4x + 4 and see what the quotient is (the remainder should be 0).