Question

solve for x.
$3x^3 + 2x^2 - 27x - 18 = 0$
$x = ?, -\frac{\quad}{\quad}, \quad$
enter the solutions in ascending order.

Explanation:

Step1: Factor by grouping

Group terms and factor out GCFs:
$(3x^3 + 2x^2) + (-27x - 18) = 0$
$x^2(3x + 2) - 9(3x + 2) = 0$
$(x^2 - 9)(3x + 2) = 0$

Step2: Factor difference of squares

Factor $x^2 - 9$ into linear terms:
$(x - 3)(x + 3)(3x + 2) = 0$

Step3: Solve for x individually

Set each factor equal to 0:

1. $x - 3 = 0 \implies x = 3$
2. $x + 3 = 0 \implies x = -3$
3. $3x + 2 = 0 \implies x = -\frac{2}{3}$

Step4: Order solutions ascendingly

Arrange values from smallest to largest:
$-3 < -\frac{2}{3} < 3$

Answer:

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