How to determine number of roots and what type for quartic equations?

Question

I want to figure out the number of roots and their types in the quartic equation $$x^4-34x^2-x+272=0$$ without actually solving it. Is there such a way to do so?

Answer 1

With this particular equation, you can let $f(x) = x^4 - 34x^2 - x + 272$ (this is just for convenience), and then calculate:

\begin{align*} f(-5) &= 52 \\ f(-4) &= -12 \\ f(0) &= 27 \\ f(4) &= -20 \\ f(5) &= 42 \end{align*}

Since this function should look "continuous" when you draw it, and it changes sign between each of -5, -4, 0, 4, 5, this implies that there is a root in between any adjacent pair of each of these numbers, so we have at least two negative real roots and two positive real roots. I suspect that's what your sign changing business is supposed to mean.

Answer 2

Descartes’ rule of signs tells us that there are either zero or two positive roots and either zero or two negative roots. That’s not much help. We can, however, examine the Sturm chain for this polynomial for a more definitive answer.

With $P=x^4-34x^2-x+272$, we have $$ P_0 = P = x^4-34x^2-x+272 \\ P_1 = P' = 4x^3-68x-1 \\ P_2 = -\operatorname{rem}(P0,P1) = 17x^2+\frac34x-272 \\ P_3 = -\operatorname{rem}(P1,P2) = \frac{4615}{1156}x+\frac{65}{17} \\ P_4 = -\operatorname{rem}(P2,P3) = \frac{1296165}{5041}. $$

There are four sign changes at $-\infty$: $(+,-,+,-,+)$ and zero at $+\infty$: $(+,+,+,+,+)$, therefore the number of real roots is equal to $4-0=4$. Combined with the results from the rule of signs, two of those roots are negative and two are positive.

Answer 3

Before delving into more general methods, it may be worth first checking whether the quartic is perhaps reducible. This can be done relatively cheaply e.g. 1, 2, 3, and it turns out that:

$$ x^4-34x^2-x+272 = (x^2 - x - 17) (x^2 + x - 16) $$

Each quadratic factor has real roots, so the quartic has $\,4\,$ real roots.