Determine k and solve the equation x^3 -kx^2 + 20x +12 = 0.
Thanks
Let $\displaystyle a,b,c$ be zero's (counting multiplicity).
One of its zeros, say $\displaystyle c$, is triple of another, say $\displaystyle a$. Thus, $\displaystyle c=3a$.
Thus, we have that,
$\displaystyle \left\{ \begin{array}{c}a+b+3a = k\\ab+3a^2+3ab=20\\ 3a^2b= -12 \end{array}\right.$
Thus,
$\displaystyle \left\{ \begin{array}{c}4a+b = k\\ 4ab+3a^2 = 20 \\ a^2b = -4 \end{array}\right.$
Now we solve for $\displaystyle a,b,k$ ...
But, I have an easier way. Solving for that equation can get messy. Instead, let us hope that its solutions are integers. Not necessarily this approach might work but let us try it, if it works then we can save a lot of work. If it does not we can try another way.
Note,
$\displaystyle a^2b=-4$.
Now, $\displaystyle a^2>0$ so $\displaystyle b<0$. It must be that $\displaystyle a^2 \mbox{ and }b$. Are factors of $\displaystyle 4$. Since $\displaystyle a^2$ must be a square and positive. It means $\displaystyle a^2 = 4 \mbox{ or }1$. Thus, the solutions are $\displaystyle (a^2,b) = (4,-1) \mbox{ or } (1,-4)$. Now take the square root to remove the square on $\displaystyle a$ to get: $\displaystyle (a,b) = (-2,-1), (2,-1), (-1,-4) , (1,-4)$.
Now which of these pairs satisfies equation 2, i.e. $\displaystyle 4ab+3a^2 = 20$.
We easily see that $\displaystyle (-2,-1)$ is the only solution.
Thus, $\displaystyle a=-2,b=-1,c=3(a)=-6$.
Thus, $\displaystyle k = -(a+b+c) = 9$.