Find all zeros of \( f(x)=4 x^{3}+16 x^{2}+15 x+9 \). Enter the zeros separated by commas. Enter exact value, not decimal approximations.

To find the zeros of the polynomial \( f(x) = 4x^3 + 16x^2 + 15x + 9 \), we can use the Rational Root Theorem, synthetic division, or factoring methods. First, let's look for possible rational roots by testing the factors of the constant term (9) divided by the leading coefficient (4). The potential rational roots are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4} \). After testing \( x = -1 \): \[ f(-1) = 4(-1)^3 + 16(-1)^2 + 15(-1) + 9 = -4 + 16 - 15 + 9 = 6 \quad (\text{not a root}) \] Now testing \( x = -\frac{3}{2} \): \[ f\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^3 + 16\left(-\frac{3}{2}\right)^2 + 15\left(-\frac{3}{2}\right) + 9 \] Calculating term by term: - \( 4 \left(-\frac{27}{8}\right) = -\frac{27}{2} \) - \( 16 \left(\frac{9}{4}\right) = 36 \) - \( 15 \left(-\frac{3}{2}\right) = -\frac{45}{2} \) - Constant = 9 Now combine: \[ f\left(-\frac{3}{2}\right) = -\frac{27}{2} + 36 - \frac{45}{2} + 9 = -\frac{72}{2} + \frac{72}{2} = 0 \] Thus, \( x = -\frac{3}{2} \) is a root. Now, divide the polynomial by \( x + \frac{3}{2} \): Perform synthetic division or long division to find the other factors: Dividing \( f(x) \) by \( x + \frac{3}{2} \) gives the quotient: \[ 4x^3 + 16x^2 + 15x + 9 = (x + \frac{3}{2})(4x^2 + 6x + 6) \] Next, let's solve \( 4x^2 + 6x + 6 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 4 \cdot 6}}{2 \cdot 4} \] \[ x = \frac{-6 \pm \sqrt{36 - 96}}{8} = \frac{-6 \pm \sqrt{-60}}{8} = \frac{-6 \pm \sqrt{60} i}{8} = \frac{-6 \pm 2\sqrt{15} i}{8} = \frac{-3 \pm \sqrt{15} i}{4} \] Thus, the zeros of \( f(x) = 4x^3 + 16x^2 + 15x + 9 \) are: \[ x = -\frac{3}{2}, \quad x = \frac{-3 + \sqrt{15} i}{4}, \quad x = \frac{-3 - \sqrt{15} i}{4} \] So, the zeros are: \[ -\frac{3}{2}, \frac{-3 + \sqrt{15} i}{4}, \frac{-3 - \sqrt{15} i}{4} \]