Polynomial Points Sum
The graph of $ y=f(x) $, where $ f(x) $ is a polynomial of degree $ 3 $, contains points $ A(2,4) $, $ B(3,9) $, and $ C(4,16) $. Lines $ AB $, $ AC $, and $ BC $ intersect the graph again at points $ D $, $ E $, and $ F $, respectively, and the sum of the $ x $-coordinates of $ D $, $ E $, and $ F $ is 24. What is $ f(0) $?
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Solution
Since the graph of $ y = f(x) $ passes through $ (2,4), $ $ (3,9), $ and $ (4,16), $ $ f(2) = 4, $ $ f(3) = 9, $ and $ f(4) = 16 $.
Let $ g(x) = f(x) - x^2 $. Then $ g(x) $ is cubic, and $ g(2) = g(3) = g(4) = 0, $ so
\[g(x) = c(x - 2)(x - 3)(x - 4)\]for some constant $ c $. Then
\[f(x) = g(x) + x^2 = cx^3 + (1 - 9c)x^2 + 26cx - 24c.\]Let $ d, $ $ e, $ $ f $ be the $ x $-coordinates of the points $ D, $ $ E, $ $ F, $ respectively. Let $ L(x) $ be the equation of the line passing through $ A, $ $ B, $ and $ D $. Then the solutions to $ f(x) = L(x) $ are $ x = 2, $ 3, and $ d $. By Vieta's formulas,
\[2 + 3 + d = -\frac{1 - 9c}{c}.\](Note that the $ x^3 $ and $ x^2 $ terms of $ f(x) - L(x) $ are the same as those in $ f(x) $.)
Similarly,
\begin{align*}
2 + 4 + e &= -\frac{1 - 9c}{c}, \\
3 + 4 + f &= -\frac{1 - 9c}{c}.\end{align*}Adding these equations, we get
\[d + e + f + 18 = -\frac{3(1 - 9c)}{c}.\]We are told that $ d + e + f = 24, $ so
\[42 = -\frac{3(1 - 9c)}{c}.\]Solving for $ c, $ we find $ c = -\frac{1}{5} $. Hence,
\[f(x) = -\frac{1}{5} (x - 2)(x - 3)(x - 4) + x^2.\]It follows that $ f(0) = \boxed{\frac{24}{5}} $.
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