00:01
So in this problem, you're being asked to write an equation that has 2, negative 2, 3, and negative 3 is solutions.
00:06
So you're absolutely right.
00:07
This is going to be a fourth degree polynomial.
00:10
So because these are our zeros, these are our solutions, they're the zeros for our function.
00:14
So to get from your solutions to their factors, you just take the opposite side.
00:18
So for example, if 2 is one of our solutions, we know that x minus 2 is a factor.
00:23
If negative 2 is a 0, then x plus 2 is a factor.
00:27
If 3 is a solution, then x minus 3 is a factor, and if negative 3 is a 0, then x plus 3 is a factor.
00:34
So now we have our function in factored form.
00:38
So if you just need an in factored form, you'll be all set.
00:41
But sometimes you need to multiply it up.
00:43
So first, i'll multiply x minus 2 by x plus 2.
00:46
We'll notice this is one of our special products.
00:49
It's the product of the sum and difference of two terms...