Real Roots and Differentiation 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.
 A: 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 $f=0$
A: It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.
A: You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each.
First observe that $x=0$ is a root for:
$f(x)=x^5 − 1102x^4 − 2015x$
Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root.
third, count negative real roots by counting sign changes in $f(-x)$ where:
$f(-x)=-x^5 - 1102x^4 + 2015x$ 
here we have the signs: (--)(-+), so we have 1 negative root.
From the above, we have 3 real roots for $f(x)$.
A: $$x^5 − 1102 \cdot x^4 − 2015 \cdot x = 0$$
This factors into,
$$\left( x^4 − 1102x^3 − 2015 \right) \cdot x = 0$$
Therefore $x=0$ is a root. Keep simplifying,
$$x^4 − 1102x^3 − 2015=0$$
Set this equal to $f(x)$,
$$f(x)=x^4 − 1102x^3 − 2015$$
$f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between $-2$ and $-1$.
The same thing holds for $f(1100)$ and $f(1110)$
Now, in the spirit of fairness, let's quickly come up with the method that will allow us to "guess" where the roots of $f(x)$ lie. We'll use Newton's Method.
We get,
$$x_{n+1}=x_{n}-{{x^4 − 1102x^3 − 2015} \over {4 \cdot x^3-2204 \cdot x^2}}$$
Now, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.
$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)
A: The Newton polygon tells us that the dominant binomials are 


*

*$x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and

*$-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\frac{2015}{1102}\approx (-1.22282549628)^3$.


Single real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to http://www.akiti.ca/PolyRootRe.html):
 0
 0.6111860961336238   +   1.0593896464200445 i
 0.6111860961336238   -   1.0593896464200445 i
 -1.2223736979388697
 1102.0000015056714

