Answer:
3
Explanation:
A solution to a function is when the graph of the function crosses the x-axis. In other words, it's a point where y = 0. So, we can set x^3 + kx^2 - 9x -18 equal to 0. Note that there is an unknown coefficient in this equation (k) that we have to solve for. We are given that one solution to this equation is -2. This means that we can plus x = -2 into the equation and solve for the unknown k:
![(-2)^3+k(-2)^2-9(-2)-18=0\\-8+4k+18-18=0\\4k=8\\k=2](https://tex.z-dn.net/?f=%28-2%29%5E3%2Bk%28-2%29%5E2-9%28-2%29-18%3D0%5C%5C-8%2B4k%2B18-18%3D0%5C%5C4k%3D8%5C%5Ck%3D2)
Therefore, the complete equation is the following:
![y=x^3+2x^2-9x-18](https://tex.z-dn.net/?f=y%3Dx%5E3%2B2x%5E2-9x-18)
As it turns out, we can use grouping to factor this polynomial. Let's start by re-ordering the terms to visualize it better:
![x^3-9x+2x^2-18=0](https://tex.z-dn.net/?f=x%5E3-9x%2B2x%5E2-18%3D0)
The first two terms have a GCF of x, and the last two terms have a GCF of 2:
![x(x^2-9)+2(x^2-9)](https://tex.z-dn.net/?f=x%28x%5E2-9%29%2B2%28x%5E2-9%29)
Now, there is a common term that we can bring to the front to complete our factoring process:
![(x^2-9)(x+2)](https://tex.z-dn.net/?f=%28x%5E2-9%29%28x%2B2%29)
First of all, note that x+2 is one of the factors, and if this was set to 0, we would get -2, which is consistent with the problem. Also, note that our other term can be factored more because it is a difference of two squares:
![(x-3)(x+3)(x+2)=0](https://tex.z-dn.net/?f=%28x-3%29%28x%2B3%29%28x%2B2%29%3D0)
If we set each term to 0, we get x = 3, x = -3, and x = -2. We are only looking for the positive solution to this equation, which is 3