```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>```