Factoring out a c Factor completely. \( 4 y^{2}-4 y-16 \)

To start, we can factor out a common term from the expression \( 4y^2 - 4y - 16 \). Here, the common factor is \( 4 \), so we can rewrite the expression as \( 4(y^2 - y - 4) \). Next, we focus on factoring the quadratic \( y^2 - y - 4 \). We look for two numbers that multiply to -4 (the constant term) and add to -1 (the coefficient of \( y \)). Since there are no such pairs of integers, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plugging in \( a = 1, b = -1, c = -4 \) gives us: \[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2} \] Thus, the fully factored form of the original expression is: \[ 4 \left( y - \frac{1 + \sqrt{17}}{2} \right) \left( y - \frac{1 - \sqrt{17}}{2} \right) \]