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So i have this equation:

$z^5-4z^4+11z^3+12z^2-42z+52=0 \text{ for }z\in\Bbb{C}$

One root is: $z=1+i$ That gives us also the 2nd root. $z=1-i$ But i am stuck with how to get other 3. I thought i could divide the equation with those two but i don't know how. Or maybe it would be possible with horner's algorithm,which i also am not sure how tio use in this situation.

Any help would be appreciated.

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  • $\begingroup$ Have you tried using the polynomial remainder theorem? $\endgroup$ – Budenn Aug 1 '15 at 11:39
  • $\begingroup$ How can apply it here though? $\endgroup$ – MathIsTheWayOfLife Aug 1 '15 at 11:40
  • $\begingroup$ By dividing the polynomial by $(z - 1 - i)(z - 1 + i)$. $\endgroup$ – Budenn Aug 1 '15 at 11:41
  • $\begingroup$ Hint: $(z-1-i)(z-1+i) = z^2-2z+2$; next, use long division. $\endgroup$ – Graham Kemp Aug 1 '15 at 11:41
  • $\begingroup$ If $z=1\pm i$ are roots, then divide the expression by the product (to be developed) of the two roots $(z-1-i)\times (z-1+i)= ???$ $\endgroup$ – Claude Leibovici Aug 1 '15 at 11:42
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Tip: $(z-1-i)(z-1+i)=z^2-2z+2$

Then, long division gives:$$\frac{z^5−4z^4+11z^3+12z^2−42z+52}{z^2-2z+2} = z^3-2 z^2+5 z+26$$

So, now, can you find a factor $(z+a)$ that divides that to produce a quadratic?

vis: find $a,b,c$ so $(z+a)(z^2+bz+c) \\ = z^3 +(a+b)z^2+ (ab+c) z+ ac \\ = z^3-2 z^2+5 z+26$

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  • $\begingroup$ Are you surr about $5z$? $\endgroup$ – MathIsTheWayOfLife Aug 1 '15 at 12:07
  • $\begingroup$ @user246608 Yes. $\endgroup$ – Graham Kemp Aug 1 '15 at 23:17
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it can be factorized into $$(z+2) \left(z^2-4 z+13\right) \left(z^2-2 z+2\right)$$

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