<s>[INST] 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?
 [/INST] 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
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b).\end{align}$$
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