Question

use the rational root theorem to solve for x.
$2x^3 - 3x^2 - 8x + 12 = 0$
$x = ?, \frac{\square}{\square}, \square$
enter the solutions in ascending order.

Explanation:

Step1: List possible rational roots

By Rational Root Theorem, possible roots are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{3}{2}$

Step2: Test $x=2$ in polynomial

$2(2)^3 - 3(2)^2 - 8(2) + 12 = 16 - 12 - 16 + 12 = 0$

Step3: Factor out $(x-2)$ via grouping

Group terms: $(2x^3 - 8x) + (-3x^2 + 12) = 2x(x^2-4) -3(x^2-4) = (2x-3)(x^2-4)$

Step4: Factor quadratic completely

$x^2-4=(x-2)(x+2)$, so full factorization: $(2x-3)(x-2)(x+2)=0$

Step5: Solve for $x$

Set each factor to 0:
$2x-3=0 \implies x=\frac{3}{2}$; $x-2=0 \implies x=2$; $x+2=0 \implies x=-2$

Step6: Order roots ascendingly

Sort: $-2, \frac{3}{2}, 2$

Answer:

$x = -2, \frac{3}{2}, 2$