Raymond B. answered 11/03/22
Math, microeconomics or criminal justice
x^4 +4x^2+ 4
= (x^2+2)^2
=(x^2+2)(x^2+2)
maybe the easiest way to see this is let x^2=z
then P(z) = z^2+4z +4 which factors into (z+2)^2
then substitute back to get P(x)= (x^2+2)^2
x^2+2=0
x^2=-2
x= +/-isqr2
zeros are isqr2, -isqr2 each with multiplicity 2
or
x= isqr2, isqr2, -isqr2, -isqr2
there are no real zeros, just 4 imaginary zeros
which also means
if you factor further you only get imaginary irrational factors
P(x)= (x-isqr2)^2(x+isqr2)^2
=(x+isqr2)(x+isqr2)(x-isqr2)(x-isqr2)
usually you stop with the integer rational factors
as the "complete" factors