Simplify the expression and state the restrictions on the variable. \( \frac{16-u^{2}}{4 u^{2}-16 u} \) Part: \( 0 / 2 \) Part 1 of 2 Simplify \( \frac{16-u^{2}}{4 u^{2}-16 u}= \)

To simplify the expression \( \frac{16-u^{2}}{4 u^{2}-16 u} \), we can start by factoring both the numerator and the denominator. The numerator \( 16 - u^2 \) can be factored using the difference of squares: \[ 16 - u^2 = (4 - u)(4 + u) \] The denominator \( 4u^2 - 16u \) can be factored by taking out the common factor of \( 4u \): \[ 4u^2 - 16u = 4u(u - 4) \] Now we can rewrite the expression: \[ \frac{(4 - u)(4 + u)}{4u(u - 4)} \] Next, we can cancel the common factor \( (4 - u) \) and \( (u - 4) \) (note: \( (4 - u) = -(u - 4) \)): \[ = \frac{-(4 + u)}{4u} \] So the simplified expression is: \[ = \frac{-(4 + u)}{4u} \] As for the restrictions on the variable \( u \), we need to identify values that make the denominator zero: 1. \( 4u \neq 0 \) implies \( u \neq 0 \) 2. \( u - 4 \neq 0 \) implies \( u \neq 4 \) Thus, the restrictions are \( u \neq 0 \) and \( u \neq 4 \).