00:01
In this problem, you're given that a polynomial has zeros of negative 3, negative 1, and 4, and it also has a function value of f of negative 2 equals negative 30.
00:15
So the first thing we're going to do is write the function in its factored form with a coefficient that we don't know what it is yet.
00:26
We're going to solve for that.
00:27
All right, so we're going to take each one of the zeros and write the factor, and the factor will be the opposite sign.
00:35
So if the zero is negative three, the factor will be x plus three.
00:44
If the zero is negative one, the factor will be x plus one.
00:50
And if the zero is four, the factor will be x minus four.
00:57
All right, so now the next thing we're going to do is use the function value.
01:05
And this says take all the xes and substitute them with a negative 2.
01:10
So we will leave the a alone.
01:12
That's what we're going to end up solving for.
01:15
So we'll have negative 2 plus 3, negative 2 plus 1, and negative 2 minus 4...