salmawisoky

To find the area of triangle $ACD$, we can use the fact that the ratio of the areas of two triangles with the same altitude is equal to the ratio of their corresponding base lengths.

Let $h$ be the altitude of both triangles $ABC$ and $ADE$ with respect to base $\overline{BE}$. Then, we have:
\[\frac{[ABC]}{[ADE]} = \frac{AB}{AD} = \frac{h}{h} = 1.\]
Since $[ABC] = 4$ and $[ADE] = 8$, we can write the equation:
\[\frac{4}{8} = \frac{AB}{AD}.\]
Simplifying, we find $AB = \frac{1}{2}AD$.

Now, let's focus on triangle $ACD$. We know that $BD = 8$ and $CE = 24$. Since $AB = \frac{1}{2}AD$, we can express $BD$ and $CE$ in terms of $AD$:
\[BD = \frac{1}{2}AB = \frac{1}{4}AD,\]
and
\[CE = 2AB = AD.\]

To find the area $[ACD]$, we need to find the base length $AC$. Using the fact that the ratio of the areas is equal to the ratio of base lengths, we have:
\[\frac{[ACD]}{[ADE]} = \frac{AC}{AD}.\]

Substituting the given values, we have:
\[\frac{[ACD]}{8} = \frac{AC}{AD}.\]

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 $k$.