Solve the equation. \[ 2 x^{3}+x^{2}=18 x+9 \]

Let's solve the equation \( 2 x^{3} + x^{2} = 18 x + 9 \) step by step. First, we can rearrange the equation to bring all terms to one side: \[ 2x^3 + x^2 - 18x - 9 = 0 \] Now, we can look for potential rational roots using the Rational Root Theorem, which suggests testing factors of the constant term (-9) over factors of the leading coefficient (2). The possible rational roots are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \). After testing these values, we start with \( x = 3 \): \[ 2(3)^3 + (3)^2 - 18(3) - 9 = 2(27) + 9 - 54 - 9 = 54 + 9 - 54 - 9 = 0 \] So, \( x = 3 \) is a root. Now we can factor the polynomial using synthetic division or long division. Dividing \( 2x^3 + x^2 - 18x - 9 \) by \( x - 3 \): Using synthetic division: ``` 3 | 2 1 -18 -9 | 6 21 9 --------------------- 2 7 3 0 ``` This gives us a quotient of \( 2x^2 + 7x + 3 \). Thus, we can express the polynomial as: \[ (2x^2 + 7x + 3)(x - 3) = 0 \] Next, we'll factor \( 2x^2 + 7x + 3 \). We need to find two numbers that multiply to \( 2 \cdot 3 = 6 \) and add to \( 7 \). The numbers are \( 6 \) and \( 1 \). Thus, we can rewrite \( 7x \) as \( 6x + x \): \[ 2x^2 + 6x + x + 3 = 0 \] Now we can factor by grouping: \[ 2x(x + 3) + 1(x + 3) = 0 \] Which gives: \[ (2x + 1)(x + 3) = 0 \] So, the complete factorization is: \[ (2x + 1)(x + 3)(x - 3) = 0 \] Now we can solve for \( x \): 1. \( 2x + 1 = 0 \) leads to \( x = -\frac{1}{2} \) 2. \( x + 3 = 0 \) leads to \( x = -3 \) 3. \( x - 3 = 0 \) leads to \( x = 3 \) Thus, the solutions to the equation \( 2x^3 + x^2 = 18x + 9 \) are: \[ x = -\frac{1}{2}, \quad x = -3, \quad x = 3 \]