 # Show that a polynomial of degree 3 has at most Oberlaudacu 2021-12-19 Answered
Show that a polynomial of degree 3 has at most three real roots.
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Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on (a,b). Then if $f\left(a\right)=f\left(b\right)$ then there exists at least one point c in (a,b) for which ${f}^{\prime }\left(c\right)=0$
Let it be f(x) a cubic polynomial:
$f\left(x\right)=p{x}^{3}+q{x}^{2}+rx+s$
The domain of the function f(x) is $\mathbb{R}$, because f(x) is a polynomial.
So, f(x) is continuos and differentiable everywhere.
We need to show that f(x) has three real roots
Suppose f(x) has four real roots $a
Find the first derivative od the function $f\left(x\right)=p{x}^{3}+q{x}^{2}+rx+s$
${f}^{\prime }\left(x\right)={\left(p{x}^{3}+q{x}^{2}+rx+s\right)}^{\prime }$
$=3p{x}^{2}+2qx+r$
${f}^{\prime }\left(x\right)$ is a quadratic function and she can have no more than two real roots.
There is a contradiction, because:
By Roller's Theorem there must be three real numbers $m,n,k\because m\in \left(a,b\right),n\in \left(b,c\right)$ and $k\in \left(c,d\right)\therefore {f}^{\prime }\left(m\right)={f}^{\prime }\left(n\right)={f}^{\prime }\left(k\right)=0$
Hence, the function f(x) has at most three real roots.

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