If $x_0$ is a real root of $p(x)=x^4+a_3 x^3 + a_2 x^2 +a_1 x + a_0$ and $p'(x_0) \ne 0$. Does $p(x)$ have at least two real roots?

Question

If $x_0$ is a real root of $p(x)=x^4+a_3 x^3 + a_2 x^2 +a_1 x + a_0$ and $p'(x_0) \ne 0$. Does $p(x)$ have at least two real roots?

I don't know what would be a good way to solve this. Any tips?

Edit: I'm in calculus 1 and this should my answer should probably not assume things about things from algebra about roots of polynomials.

Answer 1

Notice that $\lim_{x\to\infty}p(x)=\lim_{x\to-\infty}p(x)=\infty$. If $p(x)$ crosses $x$-axis once it must do it at least twice.

Answer 2

Since $x_0$ is a root of $p$ with $p'(x_0)\neq 0$, it has multiplicity $1$. Thus $p(x)=(x-x_0)q(x)$ where $q$ has degree $3$ and $x_0$ is not one of its root.

Since $q$ has degree $3$ it has a real root $x_1\neq x_0$, and you're done.

Answer 3

Complex roots are always complex conjugate pairs. Now since your polynomial is of degree 4, this means that it should have four roots. Since there are 3 remaining unknown roots ($x_0$ given to be real) there is only room for one pair should there be one therefore one of the other three roots must be real also. The condition $p'(x_0)\neq 0$ also rules out the possibility of the other root being equal to $x_0$ since the curve would not be tangent to the $x$-axis

Answer 4

There is several ways, for example:

• As $(x-x_0)$ divides $p$ you have: $$p(X)=(X-x_0) Q(X)$$ where $Q$ is a polynomial of degree $3$. But any polynomial of degree $3$ as a real root and $x_0$ is not a root of $Q$.
• As $p(x_0)=0$ and $p'(x_0)=0$ you know that the sign of $p$ change near $x_0$. Without loss of generality we can suppose that for some $\epsilon >0$, $p(x)< 0$ for $x \in (x_0-\epsilon,x_0)$. So $p(x_0-\epsilon)<0$ but $\lim_{x \to - \infty} p(x)=+\infty$. So as $p$ is continuous there exists $x_1 \in (-\infty,x_0-\epsilon)$ such that $p(x_1)=0$.

Answer 5

Here's a rather elementary proof. If $ z=a+bi \in \mathbb{C} $ we denote the complex conjugate by $ \overline{z} = a - bi $. We denote the roots of $ p(x) $ by $ x_0, x_1, x_2, x_3 $ where $ x_0 \in \mathbb{R} $ and $ x_i $ are the other zeros - possibly complex.

Now you can check for yourself that for any two complex numbers the following holds: \begin{align*} \overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2} \\ \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \end{align*} This means that for any complex number $ z $ we have \begin{align*} \overline{p(z)} = \overline{z^4 + a_3z^3 + a_2z^2 +a_1z+a_0} = \overline{z}^4 + a_3 \overline{z}^3 + a_2\overline{z}^2 + a_1\overline{z} +a_0 = p(\overline{z}) \end{align*} This implies that whenever we have any root $ z_0 $ of $ p $, then $ \overline{z_0} $ is also a root of $ p $. Hence complex roots come in pairs.

Now we can deduce your desired statement. If any of $ x_1, x_2, x_3 $ is real, we're done (because $ x_0 $ is a single root). If wlog $ x_3 $ is complex, so is wlog $ x_2 $. Now $ x_1 $ cannot be complex, because otherwise $ x_0 $ would also be comlex. Hence $ x_1 $ must be real. Since $ p'(x_0) \neq 0 $, $ x_0 $ must be a single root, in particular $ x_1 \neq x_0 $ is another real root of $ p $.