Question

Find the fourth degree polynomial with real coefficients having the zeros: 0, 1, 1+i

Answer 1

complex zeros always occur in pairs.
so zeros are 0,1,1+i,1-i
if a,b,c,d are zeros of a polynomial ,then polynomial is
f(x)=(x-0)(x-1)(x-(1+i))(x-(1-i))
solve it.

Answer 2

I know that much. I just need to know the next step after that.

Answer 3

What happens to the imaginary number? How do I solve with it?

Answer 4

\[(x-1-i)(x-1+i)=\left\{ \left( x-1 \right) -i\right\}\left\{ \left( x-1 \right)+i \right\}=\left( x-1 \right)^2-i^2=\left( x-1 \right)^2+1\]

Answer 5

\[=x^2-2x+1+1=x^2-2x+2\]

Answer 6

Apparently the correct answer on the paper is x^4-3x^3+4x^2-2x

Answer 7

\[f \left( x \right)=x(x-1)(x^2-2x-2)\]
simplify it.

Answer 8

I'm so confused as to where you got that. I don't understand this at all.

Answer 9

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