<s>[INST] Determine how many roots are real, and finding all roots of a quintic: $-2y^5 +4y^4-2y^3-y=0$ 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. 

 [/INST] 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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