Factor each polynomial. \( \begin{array}{ll}-8 q^{4}+4 q^{6}-16 q^{3} & \text { c. } 4\left(-2 q^{4}+q^{6}-4 q^{3}\right) \\ \text { a. } 4 q^{3}\left(-2 q+q^{3}-4\right) & \text { d. } 4 q^{2}\left(-2 q^{2}+q^{4}-4 q\right) \\ \text { b. } 4 q^{3}\left(-2 q+4 q^{6}-16 q^{3}\right) & \text {. }\end{array} \)

We begin with the polynomial \[ -8q^4 + 4q^6 - 16q^3. \] It is easiest to first rearrange the terms in decreasing order of degree: \[ 4q^6 - 8q^4 - 16q^3. \] **Step 1. Factor Out the Greatest Common Factor (GCF):** All the terms have a factor of \(4q^3\) (note that \(4\) divides each coefficient and the smallest power of \(q\) is \(q^3\)). Factoring \(4q^3\) gives: \[ 4q^6 - 8q^4 - 16q^3 = 4q^3\left(\frac{4q^6}{4q^3} - \frac{8q^4}{4q^3} - \frac{16q^3}{4q^3}\right) = 4q^3\bigl(q^3 - 2q - 4\bigr). \] **Step 2. Factor the Cubic inside the Parentheses:** We now focus on factoring the cubic \[ q^3 - 2q - 4. \] First, check for possible rational roots using the Rational Root Theorem. The possible factors of the constant term (4) are \(\pm 1, \pm 2, \pm 4\). Testing \(q = 2\): \[ 2^3 - 2\cdot 2 - 4 = 8 - 4 - 4 = 0. \] Since \(q=2\) is a root, \((q-2)\) is a factor. Divide \(q^3 - 2q - 4\) by \((q-2)\) (by synthetic division or polynomial long division) to find the quadratic factor: \[ q^3 - 2q - 4 = (q-2)(q^2 + 2q + 2). \] **Step 3. Write the Final Factored Form:** Substitute the factorization of the cubic back into the expression: \[ 4q^3\bigl(q^3 - 2q - 4\bigr) = 4q^3 (q-2)(q^2 + 2q + 2). \] Thus, the fully factored form of the polynomial is \[ 4q^3 (q-2)(q^2+2q+2). \]