Can you show that $z^4 + 64$ can be factorised into two real quadratic factors of the form $z^2 + az + 8$ and $z^2 + bz + 8$ but cannot be factorised into two real quadratic factors of the form $z^2 + bz + 16$ and $z^2 + bz + 4$ ?

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Answer 1 · 2017-02-26T17:19:34

$z^4+64=(z^2-4z+8)( z^2+4z+8)$

The factors of $z^4+64=(z-z_0)(z-z_1)(z-z_2)(z-z_3)$

are obtained solving

$z^4=-64=2^6 e^(i pi +i2k pi)$

Here $e^(i pi) = cos(pi)+isin(pi)=-1$ (de Moivre's identity)

so

$z = root(4)(2^6) e^(i (pi/4+k pi/2))$ obtaining for $k=0,1,2,3$

$z_0 = root(4)(2^6)(1 + i)/sqrt[2]$
$z_1 = - root(4)(2^6)(1 + i)/sqrt[2]$
$z_2 =- root(4)(2^6)(1 + i)/sqrt[2]$
$z_0 = root(4)(2^6)(1 - i)/sqrt[2]$

The arrangements

$(z-z_0)(z-z_3) = z^2-4z+8$
$(z-z_1)(z-z_2) = z^2+4z+8$

are the only to give trinomials with real coefficients.

The answer is $z^4+64=(z^2-4z+8)( z^2+4z+8)$

Another way to solve this problem is by grouping coefficients in

$z^4 + 64 - (z^2 + a z + 8) (z^2 + b z + 8)$ and solving

${(16 + a b=0), (a + b=0):}$ giving $a = 4, b= -4$

or

$z^4 + 64 - (z^2 + b z + 16) (z^2 + b z + 4)$ and solving

${(b=0), (20 +b^2=0):}$ without solution.

Can you show that z^4 + 64z^4 + 64 can be factorised into two real quadratic factors of the form z^2 + az + 8z^2 + az + 8 and z^2 + bz + 8z^2 + bz + 8 but cannot be factorised into two real quadratic factors of the form z^2 + bz + 16z^2 + bz + 16 and z^2 + bz + 4z^2 + bz + 4?