# Using Descartes rule of signs to determine number of real roots of a polynomial

$f(x)=x^5+2x^4-12x^3+134x^2-440=0$

To determine number of real roots of equation, using Descartes rule of signs ,

Sign changes in $f(x)=3$

Sign changes in $f(-x)=2$

Number of Positive roots $=$ either $1,3$

Number of Negative roots $=$ either $2,0$

This gives 4 possible combinations . How can I be able to determine the combination which results in exact answer ?

Is there an alternate way ?

• in These intervals $$[[-{\frac{3003}{512}},-{\frac{6005}{1024}}],[-{\frac{1827}{1024}},-{ \frac{913}{512}}],[{\frac{1801}{1024}},{\frac{901}{512}}]]$$ – Dr. Sonnhard Graubner Oct 3 '17 at 16:43

$f'(x)=x(5x^3+8x^2-36x+268)$
$5x^3+8x^2-36x+268$ has a discriminant of $-5.5\times10^{7}$ so there are only $2$ real roots of $f'(x)=0$, including $x=0$.
$f'(-6)<0<f'(-4)$ so the other root of $f'(x)=0$ is in $(-6,-4)$. Then, $f(x)$ is strictly increasing over $(0,\infty)$ so $f(x)$ has only $1$ positive root.
Since $f(x)$ is an odd degree polynomial, $f(x)$ approaches $-\infty$ as $x$ approaches $-\infty$. But $f(-4)>0$ so there's at least $1$ negative root.