Question

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)` ?

Answer 1

`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`.