Question 752756
18x^4               *    6x^2+19x+3
x^3+3x^2-9x-27           18x^2+3x
<pre>
{{{(18x^4)/(x^3+3x^2-9x-27)}}}{{{""*""}}}{{{(6x^2+19x+3)/(18x^2+3x)}}}

You must learn to factor in order to do this kind of problem.

{{{(18x^4)/(x^3+3x^2-9x-27)}}}{{{""*""}}}{{{(6x^2+19+3)/(18x^2+3x)}}}

You must learn to factor the lower left denominator by grouping:

x³+3x²-9x-27
x²(x+3)-9(x+3)
(x+3)(x²-9)
(x+3)(x-3)(x+3)

You must learn to factor the upper right numerator either by the
AC method or by the guess and check method:

6x²+19x+3
(6x+1)(x+3)

You must learn to factor the lower right denominator by taking out
a common factor:

18x²+3x
3x(6x+1)


{{{(18x^4)/((x-3)(x-3)(x+3))}}}{{{""*""}}}{{{((6x+1)(x+3))/(3x(6x+1))}}}

Indicate the multiplication of numerators and denominators
making just one fraction:
        
{{{(18x^4(6x+1)(x+3))/((x+3)(x-3)(x+3)3x(6x+1))}}}

Now you can cancel what will cancel.  You can do it all in one step,
but I'll just cancel one thing at a time:

Cancel the 3 on the bottom into the 18 on top, getting 6 on top:

{{{(""^6cross(18)x^4(6x+1)(x+3))/((x+3)(x-3)(x+3)cross(3)x(6x+1))}}}

{{{(6x^4(6x+1)(x+3))/((x+3)(x-3)(x+3)x(6x+1))}}}

Cancel the x on the bottom into the x<sup>4</sup> on top, getting x<sup>3</sup>
on top:

    ³
{{{(6x^cross(4)*(6x+1)(x+3))/((x+3)(x-3)(x+3)cross(x)(6x+1))}}}

{{{(6x^3(6x+1)(x+3))/((x+3)(x-3)(x+3)(6x+1))}}}

Cancel the (6x+1)'s:


{{{(6x^3(cross(6x+1))(x+3))/((x+3)(x-3)(x+3)(cross(6x+1)))}}}


{{{(6x^3(x+3))/((x+3)(x-3)(x+3))}}}

Cancel the (x+3)'s

{{{(6x^3(cross(x+3)))/((x+3)(x-3)(cross(x+3)))}}}

{{{(6x^3)/((x+3)(x-3))}}}

You can leave it like that, since nothing else will cancel.
As I said above you can do all that canceling in one step.

Edwin</pre>