TITLE: Finding roots of a function in an interval
QUESTION [6 upvotes]: Does the equation $x^3-12x+2=0$ have three solutions in the interval $[-4,4]$?

We know that this is a continuous function because it's a polynomial, and so we can use the Intermediate Value Theorem to do this problem:  

If $f(x)$ is continuous on $[a,b]$, let $M$ be any number between $f(a)$ and $f(b)$. Then there exists a number $c$ such that:   
1) $a<c<b$
  2) $f(c)=M$  

If we define $M=0$, $a=-4$ and $b=4$, then $-4<c<4$ and $f(-4)<0<f(4)$.  
Compute $f(-4)$ and $f(4)$ to show that $f(-4)<0<f(4)$
$\implies -14=f(-4)<0<f(4)<18$.  
So I've shown that $M$ is between $f(-4)$ and $f(4)$, and thus there must be some number $c$ such that $f(c)=M=0$.  
Is all of this correct?
How do I show that there are three solutions (roots) in this interval?
Thank you.

REPLY [5 votes]: No quadratic equation can have three roots in any interval.
I expect that the correct equation is $x^3-12x+2=0$, which does have three roots in $[-4,4]$ because $f(-4)<0$, $f(0)>0$, $f(1)<0$, $f(4)>0$. By the Intermediate Value Theorem, each sign variation gives you at least one root in that interval.