Editing Rational root theorem (section)

==Examples==

===First===

In the polynomial
<math display="block">2x^3+x-1,</math>
any rational root fully reduced should have a numerator that divides 1 and a denominator that divides 2. Hence the only possible rational roots are ±1/2 and ±1; since neither of these equates the polynomial to zero, it has no rational roots.

===Second===
In the polynomial
<math display="block">x^3-7x+6</math>
the only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ±1, ±2, ±3, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots (in fact these are its only roots since a cubic polynomial has only three roots).

===Third===

Every rational root of the polynomial
<math display="block">P=3x^3 - 5x^2 + 5x - 2 </math>
must be one of the 8 numbers
<math display="block">\pm 1, \pm2, \pm\tfrac{1}{3}, \pm \tfrac{2}{3} .</math>
These 8 possible values for {{mvar|x}} can be tested by evaluating the polynomial. It turns out there is exactly one rational root, which is <math display=inline>x=2/3.</math> 

However, these eight computations may be rather tedious, and some tricks allow to avoid some of them.

Firstly, if <math>x<0,</math> all terms of {{mvar|P}} become negative, and their sum cannot be 0; so, every root is positive, and a rational root must be one of the four values <math display=inline>1, 2, \tfrac{1}{3},  \tfrac{2}{3} .</math>

One has <math>P(1)=3-5+5-2=1.</math> So, {{math|1}} is not a root. Moreover, if one sets {{math|1=''x'' = 1 + ''t''}}, one gets without computation that <math>Q(t)=P(t+1)</math> is a polynomial in {{mvar|t}} with the same first coefficient {{math|3}} and constant term {{math|1}}.<ref>{{cite journal |last=King |first=Jeremy D. |title=Integer roots of polynomials |journal=Mathematical Gazette |volume=90 |date= November 2006 |pages=455–456 |doi=10.1017/S0025557200180295 |doi-access=free }}</ref> The rational root theorem implies thus that a rational root of {{mvar|Q}} must belong to <math display=inline>\{\pm1, \pm\frac 13 \},</math> and thus that the rational roots of {{mvar|P}} satisfy <math display=inline>x = 1+t \in \{2, 0, \tfrac{4}{3}, \tfrac{2}{3}\}.</math> This shows again that any rational root of {{mvar|P}} is positive, and the only remaining candidates are {{math|2}} and {{math|2/3}}.

To show that {{math|2}} is not a root, it suffices to remark that if <math>x=2,</math> then <math>3x^3</math> and <math>5x-2</math> are multiples of {{math|8}}, while <math>-5x^2</math> is not. So, their sum cannot be zero.

Finally, only <math>P(2/3)</math> needs to be computed to verify that it is a root of the polynomial.