SOLUTION: find the root of the polynomial p(x)=x^4 + 4x^3 +6x^2 + 4x + 5 = 0 given that one of the roots is x = -i

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Question 977119: find the root of the polynomial p(x)=x^4 + 4x^3 +6x^2 + 4x + 5 = 0 given that one of the roots is x = -i
Found 2 solutions by josgarithmetic, anand429:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
About a week ago, answered.

Divide p(x) by %28x-%28-i%29%29%28x-%28i%29%29 or %28x%2Bi%29%28x-i%29=x%5E2%2B1. The next two roots will be easy to identify.

Answer by anand429(138) About Me  (Show Source):
You can put this solution on YOUR website!
Given
x%5E4+%2B+4x%5E3+%2B6x%5E2+%2B+4x+%2B+5+=+0
Since x= -i is one of the roots,
i.e x%5E2+=+i%5E2+=+-1
so x%5E2+%2B+1 is a factor (since imaginary roots occur in conjugate pairs, hence x=i is also a factor, hence this conclusion)
So arranging the equation to get above factor common, we get,
x%5E4+%2B+x%5E2+%2B+4x%5E3+%2B+4x+%2B+5x%5E2+%2B+5+=+0
=>%28x%5E2%2B4x%2B5%29%2A%28x%5E2%2B1%29=0
So we will find the roots of x%5E2%2B4x%2B5 using quadratic roots formula,
x=+%28-4%2Bsqrt%284%5E2-4%2A1%2A5%29%29%2F%282%2A1%29 and x=+%28-4+-+sqrt%284%5E2-4%2A1%2A5%29%29%2F%282%2A1%29
i.e x=+-2%2Bi and x=+-2+-i
So, the roots are i, -i, -2+i and -2-i