SOLUTION: p(x)=2x^3-5x^2+6x-2; 1+i use the given zero to find the remaining zeros of each polynomial function....

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Question 310540: p(x)=2x^3-5x^2+6x-2; 1+i
use the given zero to find the remaining zeros of each polynomial function....
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If %281%2Bi%29 is a zero, it's complex conjugate highlight_green%281-i%29 is also a zero.
So then the quadratic formed from those two roots is
%28x-%281%2Bi%29%29%28x-%281-i%29%29=x%5E2-2x%2B2
Using polynomial long division, divide the original polynomial by this quadratic,
%282x%5E3-5x%5E2%2B6x-2%29%2F%28x%5E2-2x%2B2%29
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.
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The first part of the factor is %282x%29
%282x%29%28x%5E2-2x%2B2%29=2x%5E3-4x%5E2%2B4x
subtracted from the original polynomial leaves a remainder of,
%282x%5E3-5x%5E2%2B6x-2%29-%282x%5E3-4x%5E2%2B4x%29=-x%5E2%2B2x-2
.
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The second part of the factor is %28-1%29
%281%29%28x%5E2-2x%2B2%29=-x%5E2%2B2x-2
which subtracted from the remainder leaves 0.
%28-x%5E2%2B2x-2%29-%28-x%5E2%2B2x-2%29
So then,
%282x%5E3-5x%5E2%2B6x-2%29%2F%28x%5E2-2x%2B2%29=2x-1
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.
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2x-1=0
2x=1
highlight_green%28x=1%2F2%29
which is the third root, which you can see by graphing the function.