Question 724943
{{{(x^2-5x-4)/(-3x^3+15x^2-12)}}}
The numerator will not factor. But the denominator will. First the greatest common factor (GCF). The GCF is 3. But for reasons I will explain shortly, I am going to factor out -3 instead:
{{{(x^2-5x-4)/((-3)(x^3-5x^2+4))}}}
By factoring out a -3, I have a positive leading coefficient in {{{x^3-5x^2+4}}}. Factoring with a positive leading coefficient is easier IMHO. Now we factor {{{x^3-5x^2+4}}}. This has too many terms for the factoring patterns or for trinomial factoring. Since I don't see how to factor this by grouping, I will resort to the only method left: Factoring by trial and error of the possible rational roots.<br>
Fortunately, 1 is always a possible rational root and 1 actually turns out to be a root:
<pre>
1  |   1   -5   0   4   <== Note the zero for the missing x term
----        1  -4  -4
      ----------------
       1   -4  -4   0
</pre>With a remainder (in the lower right corner) of zero, 1 is a root and (x-1) is a factor. Also, the rest of the bottom row tells us what the other factor is. The "1 -4 -4" translates into {{{x^2-4x-4}}}. So now our denominator is:
{{{(x^2-5x-4)/((-3)(x-1)(x^2-4x-4))}}}
The {{{x^2-4x-4}}} factor will not factor further so we are finished factoring the denominator.<br>
With the fraction fully factored we can see that there are no common factors we could cancel. So the fraction will not reduce/simplify.<br>
P.S. If the numerator had been {{{x^2-5x+4}}} then it would factor and we could reduce the fraction:
{{{((x-4)(x-1))/((-3)(x-1)(x^2-4x-4))}}}
{{{((x-4)cross((x-1)))/((-3)cross((x-1))(x^2-4x-4))}}}
{{{(x-4)/((-3)(x^2-4x-4))}}}
or
{{{(x-4)/(-3x^2+12x+12)}}}