Answer:
B
Step-by-step explanation:
We have the two functions:
![f(x)=2x^3+6x^2+4x\text{ and } \\g(x)=x^2+3x+2](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E3%2B6x%5E2%2B4x%5Ctext%7B%20and%20%7D%20%5C%5Cg%28x%29%3Dx%5E2%2B3x%2B2)
First, let's factor each of the functions:
![f(x)=2x^3+6x^2+4x](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E3%2B6x%5E2%2B4x)
Factor out a 2x:
![f(x)=2x(x^2+3x+2)](https://tex.z-dn.net/?f=f%28x%29%3D2x%28x%5E2%2B3x%2B2%29)
Factor:
![f(x)=2x(x+2)(x+1)](https://tex.z-dn.net/?f=f%28x%29%3D2x%28x%2B2%29%28x%2B1%29)
Factor our second function:
![g(x)=x^2+3x+2](https://tex.z-dn.net/?f=g%28x%29%3Dx%5E2%2B3x%2B2)
Factor:
![g(x)=(x+2)(x+1)](https://tex.z-dn.net/?f=g%28x%29%3D%28x%2B2%29%28x%2B1%29)
Now, let's see what happens if we add them together. This will yield:
![f(x)+g(x)=2x(x+2)(x+1)+(x+2)(x+1)](https://tex.z-dn.net/?f=f%28x%29%2Bg%28x%29%3D2x%28x%2B2%29%28x%2B1%29%2B%28x%2B2%29%28x%2B1%29)
We can rewrite this as:
![f(x)+g(x)=2x((x+2)(x+1))+1((x+2)(x+1))](https://tex.z-dn.net/?f=f%28x%29%2Bg%28x%29%3D2x%28%28x%2B2%29%28x%2B1%29%29%2B1%28%28x%2B2%29%28x%2B1%29%29)
Using grouping (or the reverse of our distribute property), we can write this as:
![f(x)+g(x)=(2x+1)((x+2)(x+1))](https://tex.z-dn.net/?f=f%28x%29%2Bg%28x%29%3D%282x%2B1%29%28%28x%2B2%29%28x%2B1%29%29)
However, we want our function to be divisible by (2x+3). We have (2x+1).
So, we can multiply our g(x) by 3. If will go back, this will yield:
![f(x)+3g(x)=2x((x+2)(x+1))+3((x+2)(x+1))](https://tex.z-dn.net/?f=f%28x%29%2B3g%28x%29%3D2x%28%28x%2B2%29%28x%2B1%29%29%2B3%28%28x%2B2%29%28x%2B1%29%29)
Now, we can do grouping again:
![f(x)+3g(x)=(2x+3)(x+2)(x+1)](https://tex.z-dn.net/?f=f%28x%29%2B3g%28x%29%3D%282x%2B3%29%28x%2B2%29%28x%2B1%29)
Since this now has a (2x+3) term, this is divisible by (2x+3).
So, the equation that is divisible by (2x+3) is f(x)+3g(x).
Our answer is B.
And we're done!