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I want to express this polynomial as a product of linear factors:

$x^5 + x^3 + 8x^2 + 8$

I noticed that $\pm$i were roots just looking at it, so two factors must be $(x- i)$ and $(x + i)$, but I'm not sure how I would know what the remaining polynomial would be. For real roots, I would usually just do use long division but it turns out a little messy in this instance (for me at least) and was wondering if there was a simpler method of finding the remaining polynomial.

Apologies for the basic question!

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If you divide $$ x^5 + x^3 + 8x^2 + 8$$ by $$(x-i)(x+i) = x^2+1$$ you will get $$x^3+8$$ which factors as $$(x^3+8) = (x+2)(x^2-2x+4)$$ which has a solution of $x=-2$

Now use quadratic formula to solve $x^2-2x+4=0$ to find other roots and factor if you wish.

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  • $\begingroup$ Thanks, a lot simpler than I thought! $\endgroup$ – Aidan Sep 5 '18 at 22:10
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If you spotted this by looking at it you have good intuition. When roots come in complex pairs you can always combine them to find a quadratic factor with real coefficients. Here $(x+i)(x-i)=x^2+1$

If you could do the first bit with intuition, I am sure you can do that division and complete the factorisation.

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Yo can also do long division for complex polynomials. But in this case, I would suggest to pull out the two factors at once, i.e., divide by their product, which is $(x^2+1)$.

$(x^5+x^3+8x^2+8):(x^2+1)= x^3+8$. You can easily factorize this polynomial. (Hint: $8=2^3$.)

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Alt. hint:   you might as well notice that $\,x=-2\,$ is a root by the rational root theorem. Then if the remaining two roots are $\,a,b\,$, by Vieta's relations $\,i+(-i)+(-2)+a+b=0 \iff a+b = 2\,$ and $\,i \cdot (-i)\cdot(-2)\cdot a \cdot b = -8 \iff a \cdot b = 4\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2 x + 4 = 0\,$.

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