Question 602449
<pre>
{{{(x^2+9x+8)/(x^2+x-6)}}}

To break into partial fractions, the numerator must have a smaller degree
than the denominator.  If the numerator has the same or larger degree, we
must first divide by long division:

          <u>            1</u>
x² + x - 6)x² + 9x +  8 
           <u>x² +  x -  6</u>
                8x + 14

 
{{{(x^2+9x+8)/(x^2+x-6)}}} = 1 + {{{(8x+14)/(x^2+x-6)}}}

So we break {{{(8x+14)/(x^2+x-6)}}} into partial fractions and add 
what we get to the 1 above:

{{{(8x+14)/(x^2+x-6)}}} = {{{(8x+14)/((x+3)(x-2))}}}

{{{(8x+14)/((x+3)(x-2))}}} = {{{A/(x+3)}}} + {{{B/(x-2)}}}

 8x + 14 = A(x - 2) + B(x + 3)

This must be an identity for all x, so we substitute 2 for x to make the
first term on the right become 0:

 8(2) + 14 = A(2 - 2) + B(2 + 3)
   16 + 14 = 5B
        30 = 5B
         6 = B 

Substitute -3 for x to make the second term on the right become 0:

 8(-3) + 14 = A(-3 - 2) + B(-3 + 3)
   -24 + 14 = -5A
        -10 = -5A
          2 = A

{{{(8x+14)/((x+3)(x-2))}}} = {{{2/(x+3)}}} + {{{6/(x-2)}}}



Now we add that to 1 and get

{{{(x^2+9x+8)/(x^2+x-6)}}} = 1 + {{{2/(x+3)}}} + {{{6/(x-2)}}}

Edwin</pre>