# Find coefficients so that polynomial has at least one rational root

I have the following problem:

Given $P(X) = X^5 + 15aX^4 + 12bX^3 -18X^2 -1$

• Find $a,b \in \Bbb Z$ so that $p$ has at least one rational root.
• Prove that for any $a, b$ the rational root is unique and simple.

I did the following:

By the Rational Roots Theorem I know that if $r = \frac{p}{q} \Rightarrow p|-1$ and $q|1 \Rightarrow |\frac{p}{q}| = 1$

Then, by replacing in $P$, I get that either $(a,b) = (a, \frac{3}{2} - \frac{5}{4}a)$ or $a = (a, -\frac{5}{3} + \frac{5}{4}a)$. For each pair a rational root would be 1 or -1 respectively.

Would that be correct? Or am I not taking other solutions into consideration?

If it is, I was thinking of deriving each polynomial and checking that 1 or -1 is not a root (proving that it is a root of multiplicity one). But what would be the simplest way of checking that it is the only rational root?

Thanks a lot.

By the Rational Roots Theorem, $1$ and $-1$ are the only possible rational roots. For a rational root must have shape $\frac{p}{q}$ where $p$ divides $1$ and $q$ divides $1$. So the only possibilities are $p=1$ or $p=-1$, $q=1$ or $q=-1$.

Now we look at conditions on $a$ and $b$.

Plug in $1$. It is a root precisely if $15a+12b=18$.

Plug in $-1$. It is a root precisely if $15a-12b=20$. This has no solution in integers. For if $a$ and $b$ are integers, the left side is divisible by $3$. But the right side is not. So there can never be a rational root other than $1$.

It should not be hard, by experimentation, to find integers $a$ and $b$ such that $15a+12b=18$. It is easier if we rewrite as $5a+4b=6$.

As to multiple roots, $1$ (the only candidate) is a root of multiplicity $\gt 1$ precisely if the derivative is equal to $0$ at $1$. That will give you an equation in $a$ and $b$ to join $15a+12b=18$.

• Great. I was starting it differently. Thank you very much, André! – John Feb 18 '14 at 22:53
• You are welcome. There are a few things left to do. Uniqueness is taken care of, multiple root not quite. You should find that the two linear equations in $a$ and $b$ have no common integer root. – André Nicolas Feb 19 '14 at 0:00