Show that a polynomial of degree 3 has at most three real roots.

Question

einfachmoipf · Accepted Answer

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) = p x^{3} + q x^{2} + r x + 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 < d ∴ f (a) = f (b) = f (c) = f (d)

Find the first derivative od the function

f (x) = p x^{3} + q x^{2} + r x + s

f^{'} (x) = {(p x^{3} + q x^{2} + r x + s)}^{'}

= 3 p x^{2} + 2 q x + 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, k ∵ m \in (a, b), n \in (b, c)

and

k \in (c, d) ∴ f^{'} (m) = f^{'} (n) = f^{'} (k) = 0

Hence, the function f(x) has at most three real roots.

Show that a polynomial of degree 3 has at most

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