Proof that a degree 4 polynomial has at least two roots 
Let $$P(x) = x^4+a_3x^3+a_2x^2+a_1x+a_0$$
  $$P(x_0) = 0$$
  $$P'(x_0) \not= 0$$
  with $x_0$ and each $a_i$ real. Prove that $P(x)$ has a at least two real roots.

I can't figure why this is true.
 A: Hint: have you tried sketching possible shapes for the polynomial, noting that the $x^4$ term dominates when $|x|$ is large.

 Hint: can you show that there is a value of $x$ for which the polynomial takes a negative value? Then use the intermediate value theorem.

A: You know that $P(x_0)=0$. Thus you can write $P(x)=(x-x_0)\cdot g(x)$ where $g$  is a polynom of odd degree. Thus, by mean value theorem, there is $r$ such that $g(r)=0$. If $x_0=r$ you can write $P(x)=(x-x_0)^2 \cdot h(x)$. But in this case you get $P'(x_0)=0$, so it is a contradiction and you get $r\neq x_0 $ and you get at least two real roots.
A: By the hyphotesis $x_0$ is a simple root of $P(x)$; $i.e.$ $(x-x_0)^2$ do not divided $P(x)$ because $P'(x)\neq0$. Then you can decompose $P(x)$ as follow
$$ P(x)=(x-x_0)(x^3+b_1x^2+b_2x+b_3)=(x-x_0)q(x).$$
Now $q(x)$ is a polynomial of degree $3$ with real coefficients; then there exists another real number $x_1$ such that $P(x_1)=0$; otherwise $P(x)$ needs to have $3$ complex roots; and this is not possible. Then $P(x)$ has al least two real different roots;
$$(x-x_0)(x-x_1)q'(x),$$ where $q'(x)$ is a polynomial of degree $2$. Now: if the discrimant $\Delta$ of $q'(x)$ is greater than $0$ you have four real roots (one of them could be equal to $x_1$), if $\Delta=0$ you have another two real roots $x_2=x_3$; if $\Delta<0$, $q'(x)$ has two conjugate complex roots; then $P(x)$ has only two real roots.
Notice that in general a polynomial with real coefficient of degree $4$ could not have real roots; for instance
$$P(x)=(x^2+1)(x^2+4);$$
so the roots $x_0,...x_3$ are complex and $x_0=\overline{x_1}$; $x_2=\overline{x_3}$; but if $P(x)$ has a real root; then it needs to has another one.
A: The idea is simple. Since coefficient of $x^{4}$ is $1$ it is clear that $P(x) \to \infty$ when $x \to \infty$ or $x \to -\infty$. Now we are given that $x_{0}$ is a root of $P(x)$ and we need to show that there is at least one more root of $P(x)$. Since $P'(x_{0}) \neq 0$, it follows that there is a point $c$ near $x_{0}$ for which $P(c) < 0$ (if $P'(x_{0}) > 0$ then $c < x_{0}$ and if $P'(x_{0}) < 0$ then $c > x_{0}$, moreover there is an interval full of such points $c$ but we just need one such $c$).
If $c < x_{0}$ then we can see that $P(c) < 0$ and $P(x) \to \infty$ as $x \to -\infty$ and hence by intermediate value theorem there is a root of $P(x)$ in $(-\infty, c)$. If $c > x_{0}$ then we can show in similar manner that there is a root of $P(x)$ in $(c, \infty)$.
A: First of all, assume by root, you mean real root here, and also assume all coefficients $a_i$ are real here.
No matter what, $P$ has 4 roots in $\mathbb{C}$, and since all $a_i$ are real, when $y$ is a root, $\bar{y}$ has to be a root too, where $\bar{y}$ is the conjugate of $y$. So $P$ can have 0,2,4 real roots, multiplicity counted. So only chance $P$ has a single real root is that this root's multiplicity larger than 1, in other words, $P(x)=(x-x_0)^2(x^2+bx+c)$. If this were true, $P'(x_0)=0$, which contradicts of what was given. So $P$ must have at least 2 real roots.
A: Since, the given polynomial has the real coefficients. So we know that if it has a complex root then its conjugate must be the root of the polynomial. Now in the given polynomial since $ P(x_0)=0, x_0\in \mathbb{R} $ also $ P'(x_0)\neq 0 $ which implies the multiplicity of $x_0$ is 1; $i.e.$ if $P(x)=(x-x_0)g(x) \implies g(x_0)\neq 0.$ the degree of $g(x)$ is 3 which is odd. So, $ g(x)$ must have a real root as the coefficients of $g(x)$ is also real. Hence, $P(x)$ has at least 2 real roots. 
