Roots of a quartic polynomial If $a, 3a, 5a, b, b + 3,$ and $b+5$ are all roots of a fourth-degree polynomial equation where $0<a<b$, compute all possible values of $a$.
By the fundamental theorem of algebra, the polynomial has at most 4 roots.
Therefore, some of the terms must be equal to one another and since $0<a<b$, $3a$ has to be equal to either $b, b+3,$ or $b+5$. The same reasoning holds for $5a$.
Can someone please help me? I don't know how to proceed from here.
 A: Since $a \ne0$, we know that $a$, $3a$ and $5a$ are distinct roots.
We can therefore say that there are four roots: $a$, $3a$, $5a$ and an unidentified $p$. We know that $a<3a<5a$, but we don't know where $p$ fits in this hierarchy.
We also know that three of the roots take the form $b$, $b+3$ and $b+5$, but we do not know which of the four roots listed already correspond to which of the new roots. We do know that $b<b+3<b+5$.
As has been said elsewhere, deal with this case by case:
Case 1.
$a=b+3$ and $3a=b+5$. This leads to a value of $a$, but does the value of $b$ satisfy the requirements?
Case 2.
$a=b+3$ and $5a=b+5$. This leads to a value of $a$, but does the value of $b$ satisfy the requirements?
Case 3.
$a=b+5$ can't work, because there would be two roots larger than $a$ and two roots smaller than $a$, giving too many roots.
Case 4.
$3a=b$ and $5a=b+5$. This leads to a value of $a$. Does everything work out?
Case 5.
$5a=b$. This also can't work, because there would be two roots larger than $5a$ and two roots smaller than $5a$, giving too many roots.
