The function is a degree 4 polynomial: \[ f(x) = \lambda(x-1)(x-2)(x-3)(x-4) + (x + 1) \] We are tasked with finding the value of \( f(5) \). \[\] Step 1: Find the value of \( \lambda \) We are given that: \[ f(0) = \lambda(-1)(-2)(-3)(-4) + 1 = 25 \] Simplifying the terms: \[ f(0) = \lambda \cdot 24 + 1 = 25 \] \[ 24\lambda + 1 = 25 \] \[ 24\lambda = 24 \] \[ \lambda = 1 \] Step 2: Evaluate \( f(5) \) \[\] Now that we know \( \lambda = 1 \), substitute this into the equation for \( f(x) \): \[ f(x) = (x-1)(x-2)(x-3)(x-4) + (x + 1) \] Now evaluate \( f(5) \): \[ f(5) = (5-1)(5-2)(5-3)(5-4) + (5 + 1) \] Simplifying: \[ f(5) = (4)(3)(2)(1) + 6 \] \[ f(5) = 24 + 6 = 30 \] Thus, the value of \( f(5) \) is \( 30 \).