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 ?