Answer:
The values of k will be:
![k=-1,\:k=3](https://tex.z-dn.net/?f=k%3D-1%2C%5C%3Ak%3D3)
Step-by-step explanation:
Let the expression of polynomial P be
![P\left(x\right)=3x^2-4kx-4k^2](https://tex.z-dn.net/?f=P%5Cleft%28x%5Cright%29%3D3x%5E2-4kx-4k%5E2)
Let the expression if the polynomial Q be
![\:Q\left(x\right)=\:\left(x+2\right)\:](https://tex.z-dn.net/?f=%5C%3AQ%5Cleft%28x%5Cright%29%3D%5C%3A%5Cleft%28x%2B2%5Cright%29%5C%3A)
Plug in Q(x) = 0
0 = x+2
x = -2
As (x+2) is a factor of 3x²-4kx-4k²
substitute x = -2 in the the polynomial
3x²-4kx-4k² = 0
![3\left(-2\right)^2-4k\left(-2\right)-4k^2\:=0](https://tex.z-dn.net/?f=3%5Cleft%28-2%5Cright%29%5E2-4k%5Cleft%28-2%5Cright%29-4k%5E2%5C%3A%3D0)
![12+8k-4k^2=0](https://tex.z-dn.net/?f=12%2B8k-4k%5E2%3D0)
Write in the standard form ax²+bx+c = 0
![-4k^2+8k+12=0](https://tex.z-dn.net/?f=-4k%5E2%2B8k%2B12%3D0)
Factor out common term -4
![-4\left(k^2-2k-3\right)=0](https://tex.z-dn.net/?f=-4%5Cleft%28k%5E2-2k-3%5Cright%29%3D0)
Factor k²-2k-3: (k+1)(k-3)
![-4\left(k+1\right)\left(k-3\right)=0](https://tex.z-dn.net/?f=-4%5Cleft%28k%2B1%5Cright%29%5Cleft%28k-3%5Cright%29%3D0)
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
![k+1=0\quad \mathrm{or}\quad \:k-3=0](https://tex.z-dn.net/?f=k%2B1%3D0%5Cquad%20%5Cmathrm%7Bor%7D%5Cquad%20%5C%3Ak-3%3D0)
solving k+1=0
k+1 = 0
k = -1
solving k-3=0
k-3=0
k = 3
Thus, the values of k will be:
![k=-1,\:k=3](https://tex.z-dn.net/?f=k%3D-1%2C%5C%3Ak%3D3)