7. $2x^3 + x^2 - 9x + 18 = 0$

Explanation:

Step1: Group terms

Group the first two terms and the last two terms: $(2x^3 + x^2) + (-9x + 18) = 0$

Step2: Factor each group

Factor out $x^2$ from the first group and $-9$ from the second group: $x^2(2x + 1) - 9(2x + 1) = 0$

Step3: Factor out common binomial

Factor out $(2x + 1)$: $(2x + 1)(x^2 - 9) = 0$

Step4: Factor difference of squares

Factor $x^2 - 9$ as a difference of squares: $(2x + 1)(x - 3)(x + 3) = 0$

Step5: Solve for x

Set each factor equal to zero:

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

Answer:

The solutions are $x = -\frac{1}{2}$, $x = 3$, and $x = -3$.