Let $P(z) = z^4-z^3+z^2+2=0$. Express $P(z)$ as a product of two real quadratic factors and hence find the other two zeros.
I am given that $P(1+i)=0$, but I don't know to begin finding the other factors.
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$\begingroup$$$P(z) = z^4-z^3+z^2+2$$$$ = \underbrace{z^4+z^3+z^2}-2z^3+2$$$$ = z^2(z^2+z+1)-2(z-1)(z^2+z+1)$$$$ = (z^2+z+1)(z^2-2z+2)$$
$\endgroup$ $\begingroup$With one root being $z=1+i$ and having real coefficients, the other root is $z=1-i$ and we have one quadratic factor $(z-(1+i))(z-(1-i))=(z-1)^2+1$.
Now you can get the other quadratic factor by dividing the original by this given factor.
$\endgroup$ $\begingroup$If you are given that $P(1+i)=0$, then its complex conjugate, $1-i$ also solves the polynomial. Thus, you have that
$$P(z)=(z-(1+i))(z-(1-i))Q(z)$$
for some quadratic polynomial $Q(z)$. Now,
\begin{eqnarray*}(z-1-i)(z-1+i)&=&((z-1)-i)((z-1)+i)\\&=&(z-1)^2-i^2\\&=&z^2-2z+1+1\\&=&z^2-2z+2\end{eqnarray*}
Thus, one of the quadratic factors is $z^2-2z+2$. You can use the long division algorithm to get the second quadratic and solve that to get the other two roots.
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