<s>[INST] Complex Equations The Equation:
$$ z^{4} -2 z^{3} + 12z^{2} -14z + 35 = 0 $$
has a root with a real part 1, solve the equation. 
When it says a real part of 1, does this mean that we could use (z-1) and use polynomial division to extract the other rots? Hence: 
$$ (z^{4} -2 z^{3} + 12z^{2} -14z + 35 ) / (z-1) $$
But i don't get the right answer, any other approaches I should try? 
 [/INST] If there's a root of that equation of the form $1+ib$ then:
$$(1+ib)^4 - 2(1+ib)^3 -12(1+ib)^2-14(1+ib) + 35 = 0\Longrightarrow \\(1 + 4ib -6b^2 -4ib^3 + b^4) - 2(1 + 3ib - 3b^2 -ib^3) + 12(1 + 2ib - b^2) - 14(1+ib) + 35 =0 \Longrightarrow b^4-12b^2+32 + i(-2b^3+8b) = 0$$
Therefore:
$$\begin{cases}b^4 - 12b^2+32 = 0\\-2b^3+8b = 0 \end{cases}$$
From the second equation we can see that $b\in \lbrace -2,0,2\rbrace$.
But $b=0$ doesn't satisfy the first equation. The other two values do.
Therefore $(1-2i)$ and $(1+2i)$ are roots of the equation.
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