How do you factor completely #f(x) = x^5-2 i x^4-(5+3 i) x^3-(7-3 i) x^2+(6+11 i) x-(1+3 i)# ?

1 Answer
George C.
Dec 14, 2016

#f(x) = (x-1-i)(x-2-i)(x+1-root(3)(2))(x+1-omega root(3)(2))(x+1-omega^2 root(3)(2))#

Explanation:

Given:

#f(x) = x^5-2ix^4-(5+3i)x^3-(7-3i)x^2+(6+11i)x-(1+3i)#

Use a Gaussian integer version of the rational roots theorem.

Any Gaussian rational zeros of #f(x)# must be expressible in the form #p/q# for Gaussian integers #p, q# with #p# a divisor of the constant term #-(1+3i)# and #q# a divisor of the coefficient #1# of the leading term.

Note that:

#abs(-(1+3i)) = sqrt(1^2+3^2) = sqrt(10) = sqrt(2)sqrt(5)#

So #1+3i# is divisible by some Gaussian integers of modulus #sqrt(2)# and #sqrt(5)#

#(1+3i)/(1+i) = ((1+3i)(1-i))/((1+i)(1-i)) = (4+2i)/2 = 2+i#

Hence the possible Gaussian rational zeros of #f(x)# are:

#+-1#, #+-i#, #+-(1+i)#, #+-(1-i)#, #+-(2+i)#, #+-(1-2i)#, #+-(1+3i)#, #+-(3-i)#

Trying each in turn, we find:

#f(1) = 1-2i-(5+3i)-(7-3i)+(6+11i)-(1+3i) = -6+6i#

#f(-1) = -1-2i+(5+3i)-(7-3i)-(6+11i)-(1+3i) = -10-10i#

#f(i) = (i)^5-2i(i)^4-(5+3i)(i)^3-(7-3i)(i)^2+(6+11i)(i)-(1+3i) = -8+4i#

#f(-i) = (-i)^5-2i(-i)^4-(5+3i)(-i)^3-(7-3i)(-i)^2+(6+11i)(-i)-(1+3i) = 20-20i#

#f(1+i) = (1+i)^5-2i(1+i)^4-(5+3i)(1+i)^3-(7-3i)(1+i)^2+(6+11i)(1+i)-(1+3i)#

#color(white)(f(1+i)) = (1+5i-10-10i+5+i)-2i(1+4i-6-4i+1)^4-(5+3i)(1+3i-3-i)^3-(7-3i)(1+2i-1)^2+(6+11i)(1+i)-(1+3i)#

#color(white)(f(1+i)) = (-4-4i)-2i(-4)-(5+3i)(-2+2i)-(7-3i)(2i)+(6+11i)(1+i)-(1+3i)#

#color(white)(f(1+i)) = 0#

So #x = 1+i# is a zero and #(x-1-i)# a factor:

#x^5-2ix^4-(5+3i)x^3-(7-3i)x^2+(6+11i)x-(1+3i)#

#=(x-1-i)(x^4 + (1 - i) x^3 - (3 + 3 i) x^2 - (7 + 3 i) x + (2 + i))#

Let #g(x) = x^4 + (1 - i) x^3 - (3 + 3 i) x^2 - (7 + 3 i) x + (2 + i)#

To cut a long story a little shorter, we find:

#g(2+i) = (2+i)^4 + (1 - i) (2+i)^3 - (3 + 3 i) (2+i)^2 - (7 + 3 i) (2+i) + (2 + i) = 0#

and

#x^4 + (1 - i) x^3 - (3 + 3 i) x^2 - (7 + 3 i) x + (2 + i)#

#=(x-2-i)(x^3+3x^2+3x-1)#

Then:

#x^3+3x^2+3x-1 = x^3+3x^2+3x+1-2#

#color(white)(x^3+3x^2+3x-1) = (x+1)^3-(root(3)(2))^3#

#color(white)(x^3+3x^2+3x-1) = ((x+1)-root(3)(2))((x+1)-omega root(3)(2))((x+1)-omega^2 root(3)(2))#

#color(white)(x^3+3x^2+3x-1) = (x+1-root(3)(2))(x+1-omega root(3)(2))(x+1-omega^2 root(3)(2))#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Related questions
See all questions in Zeros
Impact of this question
1652 views around the world
You can reuse this answer
Creative Commons License