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Find the four roots of the polynomial $z^{4}+16$ and use these to factor $z^{4}+16$ into two quadratic polynomials with real coefficients.
in Mathematics by Gold Status (10,261 points) | 11 views

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Solution. The four roots of $z^{4}+16$ are given by
\sqrt[4]{-16} &=\sqrt[4]{16 e^{\pi i}}=\sqrt[4]{16} e^{\pi i / 4} e^{2 m \pi i / 4} \\
&=2 e^{\pi i / 4}, 2 e^{3 \pi i / 4}, 2 e^{5 \pi i / 4}, 2 e^{7 \pi i / 4}
for $m=0,1,2,3$. We see that these roots appear in conjugate pairs:
2 e^{\pi i / 4}=\overline{2 e^{7 \pi i / 4}} \text { and } 2 e^{3 \pi i / 4}=\overline{2 e^{5 \pi i / 4}}
This gives the way to factor $z^{4}+16$ into two quadratic polynomials of real coefficients:
z^{4}+16 &=\left(z-2 e^{\pi i / 4}\right)\left(z-2 e^{3 \pi i / 4}\right)\left(z-2 e^{5 \pi i / 4}\right)\left(z-2 e^{7 \pi i / 4}\right) \\
&=\left(\left(z-2 e^{\pi i / 4}\right)\left(z-2 e^{7 \pi i / 4}\right)\right)\left(\left(z-2 e^{3 \pi i / 4}\right)\left(z-2 e^{5 \pi i / 4}\right)\right) \\
&=\left(z^{2}-2 \operatorname{Re}\left(2 e^{\pi i / 4}\right) z+4\right)\left(z^{2}-2 \operatorname{Re}\left(2 e^{3 \pi i / 4}\right) z+4\right) \\
&=\left(z^{2}-2 \sqrt{2} z+4\right)\left(z^{2}+2 \sqrt{2} z+4\right)
by Gold Status (10,261 points)

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