Editing Factorization (section)

===Rational roots===
For polynomials with rational number coefficients, one may search for roots which are rational numbers. Primitive part-content factorization (see [[#Primitive part-content factorization|above]]) reduces the problem of searching for rational roots to the case of polynomials with integer coefficients having no non-trivial [[greatest common divisor|common divisor]].

If <math>x=\tfrac pq</math> is a rational root of such a polynomial 
:<math>P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n,</math>
the factor theorem shows that one has a factorization
:<math>P(x)=(qx-p)Q(x),</math>
where both factors have integer coefficients (the fact that {{mvar|Q}} has integer coefficients results from the above formula for the quotient of {{math|''P''(''x'')}} by <math>x-p/q</math>).

Comparing the coefficients of degree {{mvar|n}} and the constant coefficients in the above equality shows that, if <math>\tfrac pq</math> is a rational root in [[reduced fraction|reduced form]], then {{mvar|q}} is a divisor of <math>a_0,</math> and {{mvar|p}} is a divisor of <math>a_n.</math> Therefore, there is a finite number of possibilities for {{mvar|p}} and {{mvar|q}}, which can be systematically examined.<ref>{{harvnb|Dickson|1922|p=27}}</ref>

For example, if the polynomial 
:<math>P(x)=2x^3 - 7x^2 + 10x - 6</math>
has a rational root <math>\tfrac pq</math> with {{math|''q'' > 0}}, then {{mvar|p}} must divide 6; that is <math>p\in\{\pm 1,\pm 2,\pm3, \pm 6\}, </math> and {{mvar|q}} must divide 2, that is <math>q\in\{1, 2\}. </math> Moreover, if {{math|''x'' < 0}}, all terms of the polynomial are negative, and, therefore, a root cannot be negative. That is, one must have 
:<math>\tfrac pq \in \{1, 2, 3, 6, \tfrac 12, \tfrac 32\}.</math>
A direct computation shows that only <math>\tfrac 32</math> is a root, so there can be no other rational root. Applying the factor theorem leads finally to the factorization
<math>2x^3 - 7x^2 + 10x - 6 = (2x -3)(x^2 -2x + 2).</math>

====Quadratic ac method====
The above method may be adapted for [[quadratic polynomial]]s, leading to the ''ac method'' of factorization.<ref>{{cite web|last=Stover|first=Christopher|url=http://mathworld.wolfram.com/ACMethod.html|title=AC Method|work=Mathworld|archive-url=https://web.archive.org/web/20141112231252/http://mathworld.wolfram.com/ACMethod.html|archive-date=2014-11-12|url-status=live|mode=cs2}}</ref>

Consider the quadratic polynomial 
:<math>P(x)=ax^2 + bx + c</math>
with integer coefficients. If it has a rational root, its denominator must divide {{math|''a''}} evenly and it may be written as a possibly [[irreducible fraction|reducible fraction]] <math>r_1 = \tfrac ra.</math> By [[Vieta's formulas]], the other root <math>r_2</math> is
:<math>r_2 = -\frac ba - r_1 = -\frac ba-\frac ra =-\frac{b+r}a = \frac sa,</math>
with <math>s=-(b+r).</math>
Thus the second root is also rational, and Vieta's second formula <math>r_1 r_2=\frac ca</math> gives
:<math>\frac sa\frac ra =\frac ca,</math>
that is 
:<math>rs=ac\quad \text{and}\quad r+s=-b.</math>
Checking all pairs of integers whose product is {{math|''ac''}} gives the rational roots, if any.

In summary, if <math>ax^2 +bx+c</math> has rational roots there are integers {{mvar|r}} and {{mvar|s}} such <math>rs=ac</math> and <math>r+s=-b</math> (a finite number of cases to test), and the roots are <math>\tfrac ra</math> and <math>\tfrac sa.</math> In other words, one has the factorization
:<math>a(ax^2+bx+c) = (ax-r)(ax-s).</math>

For example, let consider the quadratic polynomial
:<math>6x^2 + 13x + 6.</math>
Inspection of the factors of {{math|1=''ac'' = 36}} leads to {{math|1=4 + 9 = 13 = ''b''}}, giving the two roots 
:<math>r_1 = -\frac 46 =-\frac 23 \quad \text{and} \quad r_2 = -\frac96 = -\frac 32,</math>

and the factorization 
:<math>
6x^2 + 13x + 6 = 6(x+\tfrac 23)(x+\tfrac 32)= (3x+2)(2x+3).
</math>