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

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

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Answer 1 · 2015-05-15T23:30:03

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

How about quadratic factors?

$x^{4} + 4 = (x^{2} + a x + b) (x^{2} + c x + d)$ $x^4+4 = (x^2+ax+b)(x^2+cx+d)$

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

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

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

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

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

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

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

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