Finding all $k$ such that $3x^4+4x^3-12x^2+k=0$ has four real roots. How to find roots of a degree-$4$ polynomial? I have been given this question after studying long division of polynomials and the factor theorem:

Find the set of values of $k$ for which the equation $3x^{4}+4x^3-12x^2+k=0$ has four real roots.

How do you find the set of values of $x$ in a polynomial of degree $4$?
I thought about using the formula $b^2-4ac$ from quadratics knowing very well it wasn't going to work, so I genuinely don't know how to go about this last question.
Could someone please help me with this question?
 A: $f(x)=3x^4 +4x^3-12x^2 +k=0$
Differentiating, $$f’(x) = 12x^3 +12x^2 -24x =0 \implies x=-2,0,1$$ If you know anything about what a quartic looks like, you can deduce that there is a minimum, maximum and minimum at $-2,0,1$ respectively. For $f(x)$ to have four roots, we need $f(-2)\le 0 , f(0)\ge 0 ,  f(1)\le 0$. That is, $$48-32-48+k\le 0 \implies k\le32 \\ k\ge 0 \\ 3+4-12+k\le 0 \implies k\le 5$$ Taking the intersection of these values, we get $$0\le k\le 5$$
A: The challenge here is to find a solution that doesn't use calculus, i.e., doesn't take the derivative. The key insight is that, as you translate a polynomial graph up or down, the number of roots changes only when you get to a translate with a double root. So what we're looking for is values of $k$ such that
$$3x^4+4x^3-12x^2+k=(x-r)^2(3x^2+bx+c)$$
for some root $r$ and coefficients $b$ and $c$.
Re-expanding the right hand side to
$$\begin{align}
(x-r)^2(3x^2+bc+c)&=(x^2-2rx+r^2)(3x^2+bx+c)\\
&=3x^4+(b-6r)x^3+(c-2br+3r^2)x^2+(br^2-2cr)x+cr^2
\end{align}$$
we see that
$$\begin{align}
b-6r&=4\\
c-2br+3r^2&=-12\\
br^2-2cr&=0
\end{align}$$
Thus $b=4+6r$ and $c=2br-3r^2-12=2(4+6r)r-3r^2-12=9r^2+8r-12$ and
$$(4+6r)r^2-2(9r^2+8r-12)r=-12r^3-12r^2+24r=-12r(r-1)(r+2)=0$$
so the number of roots changes when $k=cr^2=(9r^2+8r-12)r^2$ with $r=0$, $1$, and $-2$, i.e, when $k=0$, $5$, and $32$.  By thinking about the general nature of quartics, we see that $3x^4+4x^3-12x^2+k$ has two roots when $k\lt0$, four roots (counting multiplicities) when $0\le k\le5$, two roots again when $5\lt k\le32$, and no roots when $k\gt32$. In particular, the answer to the question posed is the range $0\le k\le5$.
Comparing this to the answers that use the derivative, the take-home lesson may be that it's well worth learning calculus!
A: More details: the derivative of $f$ is a cubic polynomial which is very easy to factor, and that has $3$ real roots. From this it is easy to infer the signs that $f'$ takes over $\mathbb{R}$ (for verification purposes: it's negative up to the first root, then positive, then negative, then positive again). Next, $f$ is a continuous function (since it is a polynomial) and goes to $+\infty$ at infinities. From this you have a pretty clear picture of what the graph of $f$ looks like. Then you want to find the possible $k$ such that $f$ will cross the abscissa line each time before it changes its variation (the intermediate value theorem will guarantee it). I did not finish the exercise, but the set of possible $k$ contains the interval $]0, 5[$.
A: No worries. Let's keep it simple. Now, if the equation has four real roots, it will cut the $x$-axis four times. Like this,

(This graph is not of given equation but general.)
So, from the graph we can conclude two facts:

*

*Value $f'(x)$ of the graph should be zero three times (say at $x=a,\,b,\,c$; $a<b<c$).

*$f(a)\le 0,\,f(b)\ge 0,\,f(c)\le 0$.

By calculating all these stuffs you get is $k\in[0,5]$.
