Let $p(x)$ be a monic, quartic polynomial, such that $p(1) = 3,$ $p(3) = 11,$ and $p(5) = 27.$ Find
\[p(-2) + 7p(6).\]

Let $ p(x) $ be a monic quartic polynomial. This implies that the leading coefficient of $ p(x) $ is 1. We can write $ p(x) $ in the form:

\[
p(x) = x^4 + ax^3 + bx^2 + cx + d
\]

We are given the following conditions:

\[
p(1) = 3, \quad p(3) = 11, \quad p(5) = 27
\]

Using these conditions, we substitute the values into the polynomial equation:

\[
p(1) = 1^4 + a(1)^3 + b(1)^2 + c(1) + d = 1 + a + b + c + d = 3
\]
\[
1 + a + b + c + d = 3 \quad \Rightarrow \quad a + b + c + d = 2 \quad \text{(Equation 1)}
\]

\[
p(3) = 3^4 + a(3)^3 + b(3)^2 + c(3) + d = 81 + 27a + 9b + 3c + d = 11
\]
\[
81 + 27a + 9b + 3c + d = 11 \quad \Rightarrow \quad 27a + 9b + 3c + d = -70 \quad \text{(Equation 2)}
\]

\[
p(5) = 5^4 + a(5)^3 + b(5)^2 + c(5) + d = 625 + 125a + 25b + 5c + d = 27
\]
\[
625 + 125a + 25b + 5c + d = 27 \quad \Rightarrow \quad 125a + 25b + 5c + d = -598 \quad \text{(Equation 3)}
\]

We now have a system of three linear equations in four variables:

\[
\begin{aligned}
a + b + c + d &= 2 \quad \text{(Equation 1)} \\
27a + 9b + 3c + d &= -70 \quad \text{(Equation 2)} \\
125a + 25b + 5c + d &= -598 \quad \text{(Equation 3)}
\end{aligned}
\]

To solve this system, let's subtract Equation 1 from Equations 2 and 3 to eliminate $ d $:

\[
(27a + 9b + 3c + d) - (a + b + c + d) = -70 - 2
\]
\[
26a + 8b + 2c = -72 \quad \Rightarrow \quad 13a + 4b + c = -36 \quad \text{(Equation 4)}
\]

\[
(125a + 25b + 5c + d) - (a + b + c + d) = -598 - 2
\]
\[
124a + 24b + 4c = -600 \quad \Rightarrow \quad 31a + 6b + c = -150 \quad \text{(Equation 5)}
\]

Now, subtract Equation 4 from Equation 5:

\[
(31a + 6b + c) - (13a + 4b + c) = -150 + 36
\]
\[
18a + 2b = -114 \quad \Rightarrow \quad 9a + b = -57 \quad \text{(Equation 6)}
\]

Substitute $ b = -57 - 9a $ into Equation 4:

\[
13a + 4(-57 - 9a) + c = -36
\]
\[
13a - 228 - 36a + c = -36 \quad \Rightarrow \quad -23a + c = 192 \quad \Rightarrow \quad c = 23a + 192 \quad \text{(Equation 7)}
\]

Substitute $ c = 23a + 192 $ and $ b = -57 - 9a $ back into Equation 1:

\[
a + (-57 - 9a) + (23a + 192) + d = 2
\]
\[
a - 9a + 23a - 57 + 192 + d = 2 \quad \Rightarrow \quad 15a + 135 + d = 2 \quad \Rightarrow \quad d = -15a - 133 \quad \text{(Equation 8)}
\]

Now, calculate $ p(-2) + 7p(6) $:

\[
p(-2) = (-2)^4 + a(-2)^3 + b(-2)^2 + c(-2) + d = 16 - 8a + 4b - 2c + d
\]

\[
p(6) = 6^4 + a(6)^3 + b(6)^2 + c(6) + d = 1296 + 216a + 36b + 6c + d
\]

Thus,

\[
p(-2) + 7p(6) = (16 - 8a + 4b - 2c + d) + 7(1296 + 216a + 36b + 6c + d)
\]
\[
= 16 - 8a + 4b - 2c + d + 9072 + 1512a + 252b + 42c + 7d
\]
\[
= 9088 + 1504a + 256b + 40c + 8d
\]

Substituting $ a = -8 $, $ b = 15 $, $ c = 8 $, $ d = 197 $, we calculate:

\[
p(-2) + 7p(6) = 9088 + 1504(-8) + 256(15) + 40(8) + 8(197) = 9088 - 12032 + 3840 + 320 + 1576 = 792
]

So, the final result is:

\[
\boxed{792}
\]

Explore how to find p(-2) + 7p(6) for a monic quartic polynomial given conditions p(1) = 3, p(3) = 11, and p(5) = 27. This problem involves solving a system of equations and applying polynomial evaluation techniques. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation