Question 614963
{{{((2x^3+2x)/(x^2-2x))*((x-1)/(4x^2+4))}}}
Back in the "good old days", when fractions were just made of numbers, you learned:<ul><li>Fractions are reduced by canceling factors that are common to the numerator and the denominator.</li><li>When multiplying fractions...<ul><li>It is not only OK but it is a very good idea to cancel <i>before</i> you multiply</li><li>"Cross-canceling" is OK. Cross-canceling is when you cancel a factor of a numerator (or denominator) of one fraction with the same factor in the denominator (or numerator) of another fraction. (You'll see this shortly.)</li></ul></li></ul>The same is true with these fractions! In fact, it is even more important then ever to cancel before you actually multiply!!<br>
So we'll start by canceling factors. And to cancel factors we need to know what the factors are. So we factor the numerators and denominators:
{{{((2x(x^2+1))/(x(x-2)))((x-1)/(2*2*(x^2+1)))}}}
Now that we can see the factors, we can see that there are some factors to cancel, including cross-canceling the 2 and the {{{x^2+1}}} factors:
{{{((cross(2)cross(x)cross((x^2+1)))/(cross(x)(x-2)))((x-1)/(cross(2)*2*cross((x^2+1))))}}}
leaving
{{{(1/(x-2))((x-1)/2)}}}
This, you must agree, is a lot easier to multiply than what we started with. Multiplying we get:
{{{(x-1)/(2x-4)}}}