Question

solve for x.
$x^3 + 3x^2 - 5x + 25 = 0$
$x = ?, pm i$

Explanation:

Step1: Test rational roots via Rational Root Theorem

Possible rational roots: $\pm1, \pm5, \pm25$. Test $x=-5$:
$$(-5)^3 + 3(-5)^2 -5(-5) +25 = -125 + 75 +25 +25 = 0$$
So $x=-5$ is a root, and $(x+5)$ is a factor.

Step2: Perform polynomial division

Divide $x^3 + 3x^2 -5x +25$ by $(x+5)$:
$$\frac{x^3 + 3x^2 -5x +25}{x+5} = x^2 -2x +5$$

Step3: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $x^2-2x+5=0$, where $a=1, b=-2, c=5$:
$$x=\frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)} = \frac{2\pm\sqrt{4-20}}{2} = \frac{2\pm\sqrt{-16}}{2} = 1\pm2i$$

Answer:

$x = -5,\ 1 \pm 2i$