$15a+6b+4c+8d=0$ implies $ax^3+bx^2+cx+d$ has a positive root Let $a,b,c,d$ be real numbers such that $15a+6b+4c+8d=0$. Show that $f(x)=ax^3+bx^2+cx+d$ has a positive root. (Komal, Problem N. 170.)
I want to try to use the intermediate value theorem, showing that $f(s)<0$ and $f(t)>0$ for some $s,t>0$, since this will imply that $f(u)=0$ for some $u>0$. But the numbers $15,6,4,8$ do not yield easily to plugging in appropriate values of $x$.

You can use Descartes' Rule of Signs which states that 
An equation f(x) = 0 cannot have more positive roots than there are changes of sign in f (x), and cannot have more negative roots than there are changes of sign in f (— x).
Since $15a+6b+4c+8d=0$ this implies that there is atleast one negative number and one positive number in the set ${a,b,c,d}$. This implies that the original cubic will have atleast one change in sign which implies that it can have one positive root. 
