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 ?

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


$5x^3+8x^2-36x+268$ 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.


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