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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

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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.

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