# Root of a polynomial with rational coefficients

I am currently learning about Direct Proofs. I am struggling trying to find a starting point to prove the Statement: For all real numbers $c$, if $c$ is a root of a polynomial with rational coefficients, then c is a root of a polynomial with integer coefficients. Based on a definition given by the book and the professor: A number is $c$ is called a root of a polynomial $p(x)$ if, and only if, $p(c) = 0$. But how can I prove the veracity of this statement using the given definition?

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Hint: use the lcm of the denominators of the coefficients of the polynominal –  Stefan Feb 20 '13 at 19:31

Example (as a hint):

$$P(x) = \frac{3}{2}x^2 + 2x - \frac{1}{3}$$

Multiply by 6:

$$6P(x) = \frac{18}{2}x^2 + 12x - \frac{6}{3}$$ $$6P(x) = 9x^2 + 12x - 2$$

If $c$ is a zero of $P(x)$, then $6P(c) = 0$, so $0 = 9c^2+12c-2$, and so $c$ is a zero of of a polynomial with integer coefficients.

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Any polynomial with rational coefficients can be made into a polynomial with integer coefficients by multiplying through by the product of the denominators of the coefficients (or the LCM of those denominators). All that remains, then, is for you to show that if a number $c$ is the root of a polynomial, then it is also the root of any polynomial formed by multiplying that polynomial by a constant.

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