Let p(x) be a monic, quartic polynomial, such that p(1) = 3, p(3) = 11, and p(5) = 29. Find p(-2) + 7p(6).
+9665 We assume \(p(x) = x^4 + ax^3 + bx^2 + cx + d\).
Since p(1) = 3, \(a + b + c + d = 2\).
Since p(3) = 11, \(27a + 9b + 3c + d = -70\).
Since p(5) = 29, \(125a + 25b + 5c + d = -596\).
This system has 3 equations but 4 variables, so there are infinitely many solutions.
Solving, we have \(\begin{cases}b = -9a - \dfrac{227}4\\c = 23a + 191\\d = -15a - \dfrac{529}4\end{cases}\).
With these coefficients, we have \(p(-2) = 16 - 8a + 4\left(-9a - \dfrac{227}4\right) - 2(23a + 191) + \left(-15a - \dfrac{529}4\right) = -\dfrac{2901}4 - 105 a\)
Also, \(p(6) = 1296 + 216a + 36\left(-9a - \dfrac{227}4\right) + 6(23a + 191) + \left(-15a -\dfrac{529}4\right) = \dfrac{1067}4 + 15 a\).
Hence,
\(\begin{array}{rcl}p(-2) + 7p(6) &=& -\dfrac{2901}4 - 105a + 7\left(15a + \dfrac{1067}4\right) \\ &=&1142\end{array}\)
