Quartic function having at least two roots

Question

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?

Answer 1

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.

Answer 2

Hint:

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.

Answer 3

$x_0$ 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 $x_0$.