+1671 Let $f$ be a cubic polynomial such that $f(0) = 0$, $f(1) = 0$, $f(2) = 0$, and $f(3)=0$. What is the sum of the coefficients of $f$?
+4 Let's consider the cubic polynomial $f(x)$ with the given conditions: $f(0) = 0$, $f(1) = 0$, $f(2) = 0$, and $f(3) = 0$.
Since $f(0) = 0$, we can conclude that $x = 0$ is a root of the polynomial. Similarly, $f(1) = 0$ implies that $x = 1$ is a root, $f(2) = 0$ implies that $x = 2$ is a root, and $f(3) = 0$ implies that $x = 3$ is a root.
Using this information, we can express the polynomial $f(x)$ in factored form as:
\[f(x) = k(x - 0)(x - 1)(x - 2)(x - 3)\]
where $k$ is some constant.
To find the sum of the coefficients of $f(x)$, we can expand the polynomial and look at the coefficient of $x^0$ (the constant term), $x^1$, $x^2$, and $x^3$.
Expanding the polynomial:
\[f(x) = k(x^4 - 6x^3 + 11x^2 - 6x)\]
The sum of the coefficients of $f(x)$ is given by:
\[\text{Sum of coefficients} = k(1 - 6 + 11 - 6) = k\]
Since we are only interested in the sum of the coefficients, we don't need to determine the exact value of $k$. From the given conditions, we know that $f(x)$ has four roots, which means it is a cubic polynomial. Hence, the sum of the coefficients of $f(x)$ is equal to $k$.
Therefore, the sum of the coefficients of $f(x)$ is $k$.
