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Look at the denominator. When will that equal 0 (remember, you can't divide by 0)? \[x^2 + 3x +2\] Factor that expression or use the quadratic equation \[(x+2) (x+1)\] Or quadratic equation: \[\frac{-3 \pm \sqrt((-3)^2 - 4(1)(2)}{(2)(1)}\] Solve to get x = -2, x = -1
Also, a hole occurs when both the numerator and denominator are equal to 0 (see http://www.sparknotes.com/math/algebra2/specialgraphs/section2.rhtml). Knowing that the denominator equal to 0 when x = -2 or -1, we see that the expression in the numerator cannot equal 0 when x = -2 or -1, so there are no holes.
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