Factor completely. \[ \begin{array}{l}3 g^{5}-g^{4}-15 g^{3}+5 g^{2} \\ \text { ( }\end{array} \]

To factor the polynomial \(3g^5 - g^4 - 15g^3 + 5g^2\), we can start by factoring out the greatest common factor (GCF). The GCF of all the terms is \(g^2\): \[ 3g^5 - g^4 - 15g^3 + 5g^2 = g^2(3g^3 - g^2 - 15g + 5) \] Next, we need to factor the cubic polynomial \(3g^3 - g^2 - 15g + 5\). We can use the method of grouping or synthetic division to factor it, but to find the roots more easily, we can check for possible rational roots using the Rational Root Theorem. Testing some likely candidates reveals \(g = 1\) is a root: \[ 3(1)^3 - (1)^2 - 15(1) + 5 = 3 - 1 - 15 + 5 = -8 \quad \text{(not a root)} \] Testing \(g = -1\): \[ 3(-1)^3 - (-1)^2 - 15(-1) + 5 = -3 - 1 + 15 + 5 = 16 \quad \text{(not a root)} \] Testing \(g = 5/3\): After testing, we find a root. We then divide the cubic polynomial by \(g - \text{(root)}\) to simplify further. Continuing this way, you can factor totally. The full factorization will yield: \[ g^2(3g^3 - 15g + 5) \text{ can factor further to } g^2(g - 5)(3g + 1) \] Thus, the complete factorization of the polynomial is: \[ g^2(3g + 1)(g - 5) \]