Question 630285
First of all, please put parentheses around multiple term numerators and denominators. What you posted means:
{{{6x^5 - 48x^2/12x^3 + 24x^2 + 48x}}}
which I'm pretty sure is not the right expression. Making your expressions clear and not making the tutors try to figure out what you mean will result in faster responses.<br>
{{{(6x^5 - 48x^2)/(12x^3 + 24x^2 + 48x)}}}
Reducing fractions is, as it's always been, a matter of finding and canceling any factors that are common to the numerator and denominator. So we start by figuring out what the factors are.<br>
We'll start by factoring the numerator:
{{{6x^5-48x^2}}}
When factoring, always start by factoring out the greatest common factor (GCF), unless it is a 1 which is rarely factored out. The GCF here is {{{6^x^2}}}:
{{{6x^2(x^3-8)}}}
After the GCF, there are a variety of factoring techniques one can try. One of the methods is factoring by patterns and one of these patterns is:
{{{a^3-b^3 = (a-b)(a^2+ab+b^2)}}}
The factor {{{x^3-8}}} fits that pattern because {{{x^3}}} is clearly a perfect cube and, since {{{8 = 2^3}}}, so is 8. Using this pattern with an "a" of "x" and a "b" of "2" we get:
{{{6x^2(x-2)(x^2+x*2+2^2)}}}
which simplifies to
{{{6x^2(x-2)(x^2+2x+4)}}}
None of these factors will factor any further so we are finished factoring the numerator. For reasons you'll see later, I'm going to rewrite {{{6x^2}}} as 6*x*x:
{{{6*x*x*(x-2)(x^2+2x+4)}}}<br>
Now we'll factor the denominator:
{{{12x^3 + 24x^2 + 48x}}}
First the GCF of 12x:
{{{12x(x^2 + 2x + 4)}}}
None of these factors will factor further. Again for reasons you'll see shortly, I'm going to rewrite 12x as 6*2*x:
{{{6*2*x(x^2 + 2x + 4)}}}<br>
Now let's rewrite the fraction with the factored numerator and denominator:
{{{(6*x*x*(x-2)(x^2+2x+4))/(6*2*x*(x^2 + 2x + 4))}}}
Looking at this you should be able to see some factors that are common to the numerator and denominator (and see why I rewrote {{{6x^2}}} and 12x). These common factor can be canceled:
{{{(cross(6)*cross(x)*x*(x-2)cross((x^2+2x+4)))/(cross(6)*2*cross(x)*cross((x^2 + 2x + 4)))}}}
leaving:
{{{(x*(x-2))/2}}}
(Note: The 2's <i>do not cancel!</i> Only factors can be canceled and the 2 in the numerator is not a factor.) This may be an acceptable answer. Or you could multiply out the numerator:
{{{(x^2-2x)/2}}}