# Ask Uncle Colin: A Four-Rooted Quartic

Dear Uncle Colin,

The remainder theorem of my textbook wants me to find the set of values of $k$ such that $3x^4 + 4x^3 - 12x^2 + k = 0$ has four distinct real roots, and I don’t know where to start!

- Quite Ugly Algebraic Relation: Targeting Its Constant

Hi, QUARTIC, and thanks for your message!

I know this is in the remainder theorem part, but it’s really a calculus question. So far as I can tell, anyway. The trick is to find the stationary points so you can sketch the graph (for arbitrary $k$) and see where the axis needs to go to get four solutions.

Considering the function $f(x) = 3x^4 + 4x^3 - 12x^2 + k$, you can differentiate to get $f’(x) = 12x^3 + 12x^2 - 24x$, or $f’(x) = 12x(x+2)(x-1)$. Clearly, there are stationary points when $x$ is -2, 0 or 1.

We *could* differentiate again and justify that they’re a minimum, a maximum and a minimum in that order, but the graph of $y=f(x)$ is a positive quartic, so it behaves much like a positive quadratic and has to go down, up, down, up.

Alternatively, we can find the values of the function at these points: the relevant points on the graph are $(-2, k-32)$, $(0,k)$ and $(1, k-5)$ and note that the middle one is higher up than the others.

To have four solutions, the graph must meet the $x$-axis between the upper minimum at $(1,k-5)$ and the maximum at $(0,k)$. That means $0 < k < 5$.

Hope that helps!

- Uncle Colin