# Consider the polynomial f(X)=x^4+1 a) Explain why f has no real roots, and why this means f mustfactor as a product of two ireducible quadratics. b) Factor f and find all of its complex roots.

Consider the polynomial $f\left(X\right)={x}^{4}+1$
a) Explain why f has no real roots, and why this means f mustfactor as a product of two ireducible quadratics.
b) Factor f and find all of its complex roots.
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Ezra Herbert

a) For the roots,
f(x)=0
$⇒{x}^{4}+1=0$
$⇒{x}^{4}=-1$
But there is no real number ehose even power gives negative real number which means f has no real roots.
Futher since f has no real roots so f does not have any linear factors.
Therefore f must factor as a product of two irreducible qudratics.
b) $f\left(x\right)={x}^{4}+1$
${x}^{1}+2{x}^{2}-2{x}^{2}$ Adding and substracting $2{x}^{2}$
$={\left({x}^{2}+1\right)}^{2}-{\left(\sqrt{2}\right)}^{2}$
$=\left({x}^{2}+1+\sqrt{2}x\right)\left({x}^{2}+1-\sqrt{2}x\right)\because {a}^{2}={b}^{2}=\left(a+b\right)\left(a-b\right)$
$=\left({x}^{2}+\sqrt{2}x+1\right)\left({x}^{2}-\sqrt{2}x+1\right)$
For the roots.
f(x)=0
$⇒\left({x}^{2}+\sqrt{2}x+1\right)\left({x}^{2}-\sqrt{2}x+1\right)=0$
$⇒\left({x}^{2}+\sqrt{2}x+1\right)\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{x}^{2}-\sqrt{2}x+1=0$
$⇒x=\left(-\sqrt{2}±\frac{\sqrt{{\left(\sqrt{2}\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)},x=-\left(-\sqrt{2}±\frac{\sqrt{{\left(\sqrt{2}\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}$
$⇒x=\frac{-\sqrt{2}±\sqrt{2-4}}{2},x=\frac{\sqrt{2}±\sqrt{2-4}}{2}$
$⇒x=\frac{-\sqrt{2}±\sqrt{-}2}{2},x=\frac{\sqrt{2}±\sqrt{-}2}{2}$
$⇒x=\frac{-\sqrt{2}±i\sqrt{2}}{2},x=\frac{\sqrt{2}±i\sqrt{2}}{2}$
$⇒x=\frac{-1±i}{\sqrt{2}},x=\frac{1±i}{\sqrt{2}}$
Therefore the comlex roots are,
$\frac{-1+i}{\sqrt{2}},\frac{-1-i}{\sqrt{2}},\frac{1+i}{\sqrt{2}},\frac{1-i}{\sqrt{2}}$