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Using a computer we can see that the only real root of $f(y)=-2y^5 +4y^4-2y^3-y=0$ is $0$. Furthermore, we know from algebra that since this polynomial lives in $\Bbb R[y]$ that the roots come in complex conjugate pairs. I.e. we knew that there were either $1,3$ or $5$ real roots of this polynomial. Furthermore, if we could guess factors, we could complete polynomial long division to break up the polynomial. Noting that $y=0$ is a root, we want the $4$ roots of:

$$-2y^4+4y^3-2y^2-1=0$$

  • How would we deduce that the remaining roots are not real?
  • How would we find these by hand.
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$2y^4 - 4y^3 + 2y^2 + 1 =0$ is nothing but $2y^2\cdot (y-1)^2 + 1 = 0$ which is sum of square of two non negative terms.

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  • $\begingroup$ Hence $(\sqrt{2}y(y-1)+i)(\sqrt{2}y(y-1)-i)$ $\endgroup$ – Jasper Loy Jul 20 '16 at 6:53
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Note that$$-2y^4+4y^3-2y^2-1=-2y^2(y-1)^2-1<0$$ for real $y$, so the remaining roots are not real. We need $$ y^2(y-1)^2=-\frac{1}{2}, $$ or $$ y(y-1)=\pm\frac{i\sqrt{2}}{2}. $$ For each choice of sign, this is a quadratic that you can solve with the usual formula. Specifically, $$ y^2-y(\pm)_1\frac{i\sqrt{2}}{2}=0\implies y=\frac{+1(\pm)_2\sqrt{1(\mp)_12i\sqrt{2}}}{2}=\frac{1}{2}(\pm)_2\sqrt{\frac{1}{4}(\mp)_1\frac{i\sqrt{2}}{2}}. $$

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