SOLUTION: How do you find the real-number solutions to the equation 3s^3+15s^2+12s=-60?

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Question 1001170: How do you find the real-number solutions to the equation 3s^3+15s^2+12s=-60?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
3s%5E3%2B15s%5E2%2B12s=-60

Divide through by 3

s%5E3%2B5s%5E2%2B4s=-20

s%5E3%2B5s%5E2%2B4s%2B20=0

Factor sē out of the first two terms on the left side.
Factor +4 out of the last two terms on the left side.

s%5E2%28s%2B5%29%2B4%28s%2B5%29=0

Factor (s+5) out of both terms on the left side.

%28s%2B5%29%28s%5E2%2B4%29=0

s%2B5=0; s%5E2%2B4=0

s=-5;  s%5E2=-4

There are no real number square roots of negative numbers, so
the equation s%5E2=-4 does not have a real number solution,
so the only real number solution is s = -5.

Edwin