Posted by Blue on Saturday, October 8, 2011 at 7:38pm.
factor the bottom
(x-3)(x-2)
factor the top with (x-3) a factor so we can cancel it
(x-3)(x+b) = x^2+x+c
x^2 - 3x +b x - 3 b = x^2 + x + c
so
b x -3 x = x so b = 4
then c = -12
so
(x^2 + x - 12)/(x^2 - 5 x + 6)
(x-3)(x+4) / [(x-3)(x-2)]
(x+4)/(x-2)
when x = 3
7/1 = 7
hmmm. looks familiar. Anyway,
denominator is (x-3)(x-2), so we want the top to be
(x-3)(x-k) so that the fraction exists everywhere except at x=3, but has a finite limit there.
x^2 - (k+3)x + 3k = x^2 + x + c
k+3 = -1
k = -4
so c = 3k = -12
x^2 + x + c = x^2 + x -12 = (x-3)(x+4)
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