Given the polynomial function: $f(x) = ax^3 + bx^2 + cx + 41$ We are provided the following conditions from the problem: 1. $f(1) = 40$ 2. $f^{\prime}(1) = 2$ 3. $f^{\prime \prime}(1) = 4$ First, calculate $f(1)$: $f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$ Simplifying, we get: $a + b + c + 41 = 40$ Therefore: $a + b + c = -1$ Next, calculate the first derivative $f^{\prime}(x)$: $f^{\prime}(x) = 3ax^2 + 2bx + c$ Given $f^{\prime}(1) = 2$: $f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$ Simplifying, we get: $3a + 2b + c = 2$ Next, calculate the second derivative $f^{\prime \prime}(x)$: $f^{\prime \prime}(x) = 6ax + 2b$ Given $f^{\prime \prime}(1) = 4$: $f^{\prime \prime}(1) = 6a(1) + 2b = 4$ Simplifying, we get: $6a + 2b = 4$ Dividing the entire equation by 2: $3a + b = 2$ We now have three equations: 1. $a + b + c = -1$ 2. $3a + 2b + c = 2$ 3. $3a + b = 2$ To solve for $a$, $b$, and $c$, follow these steps: First, subtract the third equation from the second equation: $(3a + 2b + c) - (3a + b) = 2 - 2$ Which simplifies to: $b + c = 0$ So, $c = -b$ Substitute $c = -b$ into the first equation: $(a + b - b = -1)$ Simplifying, we get: $a = -1$ Now substitute $a = -1$ into the third equation: $3(-1) + b = 2$ Which simplifies to: $-3 + b = 2$ Therefore: $b = 5$ Next, since $c = -b$: $c = -5$ Finally, we need to find $a^2 + b^2 + c^2$: $a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$ Simplifying, we get: $1 + 25 + 25 = 51$ Therefore, the answer is: Option B: 51