# Proof: Cubic Polynomial with Rational Coefficients and Roots

Can some explain this Proposition and proof, please? Maybe breakdown it down for me?

Information for the proposition and proof: Let us call polynomials with integer (respectively, rational) coefficients and roots Z-polynomials (respectively, Q-polynomials). To get the complete description of all cubic Q-polynomials $$h$$ with rational roots of $$h'$$ and $$h''$$ we need the following result of algebraic folk-lore: for a polynomial $$f$$ and number α $$\neq$$ 0, define a new polynomial fα by $$f_{α}(x)$$ = $$f(αx)$$,then $$r$$ is a root of $$f$$ , if and only if $$r$$ α is a root of fα. Then the following proposition is the key to a full description:

Proposition. $$f$$ is a Q-polynomial, if and only if there are rational numbers α, β $$\neq$$ 0 such that $$βf_{α}$$ is a monic Z -polynomial.

Proof. Suppose $$f$$ is a Q-polynomial of degree n. Let γ be the least common multiple of the denominators of all of its roots $$x_{1}$$, $$x_{2}$$,..., $$x_{n}$$. Let α = $$\frac{1}{γ}$$ . By folk-lore, $$f_{α}$$ is a polynomial with integer roots $$γx_{1}$$, $$γx_{2}$$,...,$$γx_{n}$$, and is therefore a rational multiple of a monic Z-polynomial. The converse is clear

Hint: I worked out a numerical example. Hope it helps.

Suppose we are given the polynomial

$$f(x)=\frac{7}{11}\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right)\left(x-\frac{1}{5}\right)$$

note that the least common multiple of the denominators of the roots is $$30$$,

therefore we replace $$x$$ with $$\frac{x}{30}$$ and factor the resulting polynomial

$$f\left(\frac{x}{30}\right)=\frac{7}{11}\frac{1}{30^3}(x-15)(x-10)(x-6)$$

this shows the integer roots mentioned in the proposition.