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Question: Factor $z^4 + 4z^2 + 6 - z.$

Here is the solution: Rewrite the given equation as $\left(z^2+2\right)^2 + 2 = z$. Observe that a solution to $z^2 + 2 = z$ is a solution of the quartic by substitution of the left hand side into itself. This means $z^2-z+2$ divides into $\left(z^2+2\right)^2-z+2 = z^4 + 4z^2 - z + 6$. Factoring it out, we obtain $\boxed{\left(z^2-z+2\right)\left(z^2+z+3\right)}.$

I have a question on this solution, specifically the 2nd line. What does it mean? Can anybody clarify? Thanks!

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It means that if $z^2+2=z$, then $(z^2+2)^2+2 = (z)^2+2 = z^2+2 = z$. So any solution of $z^2+2=z$ is also a solution of $(z^2+2)^2+2 = z$.

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We can use this idea in another form.

Let $z^2+2=w$ or $2=w-z^2$.

Thus, $$w^2+2=z,$$ which gives the needed factorization: $$z^4+4z^2+6-z=(z^2+2)+2-z=w^2+w-z^2-z=$$ $$=(w-z)(w+z)+w-z=(w-z)(w+z+1)=(z^2-z+2)(z^2+z+3).$$

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