So, for example, if I had a polynomial of order $5$, such as $$(10x^5 -3x^3 + 120x - 30) = 0$$ how would I figure out exactly how many real roots it has?
I understand that it can have maximum $5$ real roots, and it can have $3$ or $1$ real roots, as complex numbers come in conjugate pairs however I cannot figure out how to the exact number.
I would need to do this without a calculator.
Thank you.
EDIT: Apologies, I gave the wrong powers, this has now been amended
On
You have a general method: Sturm's algorithm, based on this result:
Let $P$ be a polynomial in $\mathbf R[x]$. Define a (finite) sequence of polynomials $P_n$ with $$P_0=P,\quad P_1=P',\qquad P_i=-(P_{i-2}\bmod P_{i-1})\quad(i\ge 2)$$ As the degrees are decreasing, we'll ultimately obtain the null polynomial.The sequence stops at the last nonzero polynomial $P_m$.
Let $[a,b]$ an interval such that $P(a)$ and $P(b)\ne 0$, and denote $\sigma(\xi)$ the number of changes of sign in the sequence $P(\xi), P_1(\xi), P_2(\xi),\dots, P_m(\xi)$. The number of distinct real roots of $P$ in the interval $[a,b]$ is equal to $$\sigma(a)-\sigma(b).$$
You also have to find bounds for the roots. If the polynomial is, explicitly, $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\qquad(a_n\ne 0),$$ a rough bound is $$\max\biggl(1,\frac{|a_0|+|a_1|+\dots+|a_{n-1}|+|a_n|}{|a_n|}\biggr).$$
^2$$f(x)=10x^5-3x^3+120x-30\implies f'(x)=50x^4-9x^2+120= 50y^2-9y+120>0,$$ as $B^2<4AC.~$ So this polynomial is an increasing function it can have at most one real root. Also it being odd ordered $f(-\infty)<0, f(\infty)>0$. So by intermediate value theorem it will have one real root. So the given polynomial equation ha only one real root.