# Thread: Math question involving polynomials

1. ## Math question involving polynomials

The profit, "P", of a video company (in thousands of dollars), is given by P(x) = -5x^2 + 550x - 5000, where x is the amount spent on advertising, in thousands of dollars.

a) Determine the amount spent on advertising that will result in a profit of \$0.

Ok so I started it here...

0 = -5x^2 + 550x - 5000
= -5(x^2 - 110x + 1000)

I'm not sure how to continue from here. I'm pretty sure that I'm on the right track. I think that you have to find when x = 0. Essentially what I'm trying to do I think is find the x-intercepts.

2. You have a quadratic equation.

3. You're on the right track. First of all you can cancel out -5 to give $x^2-110x+1000=0$ and solve the resulting quadratic using your favourite method .

4. What do you mean by "favourite method"? Oh wait a minute I think I know what to do. I'll use the formula that I just saw on the wiki link you gave me "emakarov". thanks.

But firstly how do I cancel out -5?

5. If you have an equation -5 * f(x) = 0 for some expression f(x), divide both sides by -5 to get f(x) = 0.

6. So it's just 0 = x^2 - 110x + 1000 ?

7. Originally Posted by Arise01
So it's just 0 = x^2 - 110x + 1000 ?
Yes.

8. but what happened to the -5?

9. Originally Posted by Arise01
but what happened to the -5?
If we have $-5(x-a)=0$ then $(x-a)=0$.

$\frac{0}{-5}=0~.$

10. So I use the quadratic formula on x^2 - 110x + 1000 to solve for the x- intercepts?

11. Originally Posted by Arise01
So I use the quadratic formula on x^2 - 110x + 1000 to solve for the x- intercepts?
Why not just factor it?
$x^2-110x+1000=(x-10)(x-100)$

12. Oh wow how could I miss that. So the x - intercepts are 10 and 100?

13. Originally Posted by Arise01
Oh wow how could I miss that. So the x - intercepts are 10 and 100?
Right on!