How to get roots in the form of a quadratic from the quartic $x^4 - 4x^2 + 16$? So I need to factor this function into quadratics:
$$x^4 - 4x^2 + 16$$
I know that there are only complex solutions to this question, however, it is still possible to obtain quadratic factors without requiring the imaginary unit to be present.
I tried by simplifying it into a quadratic by replacing $x^2 =a$.
But I just end up getting (after completing the square) $$(x^2-2)^2+12$$
which is not in factored format. So could someone help me?

We see that, the roots of the equation are not real numbers.  So, if we want to factor polynomials whose coefficients are real numbers, and if that's possible, then we have
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b) \end{align} \tag 1$$
This factorization uses the following fact:
If $x_1,x_2$ are the root of the biquadratic, then $-x_1,-x_2$ are also roots of the biquadratic. So, you have
$$\begin{align}x^4-4x^2+16&=(x^2+ax+b)(x^2-ax+b)\\
&=(x^2+b)^2-a^2x^2\\
&=x^4+x^2(2b-a^2)+b^2 \end{align}$$
This implies,
$$\begin{align}\begin{cases}2b-a^2=-4\\b^2=16\end{cases}&\implies a^2=12,\thinspace b=4\\
&\implies a=±2\sqrt 3,\thinspace b=4.\end{align}$$

Explanation: $~(1)$
Since the roots of the polynomial are not real numbers, it is not possible to factor the coefficients with real numbers such that
$$x^4-4x^2+16=(x-a)(x^3+bx^2+cx+d)$$
That is, the polynomial $x^4-4x^2+16$ will be factored with real coefficients as follows:
$$x^4-4x^2+16=(x^2+ax+b)(x^2+mx+n)$$
Then, let $x_1, \thinspace x_2$ are the root of the polynomial $x^2+ax+b$. So, the roots of the second multiplier polynomial must be $-x_1,\thinspace -x_2$. Based on the Vieta's formulas we immediately obtain the coefficients of the second multiplier polynomial as follows:
$$\begin{align}m&=-\left(-x_1+(-x_2)\right)\\
&=x_1+x_2\\
&=-a\end{align}$$
and
$$\begin{align}n&=-x_1\times (-x_2)\\
&=x_1x_2\\
&=b\end{align}$$
Thus, our polynomial will be factored as
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b).\end{align}$$
