Question 1175007
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First, get used to using parentheses where necessary.  The expression as you show it,<br>
x^2-x/3x^2+27x-30*x^2-100/x^2-10x<br>
means this:<br>
{{{x^2-x/3x^2+27x-30*x^2-100/x^2-10x}}}<br>
That's clearly not what you meant.<br>
The expression you are working with is<br>
((x^2-x)/(3x^2+27x-30))*((x^2-100)/(x^2-10x))<br>
which means this:<br>
{{{((x^2-x)/(3x^2+27x-30))*((x^2-100)/(x^2-10x))}}}<br>
Multiplying rational functions is just like multiplying numerical fractions.  If you had, for example,<br>
{{{(15/8)*(16/3)}}}<br>
you could multiply the numerators and multiply the denominators and simplify the resulting fraction:<br>
{{{(15/8)*(16/3) = 240/24 = 10}}}<br>
But it would be easier if you simplified the product before doing any multiplication:<br>
{{{(15/8)*(16/3) = (15*16)/(8*3) = (15/3)(16/8) = 5*2 = 10}}}<br>
When multiplying rational fractions you DEFINITELY want to factor the expressions and simplify where possible before performing any multiplication.<br>
{{{((x^2-x)/(3x^2+27x-30))*((x^2-100)/(x^2-10x))}}}<br>
{{{(((x)(x-1))/(3(x+10)(x-1)))*(((x+10)(x-10))/((x)(x-10)))}}}<br>
All the factors cancel except for the 3 in the denominator.<br>
So if you got 3 for the answer, you made a mistake very popular among beginning algebra students.  You saw that the only factor left was "3", so that was the answer.  But the 3 that is left is in the denominator.  You don't see it, but there is also a "1" left in the numerator, making the answer 1/3 instead of 3.<br>
So it looks as if you did all the factoring and canceling of common factors correctly; you just didn't realize what the final simplified form was.<br>