# I need help proving a statement about rational roots

I have no idea where to start...this is the statement:

If a polynomial of degree not greater than 5 with rational coefficients has multiple roots, it has also a rational root, except in the case when the degree is 4 and the polynomial is a perfect square.

Any help would be appreciated.

• It seems like the statement is false? Take $x^2 + 1.$ It has rational coefficients, degree no greater than $5$, and has multiple roots, but none rational ($\pm i$). It's neither degree $4$ nor a perfect square. – John Jul 6 '14 at 20:41
• But it doesn't have multiple roots, one is $i$ and the other $-i$ – CIJ Jul 6 '14 at 20:51
• Ahh, OK. "Has multiple roots" means "has roots with multiplicity greater than one," not "has more than one root." That's where I was confused. – John Jul 6 '14 at 22:20

Outline: Suppose the polynomial $P(x)$ is irreducible over the rationals. Then there are no multiple roots. For if $\alpha$ is a multiple root of $P(x)$, then $\alpha$ is a root of $P'(x)$, and hence of $\gcd(P(x),P'(x))$. Thus the gcd $D(x)$ divides $P(x)$.
If the degree of $P(x)$ is $\le 3$, reducibility forces a rational root. The only way we could fail to have a rational root is if $P(x)$ is a product of two irreducible quadratics, or the product of an irreducible quadratic and an irreducible cubic.
If $P(x)$ is the product of two irreducible quadratics, each with roots $\alpha$ and $\beta$, then $P(x)$ is almost a perfect square. (The problem has a mistake: $2(x^2+1)^2$ and $-(x^2+1)^2$ are not perfect squares. But $P(x)$ must be a constant times a perfect square.)
Finally, we show that irreducible quadratic and irreducible cubic is impossible. For let $\alpha$ be a root of the irreducible quadratic. Then since $\alpha$ is a root of the cubic, it is a root of their gcd. Thus the quadratic divides the cubic.