Question 94240
 Can you please help me. The problem says perform the indicated operation and simplify the result. Leave answers in factored form: 
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A.{{{12/((x^2-x))}}} *{{{((x^2-1))/((4x-2))}}}
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Step 1 is to see what can be factored, often you will have the same factors
and can see some that can be factored out, which simplifies things.
Note that (x^2-1) is the difference of squares
In this one we can do this:
{{{12/(x(x-1))}}} *{{{((x-1)(x+1))/(2(2x-1))}}}
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Cancel the (x-1)'s. The 2 will cancel into the 12. Leaving us with:
{{{6/x}}} *{{{((x+1))/((2x-1))}}} = {{{(6(x+1))/(x(2x-1))}}}
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 b. {{{x/((x-3))}}} - {{{((x+1))/((x^2+5x-24))}}}
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{{{x/((x-3))}}} - {{{((x+1))/((x+8)(x-3))}}} 
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Common denominator would be (x-3)(x+8), so we can have:
{{{(x(x+8) - (x+1))/((x-3)(x+8))}}} = {{{(x^2 + 8x - x - 1)/((x-3)(x+8))}}} = {{{(x^2 + 7x - 1)/((x-3)(x+8))}}}
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the final numerator won't factor so that's about all we can do with it
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 c. Here we want to place the second expression over a common denominator 1st.
{{{1/h}}}*{{{(1/((x+h)) - 1/x)}}} = {{{1/h}}}*{{{(1x - 1(x+h))/(x(x+h))}}} = {{{1/h}}}*{{{(1x - 1x - h)/(x(x+h))}}} = {{{1/h}}}*{{{(-h)/(x(x+h))}}}
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Cancel h into -h and you have:
{{{(-1)/(x(x+h))}}} = {{{-1/(x(x+h))}}} 
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 d. Multiply what's inside the brackets, remember a minus outside the brackets
changes the signs inside the brackets.
{{{(5(4x+1)-4(5x-2))/((5x-2)^2)}}} = {{{(20x + 5 - 20 x + 8)/((5x-2)(5x-2))}}} = {{{(13)/((5x-2)(5x-2))}}}
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If possible could you show me the steps and rules used to complete these
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The rules are much the same as for dealing with numerical fractions, 
Remember the rule about a negative outside the brackets.
When factoring, look for the difference of squares for example:
(x^2 - 4) factors to (x+2)(x-2); and 4x^2 - 9) factors to (2x+3)(2x-3)
Use a step-by-step approach
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Any questions about what we did here?