Consider the quartic real polynomial function $P(x)=x^{4}+bx^{3}+cx^{2}+dx+e$.

Given that:

$$P(x_0) = 0$$
$$P'(x_0) \neq 0$$

prove that $P(x)$ has at least $2$ (distinct) roots.

I understood that because of the $x^{4}$ term, $P(x)$ is going to tend to $\infty$ when $x \to \pm \infty$ at either end.

Because of the facts of $P(x_0) = 0$ and $P'(x_0) \neq 0$, so $x_0$ is one root of the function but not its minimum or maximum point.

But how can I explain that there is another root?

  • $\begingroup$ With != you mean $\ne\;$ as in some progamming languages? $\ne$ is \ne in Latex. $\endgroup$ – gammatester Aug 17 '16 at 12:11
  • $\begingroup$ i edit the question according your comments $\endgroup$ – motis10 Aug 17 '16 at 12:20

Assuming real roots and and coefficients, and $P'(x_0)\ne 0\,$ you can write $$P(x)=(x-x_0)P_3(x)$$ where $P_3$ is a cubic with $P_3(x_0) \ne 0.$ Since this cubic has at least a real root $x_1 \ne x_0,\;$ you have at least two real roots.

  • $\begingroup$ Im not sure i can proof P(x) with assumption that p_3 without to proof it also. $\endgroup$ – motis10 Aug 17 '16 at 12:24
  • $\begingroup$ $P_3$ goes to $\pm \infty$. Use intermediate value theorem. $\endgroup$ – gammatester Aug 17 '16 at 12:26
  • $\begingroup$ @motis10: It is a worthwhile observation that any odd degree real polynomial has at least one real root. So in particular a cubic real polynomial has at one real root. This can be proved using the Intermediate Value Theorem. $\endgroup$ – hardmath Aug 17 '16 at 12:28
  • $\begingroup$ But first why this is true? $P(x)=(x−x0)P3(x)$ and how can i prove it using the Intermediate Value Theorem $\endgroup$ – motis10 Aug 17 '16 at 12:32
  • $\begingroup$ Use Taylor series for $P$ (or long division of polynomials). $P_3$ takes positive and negative values (from the asymptotics for $\infty$), therefore there is a zero. $\endgroup$ – gammatester Aug 17 '16 at 12:35


1) remember that if $x=\alpha$ is a complex root (not a real number) of a polynomial also the conjugate $x=\bar \alpha$ is a root.

2) note that $x_0$ is a root of your polynomial and from $P'(x_0)\ne 0$ we know that it is not a double root.

So, since the given polynomial has $4$ roots in $\mathbb{C}$, and one is real, at least one other root have to be real.

  • $\begingroup$ I didn't find the answer from the hint. :( $\endgroup$ – motis10 Aug 17 '16 at 12:43
  • $\begingroup$ The non-real roots of a polynomial are always pairs of complex conjugate numbers, then a degree 4 polynomial can have a pair of complex roots and two real roots (possibly coincident) or two pairs of complex roots. Since in this case we know for sure that there is a real root (single), there must be another real root. $\endgroup$ – Emilio Novati Aug 17 '16 at 12:54

$x_0$ is not a double roots and the cubic polynomial $$Q(x)=\frac{P(x)}{x-x_0}\in \Bbb C[x]$$ has a zero, $x_1$, in $\Bbb C$ distinct of $x_0$.


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