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#?

Question

5064 views around the world

You can reuse this answer
Creative Commons License

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.