How do you factor completely $x^2 + 16$ ?

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Answer 1 · 2017-01-15T04:46:49

No factor in $RR$ .
$x = 4i$ multiplicity 2 or $x = -4i$ multiplicity 2 in $CC$

In $RR$ ,

$x^2 + 16 = 0$

$=> x^2 = -16$

No solution in $RR$ since $x^2 >= 0$ in $RR$ .

In $CC$ ,

$x^2 = -16$

$=> x = sqrt(-16)$

$=> x = sqrt16*sqrt(-1)$

$=> x = +-4i$

x = 4i multiplicity 2 or x = -4i multiplicity 2

How do you factor completely x^2 + 16x^2 + 16?