Question 89827
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Hi, good afternoon. 
Can you please help me with these two problems.
I do not seem to understand how to: 
Divide and simplify: {{{(3x^2+13x+4)/(16-x^2)}}}÷{{{(3x^2-5x-12)/(3x-12)}}} 
Simplify by rationalizing the denominator: 
{{{(sqrt(11)-sqrt(5)) / (sqrt(11)+sqrt(5))}}}
Thank you so much,
Chocolatechrrys 

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{{{(3x^2+13x+4)/(16-x^2)}}}÷{{{(3x^2-5x-12)/(3x-12)}}} 

Let's get each polynomial in order of descending powers of x.
Only one isn't, namely {{{16-x^2}}}. So write the {{{16-x^2}}} 
in order of descending powers of x,  that is, {{{-x^2+16}}}

Now factor -1 out of that {{{-1(x^2-16)}}}

{{{(3x^2+13x+4)/(-1(x^2-16))}}}÷{{{(3x^2-5x-12)/(3x-12)}}} 

We can erase the -1 and put a negative sign in front of the
first fraction

{{{-(3x^2+13x+4)/(x^2-16)}}}÷{{{(3x^2-5x-12)/(3x-12)}}} 

Factor everything:

{{{-((x+4)(3x+1))/((x-4)(x+4))}}}÷{{{((3x+4)(x-3))/(3(x-4))}}} 


Invert the second fraction and change the ÷ to multiplication

{{{-((x+4)(3x+1))/((x-4)(x+4))}}}·{{{(3(x-4))/((3x+4)(x-3))}}}

Indicate the multiplication of the numerators and denominator
all as one fraction:

{{{-(3(x+4)(3x+1)(x-4))/((x-4)(x+4)(3x+4)(x-3))}}}

The {{{(x+4)}}}'s will cancel, and also the {{{(x-4)}}}'s

So we end up with

{{{-(3(3x+1))/((3x+4)(x-3))}}}

You can leave that just asit is, or if you like,
you can multiply the numerator and denominator out:

{{{-(9x+3)/(3x^2-5x-12)}}}

As you see, we could have gotten by without
factoring the {{{3x^2-5x-12}}} but we had no way
or knowing in advance that neither of its factors
would cancel, and they might have.

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{{{(sqrt(11)-sqrt(5)) / (sqrt(11)+sqrt(5))}}}

Put parentheses around the top and bottom:

{{{((sqrt(11)-sqrt(5))) / ((sqrt(11)+sqrt(5)))}}}

Form the conjugate of the denominator by
copying it over and changing the sign of
only the second term.  So the conjugate
of denominator {{{sqrt(11)+sqrt(5)}}} is
{{{sqrt(11)-sqrt(5)}}}. Now multiply by the 
fraction 

{{{((sqrt(11)-sqrt(5)))/((sqrt(11)-sqrt(5)))}}}, which just equals 1.

{{{((sqrt(11)-sqrt(5))) / ((sqrt(11)+sqrt(5)))}}}·{{{((sqrt(11)-sqrt(5)))/((sqrt(11)-sqrt(5)))}}}

Indicating the multiplication of numerators and 
denominators:

{{{(  (sqrt(11)-sqrt(5))(sqrt(11)-sqrt(5))  ) / ( (sqrt(11)+sqrt(5))(sqrt(11-sqrt(5))))   }}}

Now use FOIL to multiply out:

{{{(  (11-sqrt(11)sqrt(5)-sqrt(5)sqrt(11)+5)  ) / (11-sqrt(11)sqrt(5)+sqrt(5)sqrt(11)-5)))   }}}

Multiplying under the radicals

{{{(11-sqrt(55)-sqrt(55)+5)/(11-sqrt(55)+sqrt(55)-5)}}}

The two middle terms in the top combine as {{{-2sqrt(55)}}},
and the two middle terms cancel in the bottom:

{{{(11-2sqrt(55)+5)/(11-5)}}}

Combining the numbers 

{{{(16-2sqrt(55))/6}}}

Factor 2 out of the top:

{{{( 2(8-sqrt(55)) )/6}}}

Cancel the 2 into the 6 and get

{{{ (8-sqrt(55)) /3}}}

Edwin</pre>