# Number of real roots of a polynomial [duplicate]

This question already has an answer here:

Find the number of real roots of the polynomial $f(x)=x^5+x^3-2x+1$

( I think similar questions have already been asked. However, I am not sure whether the exact question has already been raised by some user. I urge the members of this mathematical community to mark my question as duplicate if it has already been asked. I tried an attempt below:-)

What I attempted:- I know a little bit of theory of equations but not all. While I was in the examination, I did not ponder much more on any theorems relating to real roots. Rather I tried to do it with some intuition.
We have $f(1)=1$,$f(0)=1$,$f(2)=1$, $f(\frac{1}{2})=\frac{5}{32}$,$f(-\frac{1}{2})=\frac{59}{32}$,$f(-2)=-35$
It just helped me to trace a root between $-1$ and $-2$. Then I tried to study the nature of the graph.

Here $f'(x)=5x^4+3x^4-2$
Let $f'(x)=5x^4+3x^2-2=0$
$\Rightarrow 5y^2+3y-2=0$
(where $x^2=y$)
$\Rightarrow (5y-2)(y+1)=0$
Therefore, $y=\frac{2}{5}$, or
$y=-1$ (It is not of our interest, as we are dealing with real roots)
Therefore $x=\sqrt{\frac{2}{5}},-\sqrt{\frac{2}{5}}$
These two values divides the real line into three disjoint subsets $(-\infty,-\sqrt{\frac{2}{5}})$, $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$

It s quite easy to notice that $f$ is decreasing in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ and increasing in $(-\infty,-\sqrt{\frac{2}{5}})$, $(\sqrt{\frac{2}{5}},\infty)$

Again, $f(-\sqrt{\frac{2}{5}})=1.911$ and $f(\sqrt{\frac{2}{5}})=0.089$

Looking at all these details it is clear that $f(x)$ comes close to zero in $(-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})$ but is not equal to zero. There is no possibility for $f(x)$ to be zero in $(\sqrt{\frac{2}{5}},\infty)$ as it is increasing here. Again, we have already discovered one root between $-1$ and $-2$. Apart from that there is of course no root as $f$ is again increasing in $(-\infty,-\sqrt{\frac{2}{5}})$.
So, te graph should look something like this

Therefore there is only one real root of the polynomial.

Is my approach correct? Is there any way to do the same thing using some more beautiful as well as advanced method ? (I have just made a rough plot of the graph in my answer sheet). I am not sure if the examiner would expect much better method than this one.

## marked as duplicate by Saad, Arnaud D., user440191, Martin Sleziak, hardmathMay 14 '18 at 10:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• $f'(x) = 5 x^4 + 3 x^2 - 2$, not $5 x^4 + 3 x^4 - 2$. – Robert Israel May 14 '18 at 2:51
• The positive value of $y$ is $y=2/5$, not $5/2$, although in subsequent lines you have used the correct value $2/5$. It appears to be only a typo,...Your analysis is correct and it is as good a method as any other, if there is any other. – DanielWainfleet May 14 '18 at 3:29
• Thanks Robert Israel and DanielWainfleet for pointing the errors. – user440191 May 14 '18 at 5:29
• For fifth degree polynomial there are 5 roots, out of which one is a real root and two complex conjugates. How do you account for the two missing roots? Are the latter complex roots repeated? – Narasimham May 14 '18 at 6:17

## 2 Answers

Sturm's theorem is a very powerful tool used to count real roots, but may be overkill here since your method also works and is arguably simpler. We take $f(x)$ and $f^\prime(x)$ and then compute successive (negated) remainders $r_1(x),r_2(x),\dots$ of polynomial long division (similar to the Euclidean algorithm) to get the sequence of polynomials $f(x), f^\prime(x), r_1(x), \dots$.

So if $f(x) = x^5 + x^3 - 2x +1$ and $f^\prime(x) = 5x^4 + 3x^2 - 2$, then $$f(x) = \frac{x}{5} \cdot f^\prime(x) - (-\frac{2}{5}x^3+\frac{8}{5}x - 1)$$ $$f^\prime(x) = (-\frac{25}{2}x)(-\frac{2}{5}x^3 + \frac{8}{5}x - 1) - (-23x^2 + \frac{25}{2}x + 2)$$ $$-\frac{2}{5}x^3 + \frac{8}{5}x - 1 = (\frac{2}{115}x + \frac{5}{529})(-23x^2 + \frac{25}{2}x + 2) - (-\frac{1531}{1058}x + \frac{539}{529})$$ $$-23x^2 + \frac{25}{2}x + 2 = (\frac{24334}{1531}x+\frac{5984577}{2343961})(-\frac{1531}{1058}x + \frac{539}{529}) - (+\frac{1409785}{2343961})$$

Now, for really large numbers, this sequence of polynomials gives the signs $+,+,-,-,-,+$, which has $2$ sign changes. For really large negative numbers, the sequence has the signs $-,+,+,-,+,+$, which has $3$ sign changes. The single additional sign change means that $f(x)$ has a single real root.

Since its a fifth degree equation it will have five roots if it has one complex root then it must have its congugate too as a root thus the number of real roots can only be one or three. you can find a real rrot using newton-Raphson method and check the remaining equation.