How can I resolve this equation: $z^4 + 2z^3 + 7z^2 − 18z + 26 = 0$, where there is a root that it's $1+ i$ I know that if one root is $1+i$, other root is $1-i$.
But I don't know how I can find the last $2$ roots.
If you could explain in much detail why I always get lost in the little things, and sorry for my English.
Thanks for your help.
 A: $(z-(1+i))(z-(1-i))=z^{2}-2z+2$. So factor the given polynomial in then form $(z^{2}-2z+2) (z^{2}+az+b)$. You can find $a$ and $b$ by comparing coefficients. Then solve the quadratic  $z^{2}+az+b=0$. [ You should get $a=4$  and $b=13$].
A: Since the polynomial has real coefficients, If $1+i$ is a root, then its complex conjugate $1-i$ is also a root. In particular, then, the polynomial is divisible by
$$(z-(1+i))(z-(1-i)) = z^2 - 2z + 2$$
Now either do the polynomial division, or equivalently solve $$z^4+2z^3+7z^2−18z+26=(z^2 -2z +2)(z^2 + ax + b)$$
By expanding the right side and comparing the coefficients of the two 4th-degree polynomials, you get $a= 4, b=13$.
The last step is to solve $$z^2 +4z +13 = 0$$
which has solutions $z=-2 \pm 3i$.
A: Since $\bigl(z-(1+i)\bigr)\bigl(z-(1-i)\bigr)=z^2-2z+2$ and since, by long division, you get that$$\frac{z^4+2z^3+7z^2-18z+26}{z^2-2z+2}=z^2+4z+13,$$the remaining roots are the roots of $z^2+4z+13$.
A: Since $1+i+1-i=2$ and $(1+i)(1-i)=2$, we got a factor $z^2-2z+2.$
Now, use $$z^4 + 2z^3 + 7z^2 − 18z + 26=$$
$$=z^4-2z^3+2z^2+4z^3-8z^2+8z+13z^2-26z+26=$$
$$=(z^2-2z+2)(z^2+4z+13)=((z-1)^2+1)((z+2)^2+9).$$
Can you end it now?
A: The best and the easiest way among all answers currently present.
The equation $z^4 + 2z^3 + 7z^2 − 18z + 26 = 0$
We know the roots, $1+i$ and $1-i$, we are sure that the product of the remaining $2$ roots  is $13$. (using product of all roots =$\small{\text{constant}/\text{coef first term}}$ for even powers).
Also, the sum of all roots by Vietta's theorem is $-2$. So, the remaining two roots should sum up to $-4$ , $\implies Re(z_{remaining})=-2$ (since they exist in conjugate pairs) and $|z|=\sqrt{13}$.
Hence we get the roots as $-2+3i$ and $-2-3i$.
