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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einfachmoipf
Answered 2021-12-20 Author has 32 answers
Suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on (a,b). Then if f(a)=f(b) then there exists at least one point c in (a,b) for which f(c)=0
Let it be f(x) a cubic polynomial:
f(x)=px3+qx2+rx+s
The domain of the function f(x) is 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<b<c<df(a)=f(b)=f(c)=f(d)
Find the first derivative od the function f(x)=px3+qx2+rx+s
f(x)=(px3+qx2+rx+s)
=3px2+2qx+r
f(x) 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,km(a,b),n(b,c) and k(c,d)f(m)=f(n)=f(k)=0
Hence, the function f(x) has at most three real roots.

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