SOLUTION: Factor the polynomial x^3+8x^2+19x+12

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Question 331300: Factor the polynomial x^3+8x^2+19x+12
Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
Factor the polynomial x%5E3%2B8x%5E2%2B19x%2B12
Since the highest degree is 3, there exist 3 roots to this poynomial
The real roots (there might be some complex roots) will be from the set
-/+1, -/+2, -/+3, -/+4,-/+6, -/+12
Use Discarte's sign rule
Number of positive roots P(x) ++++ no sign change, so no positive roots
number of negative roots P(-x) -+-+ 3 sign changes, so either 3 negative roots or 1 negative root and 2 complex roots
Since we know that at least one of the roots is negative, the roots are from
the set -1,-2,-3,-4,-6,-12
so experiment with -1 and see P%28-1%29=%28-1%29%5E3+%2B+8%28-1%29%5E2%2B19%28-1%29+%2B12+=+-1%2B8-19%2B12=0 so -1 is a root.
Now apply either long division or synthetic division to reduce the polynomial by one degree yield x%5E2%2B7x%2B12 this can be easily factored into (x+3)(x+4)
so the original polymial can be factord into
x%5E3%2B8x%5E2%2B19x%2B12+=+%28x%2B1%29%2A%28x%2B3%29%2A%28x%2B4%29