Prove that the equation $x^5 − 1102x^4 − 2015x = 0$ has at least three real roots.
So do I sub in values of negative and positive values of $x$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $x$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.
Best Answer
It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $ f(x) := x^4-1102x^3-2015$, we are done.
Observe, $f(0) < 0$ and $f(-2) > 0 $, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.
Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$