MHB How Do You Factor \(x^8+4x^2+4\) into Non-Constant Polynomials?

[anemone]

Here is this week's POTW:

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Factor $x^8+4x^2+4$ into two non-constant polynomials with integer coefficients.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. Opalg
3. kaliprasad

Solution from Olinguito:

Spoiler

$$x^8+4x^2+4$$
$=\ (x^8+4x^6+6x^4+4x^2+1)-4x^6-6x^4+3$

$=\ (x^2+1)^4-4x^6-6x^4+3$

$=\ [(x^2+1)^4+2(x^2+1)^2+1]-4x^6-8x^4-4x^2$

$=\ [(x^2+1)^2+1]^2-4x^2(x^4+2x^2+1)$

$=\ [(x^2+1)^2+1]^2-4x^2(x^2+1)^2$

$=\ ([(x^2+1)^2+1]-2x(x^2+1))([(x^2+1)^2+1]+2x(x^2+1))$

$=\ (x^4-2x^3+2x^2-2x+2)(x^4+2x^3+2x^2+2x+2)$.

Alternate solution from Opalg:

Spoiler

If $x$ is a solution of the equation $x^8 + 4x^2 + 4 = 0$ then so is $-x$. So the eight (complex) roots of that equation can be split into two groups of four, with each root in the second group being the negative of one of the roots in the first group. If $x^4 + ax^3 + bx^2 + cx + d = 0$ is the equation whose roots are the first group of four, then (replacing $x$ by $-x$) $x^4 - ax^3 + bx^2 - cx + d = 0$ is the equation whose roots are the second group. Therefore $$(x^4 + ax^3 + bx^2 + cx + d)(x^4 - ax^3 + bx^2 - cx + d) = x^8 + 4x^2 + 4.$$ Now compare the coefficients of powers of $x$ on both sides of that equation: $$ x^6: \quad 2b-a^2 = 0,$$ $$ x^4: \quad 2d - 2ac +b^2 = 0,$$ $$x^2: \quad 2bd - c^2 = 4,$$ $$ x^0:\quad d^2 = 4.$$ Those equations have a solution $a=b=c=d=2$, and the coefficients of the odd powers of $x$ then agree automatically. Therefore $$x^8 + 4x^2 + 4 = (x^4 + 2x^3 + 2x^2 + 2x + 2) (x^4 - 2x^3 + 2x^2 - 2x + 2).$$