Find the values of k for which the equation $x^3-9x^2 +24x +k =0$ may have multiple roots and solve the equation in each case. This question is from Theory of Equations. Help please.
A polynomial $\;f(x)\;$  over a field has a multiple root $\;\alpha\;$ iff $\;f(\alpha)=f'(\alpha)=0\;$. Now, here we have

$$f(x)=x^3-9x^2+24x+k\implies f'(x)=3x^2-18x+24=0\iff$$

$$x^2-6x+8=(x-2)(x-4)=0\iff x=2,4$$

So it must either $\;x=2,\,4\;$ is a multiple root. Now check for what values of $\;k\;$ this happens.
One way to solve the problem is to find the roots of the derivative of the polynomial $p(x)$, because if $p(x)$ has multiple roots, then $p'(x)$ will have the same roots. To see this, say $a$ is a root of $p(x)$ with multiplicity $n$, then we can write $p(x)=(x-a)^ng(x)$ where $g(a)\not=0.$ Then $p'(x)=n(x-a)^{n-1}g(x)+(x-a)^ng'(x)$, i.e. $a$ is also a root of $p'(x)$.

So to solve your problem, you can take the derivative of your polynomial and then find the roots and then put those roots back to the original polynomial to calculate the values of $k.$