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?
 A: 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
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b).\end{align}$$
A: The polynomial $f=x^4-4x^2+16$ cannot be factored over $\Bbb Q$, since it is irreducible. Over the real numbers it factorizes into
$$
f=\left(x^2 + xa + \frac{a^2 - 4}{2}\right)\left(x^2 - xa + \frac{a^2 - 4}{2}\right)
$$
with $a=2\sqrt{3}$. One can see this by writing it as $$(x^2+ax+b)(x^2+cx+d)$$
and compare coefficients.
A: Write
$$x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b)$$
which gives the equations $b^2=16,2b-a^2=-4$. This solves to $b=4,a=\sqrt{12}$, so
$$x^4-4x^2+16=(x^2+\sqrt{12}x+4)(x^2-\sqrt{12}x+4)$$
A: ${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16} \\ $
${x}^{\mathrm{4}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{16}\:−\mathrm{12}{x}^{\mathrm{2}}  \\ $
$\left({x}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{12}{x}^{\mathrm{2}}  \\ $
$\left({x}+\mathrm{4}+\sqrt{\mathrm{12}}{x}\right)\left({x}+\mathrm{4}−\sqrt{\mathrm{12}}{x}\right) \\ $
$\left(\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){x}+\mathrm{4}\right)\left(\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){x}+\mathrm{4}\right) \\ $
A: (a little theory to lead into where you left off )
Since the quartic polynomial $ \ z^4 - 4z^2 + 16 \ $ has real coefficients, we know that its zeroes form two complex-conjugate pairs (since $ \ (-4)^2 - 4·1·16 \ < \ 0 \ ) $ .  So its factorization is
$$ ( z - \alpha - i \beta)·( z - \alpha + i \beta)·( z - \gamma - i \delta)·( z - \gamma + i \delta) $$ $$ = \ \ ( \ z^2 - 2·\alpha·z + [\alpha^2 + \beta^2] \ )·( \ z^2 - 2·\gamma·z + [\gamma^2 + \delta^2] \ ) \ \ . $$
But this polynomial is also biquadratic, and therefore is an even function.  Since there are no terms with odd powers of $ \ x \ \ , $ we must have
$$ -2·(\alpha + \gamma)·z^3 \ = \ 0 \ \ \Rightarrow \ \ \gamma \ = \ -\alpha  \ \ , $$
$$ -2·( \ \alpha·[\gamma^2 + \delta^2] \ + \ \gamma·[\alpha^2 + \beta^2] \ )·z \ \ = \ \ 0 $$ $$ \Rightarrow \ \ \alpha·[\alpha^2 + \delta^2] \ - \ \alpha·[\alpha^2 + \beta^2]   \ \ = \ \ 0  \ \  \Rightarrow \ \   \beta^2 \ \ = \ \ \gamma^2 \ \ . $$
We find then that the four complex zeroes are the vertices of a rectangle in the Argand diagram as $ \ z \ = \ \pm \alpha \ \pm \ i·\beta \ \ . $  (This is consistent with the expectation that the sum of the zeroes is zero and the sum of the reciprocals of the zeroes is $ \ -\frac{0}{16} \ \ . ) \ $  This makes the factorization
$$ \ ( \ z^2 - 2·\alpha·z + [\alpha^2 + \beta^2] \ )·( \ z^2 + 2·\alpha·z + [\alpha^2 + \beta^2] \ ) $$ $$ = \ \ z^4 \ - \ 2·( \ [\alpha^2 + \beta^2] - 2·\alpha^2 \ )·z^2 \ + \ [\alpha^2 + \beta^2]^2 $$
$$ = \ \ z^4 \ - \ 2·( \alpha^2 - \beta^2)·z^2 \ + \ [\alpha^2 + \beta^2]^2  \ \ . \quad \quad \mathbf{[1]} $$
(If this seems a bit reminiscent of "the difference-of-two-squares", you wouldn't be far off...)
Now, your "completion-of-the-square" is $ \ z^4 - 4z^2 + 16 \ = \ (z^2 - 2)^2 \ + \ 12 \ \ . $ If we complete-the-square for equation $ \ \mathbf{1} \ $ , we have
$$   [ \ z^2 \ - \  ( \alpha^2 - \beta^2) \ ]^2  \ + \ (\alpha^2 + \beta^2)^2 \ - \ ( \alpha^2 - \beta^2)^2 \ \ = \ \ [ \ z^2 \ - \  ( \alpha^2 - \beta^2) \ ]^2  \ + \ 4·\alpha^2 · \beta^2  \ \  . \ \ \mathbf{[2]} $$
By comparing  corresponding terms in equations $ \ \mathbf{1} \ $ and $ \ \mathbf{2} \ $, we find
$$   ( \alpha^2 + \beta^2 )^2 \ \ = \ \ 16 \ \ \Rightarrow \ \ \alpha^2 + \beta^2 \ \ = \ \ 4 \ \ \ , \ \ \ \alpha^2 - \beta^2 \  =  \ 2 \ \ \Rightarrow \ \   \alpha^2 \ \ = \ \ 3 \ \ \ , \ \ \ \beta^2 \ \ = \ \ 1 \ \ $$
(this is consistent with $  \ 4·\alpha^2 · \beta^2  \ = \ 12 \ ) \ . $
Thus we obtain the factorization
$$ z^4 \ - \ 4z^2 \ + \ 16 \ \ = \ \ ( \ z^2 \ - \ 2 \sqrt3·z \ +  \ 4 \ ) \ · \ ( \ z^2 \ +  \ 2 \sqrt3·z \ +  \ 4 \ ) $$
with the zeroes of the polynomial being $ \ z \ = \ \pm \sqrt3 \ \pm \ i \ \ . $
[Since it is asked about in the comments elsewhere, here is a problem where the biquadratic polynomial has two real and two complex (in fact, purely imaginary) zeroes.]
