The problem asks to show that $x - 2i$ is a factor of the polynomial $f(x) = x^4 + 6x + 8$.

1. Problem Description

The problem asks to show that $x - 2i$ is a factor of the polynomial $f(x) = x^4 + 6x + 8$.

2. Solution Steps

To show that $x - 2i$ is a factor of $f(x)$, we need to show that $f(2i) = 0$.

$f(x) = x^4 + 6x + 8$

Substitute $x = 2i$ into the polynomial:

$f(2i) = (2i)^4 + 6(2i) + 8$

$f(2i) = (2^4)(i^4) + 12i + 8$

Since $i^2 = -1$, then $i^4 = (i^2)^2 = (-1)^2 = 1$.

$f(2i) = 16(1) + 12i + 8$

$f(2i) = 16 + 12i + 8$

$f(2i) = 24 + 12i$

Since $f(2i) = 24 + 12i \neq 0$, then $x - 2i$ is not a factor of $f(x) = x^4 + 6x + 8$.

However, if $f(x) = x^4 + 14x^2 + 25$, and we need to prove that $x-2i$ is a factor. Then

$f(2i) = (2i)^4 + 14(2i)^2 + 25$

$f(2i) = 16i^4 + 14(4i^2) + 25$

$f(2i) = 16(1) + 56(-1) + 25$

$f(2i) = 16 - 56 + 25$

$f(2i) = 41 - 56$

$f(2i) = -15 \neq 0$

$x - 2i$ is not a factor of $f(x) = x^4 + 14x^2 + 25$.

There may be a typo in the problem. Let's try a different polynomial.

Let $f(x) = x^4 + 4x^2 + 4$. Then

$f(2i) = (2i)^4 + 4(2i)^2 + 4$

$f(2i) = 16i^4 + 4(4i^2) + 4$

$f(2i) = 16(1) + 16(-1) + 4$

$f(2i) = 16 - 16 + 4 = 4 \neq 0$.

Let's consider the conjugate $x + 2i$. If $x-2i$ is a factor, $x+2i$ should also be a factor (because the coefficients of the polynomial are real numbers). So $(x-2i)(x+2i) = x^2 + 4$ should be a factor.

Let's perform polynomial long division to see if this is a factor.

$(x^4 + 6x + 8) / (x^2 + 4)$

$x^4 + 0x^3 + 0x^2 + 6x + 8 = (x^2 + 4)(x^2 - 4) + 6x + 24$

So $x^2+4$ is not a factor of $f(x) = x^4 + 6x + 8$.

3. Final Answer

$x-2i$ is not a factor of $f(x) = x^4 + 6x + 8$.

$f(2i) = 24 + 12i$.