# How does the rational root test work?

I've doing some research in wikipedia, where appears this example:$$3x^{3} - 5x^2 + 5x - 2 = 0$$the rational solution must be among the numbers symbolically indicated by: $$±\frac{1,2}{1,3}$$

So I get that the rational solution must be among $±\frac{2}{3}$; but, why it must be among $±\frac{1}{1}$?

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The numbers with denominator $1$ are also rational numbers. Example: Use the method on $2x^2 + 5x+3$. If you leave out checking numbers with denominator $1$, you will miss the rational root $-1$. – André Nicolas Oct 28 '11 at 18:34
@AndréNicolas know of a well written proof of this? Having trouble finding one. – yiyi Sep 1 '12 at 16:03
@MaoYiyi: Of the Rational Root Theorem? Don't have a reference handy, must be in many places. I suggest asking for a proof on MSE. The proof only takes a paragraph (or two, for me). – André Nicolas Sep 1 '12 at 17:50

If you have a factor $ax+b$ that divides your polynomial, with $a,b$ integers, then $a$ must divide the leading term and $b$ must divide the constant term. But $1$ divides any number. So in your example, $a$ can be any of $-3,-1,1,3$ as these all divide $3$, which is the coefficient of $x^3$. Similarly, $b$ can be any of $-2,-1,1,2$ as these all divide the constant. So rational solutions can be any of $\pm 1, \pm 1/3, \pm2/3, \pm 2$