To solve the equation:

$4x^3 + 8x^2 - 9x - 18 = 0$

we can proceed with the following steps:

1. Factor the polynomial using the Rational Root Theorem to check for possible rational roots.

2. Test possible rational roots. The Rational Root Theorem suggests that any rational root must be a factor of the constant term (-18) divided by a factor of the leading coefficient (4). Possible rational roots are therefore: $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4}, \pm \frac{18}{4}$

3. Test each of these values to find a root. Let's check if $x = -3$ is a root:

   $4(-3)^3 + 8(-3)^2 - 9(-3) - 18 = 0$

   Simplifying,

   $4(-27) + 8(9) + 27 - 18 = -108 + 72 + 27 - 18 = 0$

   Thus, $x = -3$ is a root.

4. Factor out $(x + 3)$. To factor $4x^3 + 8x^2 - 9x - 18$ by $(x + 3)$, we can use synthetic division.

   • After dividing, the quotient is $4x^2 - 9$.

5. Factor the resulting quadratic $4x^2 - 9$:

   $4x^2 - 9 = (2x + 3)(2x - 3)$

6. Rewrite the polynomial in fully factored form:

   $4x^3 + 8x^2 - 9x - 18 = (x + 3)(2x + 3)(2x - 3)$

7. Solve for $x$ by setting each factor to zero:

   $x + 3 = 0 \Rightarrow x = -3$ $2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$ $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$

Solution

The solutions to the equation are: $x = -3, \quad x = -\frac{3}{2}, \quad x = \frac{3}{2}$

Would you like more details on any step?

Here are some related questions:

1. How can synthetic division help solve polynomial equations?
2. What is the Rational Root Theorem, and why is it useful in polynomial equations?
3. How do you factor quadratic expressions like $4x^2 - 9$?
4. Can the same method apply to higher-degree polynomials?
5. Why do we test specific values when looking for rational roots?

Tip: When solving polynomial equations, always check for any possible common factors first—it can simplify the equation quickly.