How do you factor $x^4 + 4$ ?

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Answer 1 · 2015-05-15T23:30:03

$x^4+4$ has no linear factors since $x^4+4 > 0$ for all real values of $x$ .

How about quadratic factors?

$x^4+4 = (x^2+ax+b)(x^2+cx+d)$

$= x^4+(a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x + bd$

Comparing coefficients of $x^3$ we must have $a+c=0$ , so $c = -a$

... $=x^4+(b+d-a^2)x^2 + a(d-b)x + bd$

Looking at the coefficients of $x$ , we either have $a = 0$ or $b = d$ .

If $a=0$ then $b+d = 0$ so $d=-b$ and $bd=-b^2$ , which would require $b^2 = -4$ - not possible for real values of $b$ .

If $b = d$ , then since $bd = 4$ , $b = d = 2$ , which would make $a^2 = b+d = 4$ , so $a=+-2$ .

Indeed $x^4 + 4 = (x^2+2x+2)(x^2-2x+2)$

How do you factor x^4 + 4x^4 + 4?