Question

factor ( x^4 + 4x^2 + 4 ) completely. all factors in your answer should have integer coefficients.

Explanation:

Step1: Rewrite the expression

We have the expression \(x^4 + 4x^2 + 4\). Notice that it resembles a perfect square trinomial. Let \(y = x^2\), then the expression becomes \(y^2 + 4y + 4\).

Step2: Factor the perfect square trinomial

A perfect square trinomial \(a^2 + 2ab + b^2=(a + b)^2\). For \(y^2 + 4y + 4\), we have \(a = y\) and \(b = 2\) (since \(2ab=2\times y\times2 = 4y\)), so \(y^2 + 4y + 4=(y + 2)^2\).

Step3: Substitute back \(y = x^2\)

Substituting \(y = x^2\) back into \((y + 2)^2\), we get \((x^2 + 2)^2\). But we can further factor \(x^2+2\) over the complex numbers? Wait, no, the problem says all factors should have integer coefficients. Wait, maybe I made a mistake. Wait, actually, \(x^4 + 4x^2 + 4=(x^2 + 2)^2\), but if we consider factoring over real numbers with integer coefficients, this is already factored? Wait, no, wait, maybe the original expression can be factored as a sum of squares? Wait, no, \(x^4 + 4x^2 + 4=(x^2 + 2)^2\), but let's check again. Wait, \(x^4 + 4x^2 + 4=(x^2)^2+2\times x^2\times2 + 2^2=(x^2 + 2)^2\). But maybe there is a way to factor it as a product of two quadratics with integer coefficients. Wait, let's try: \(x^4 + 4x^2 + 4=(x^2 + ax + b)(x^2 + cx + d)\). Expanding the right side: \(x^4+(a + c)x^3+(ac + b + d)x^2+(ad + bc)x+bd\). Comparing coefficients: \(a + c = 0\) (since there is no \(x^3\) term), \(ac + b + d = 4\), \(ad + bc = 0\) (no \(x\) term), \(bd = 4\). From \(a + c = 0\), \(c=-a\). Then \(ad + bc=ad - ab=a(d - b)=0\). So either \(a = 0\) or \(d = b\). If \(a = 0\), then \(c = 0\), and \(ac + b + d=b + d = 4\), \(bd = 4\). Solving \(b + d = 4\) and \(bd = 4\), we get \(b = d = 2\). So \((x^2 + 2)(x^2 + 2)=(x^2 + 2)^2\), which is the same as before. So the factored form is \((x^2 + 2)^2\). Wait, but maybe the problem expects a different approach? Wait, no, the expression \(x^4 + 4x^2 + 4\) is a perfect square trinomial, so factoring it gives \((x^2 + 2)^2\).

Answer:

\(\boxed{(x^2 + 2)^2}\)