SOLUTION: Degree 3; zeros: -6, -3-i

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Question 1152241: Degree 3; zeros: -6, -3-i
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


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If a polynomial has real coefficients, then complex roots occur in conjugate pairs.

Given that -3-i is a zero, -3+i is another root.

So if the polynomial is degree 3 with two roots -6 and -3-i, then the three roots are -6, -3-i, and -3+i.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Degree 3;
zeros:
x%5B1%5D=-6,
x%5B2%5D=-3-i-> complex zeros always com in pairs, so you also have
x%5B3%5D=-3%2Bi
f%28x%29=%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29
f%28x%29=%28x-%28-6%29%29%28x-%28-3-i%29%29%28x-%28-3%2Bi%29%29
f%28x%29=%28x%2B6%29%28x%2B3%2Bi%29%28x%2B3-i%29
f%28x%29=%28x%2B6%29%28x%5E2%2B3x-xi%2B3x%2B9-3i%2Bxi%2B3i-i%5E2%29

f%28x%29=%28x%2B6%29%28x%5E2%2B6x%2B9-i%5E2%29....i%5E2=-1
f%28x%29=%28x%2B6%29%28x%5E2%2B6x%2B9%2B1%29
f%28x%29=%28x%2B6%29%28x%5E2%2B6x%2B10%29
f%28x%29=x%5E3%2B12x%5E2%2B46x%2B60

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+80%2C+x%5E3%2B12x%5E2%2B46x%2B60%29+