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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?

Rodrigo de Azevedo
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Aiden Chow
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3 Answers3

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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.

amd
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  • My sincere compliments. I not know the Sturm chain for the polynomials. – Sebastiano Jan 18 '20 at 23:41
  • Sorry, I haven't learned calculus yet so I don't understand how to do derivatives. Also, I don't know what negative rem is. Can you please explain? Thanks. – Aiden Chow Jan 19 '20 at 18:46
  • Never mind, I figured out what rem meant. But I did a bit of research and I still can't understand how to do Euclidean division. Can someone help? – Aiden Chow Jan 19 '20 at 18:57
  • @AidenChow Do you know how to divide one polynomial by another? It’s really the same process as the “long division” that you were taught in elementary school. – amd Jan 19 '20 at 23:01
  • Oh that's the name for polynomial division. I never heard of it called that way. Thanks! – Aiden Chow Jan 20 '20 at 19:15
  • You could’ve simply looked up “Euclidean division of polynomials.” That takes you right to the “Polynomial long division” article in Wikipedia. – amd Jan 20 '20 at 19:43
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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.

Izaak van Dongen
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    Someone before stated that i should use descartes' rule of signs. That was what I was referring to. I think he/she deleted their comment. Thanks for helping. – Aiden Chow Jan 18 '20 at 20:21
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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.

dxiv
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