Question

If \( P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e \) is a polynomial such that: \[ P(0) = 1, \quad P(1) = 2, \quad P(2) = 5, \] \[ P(3) = 10, \quad P(4) = 17, \] then find the value of \( P(5) \)=

• A. \( 26 \)
• B. \( 146 \)
• C. \( 126 \)
• D. \( 76 \)

Hint:

For polynomial sequences, analyze first and second differences to identify patterns, then use extrapolation to determine higher-order values.

Answer 1

The Correct Option is B

The given polynomial is:

P(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e

We are given the following conditions:

• P(0) = 1
• P(1) = 2
• P(2) = 5
• P(3) = 10
• P(4) = 17

Substitute these values into the polynomial equation to create a system of linear equations:

1. For P(0) = 1:

Substitute x = 0 into the polynomial:

P(0) = 0 + a(0)⁴ + b(0)³ + c(0)² + d(0) + e = 1

So, e = 1.

2. For P(1) = 2:

Substitute x = 1:

P(1) = 1⁵ + a(1)⁴ + b(1)³ + c(1)² + d(1) + e = 2

Which simplifies to:

1 + a + b + c + d + e = 2

Substitute e = 1:

1 + a + b + c + d + 1 = 2

So, a + b + c + d = 0.

3. For P(2) = 5:

Substitute x = 2:

P(2) = 2⁵ + a(2)⁴ + b(2)³ + c(2)² + d(2) + e = 5

Which simplifies to:

32 + 16a + 8b + 4c + 2d + e = 5

Substitute e = 1:

32 + 16a + 8b + 4c + 2d + 1 = 5

Which simplifies to:

16a + 8b + 4c + 2d = -28

4. For P(3) = 10:

Substitute x = 3:

P(3) = 3⁵ + a(3)⁴ + b(3)³ + c(3)² + d(3) + e = 10

Which simplifies to:

243 + 81a + 27b + 9c + 3d + e = 10

Substitute e = 1:

243 + 81a + 27b + 9c + 3d + 1 = 10

Which simplifies to:

81a + 27b + 9c + 3d = -234

5. For P(4) = 17:

Substitute x = 4:

P(4) = 4⁵ + a(4)⁴ + b(4)³ + c(4)² + d(4) + e = 17

Which simplifies to:

1024 + 256a + 64b + 16c + 4d + e = 17

Substitute e = 1:

1024 + 256a + 64b + 16c + 4d + 1 = 17

Which simplifies to:

256a + 64b + 16c + 4d = -1008

Now solve this system of equations:

• Equation 1: a + b + c + d = 0
• Equation 2: 16a + 8b + 4c + 2d = -28
• Equation 3: 81a + 27b + 9c + 3d = -234
• Equation 4: 256a + 64b + 16c + 4d = -1008

After solving the system, we get the value of P(5) = 146.

Answer 2

Given the polynomial:
\[ P(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e \] with conditions:
\[ P(0) = 1, \quad P(1) = 2, \quad P(2) = 5, \quad P(3) = 10, \quad P(4) = 17, \] find \( P(5) \).

Step 1: Write the polynomial values:
\[ P(0) = e = 1 \]

Step 2: Define the sequence:
\[ y_n = P(n) \] Given:
\[ y_0 = 1, \quad y_1 = 2, \quad y_2 = 5, \quad y_3 = 10, \quad y_4 = 17 \]

Step 3: Calculate the first differences \( \Delta y_n = y_{n+1} - y_n \):
\[ \Delta y_0 = y_1 - y_0 = 2 - 1 = 1 \]
\[ \Delta y_1 = y_2 - y_1 = 5 - 2 = 3 \]
\[ \Delta y_2 = y_3 - y_2 = 10 - 5 = 5 \]
\[ \Delta y_3 = y_4 - y_3 = 17 - 10 = 7 \]

Step 4: Calculate the second differences \( \Delta^2 y_n = \Delta y_{n+1} - \Delta y_n \):
\[ \Delta^2 y_0 = 3 - 1 = 2 \]
\[ \Delta^2 y_1 = 5 - 3 = 2 \]
\[ \Delta^2 y_2 = 7 - 5 = 2 \]

Step 5: Since the second differences are constant (equal to 2), the sequence \( y_n \) fits a quadratic polynomial:
\[ P(n) = A n^2 + B n + C \]

Step 6: Use the given points to find \( A, B, C \):
From \( P(0) = C = 1 \).
Using \( P(1) = A(1)^2 + B(1) + 1 = 2 \Rightarrow A + B + 1 = 2 \Rightarrow A + B = 1 \).
Using \( P(2) = 4A + 2B + 1 = 5 \Rightarrow 4A + 2B = 4 \).

Step 7: Solve the system:
From \( A + B = 1 \Rightarrow B = 1 - A \).
Substitute into second equation:
\[ 4A + 2(1 - A) = 4 \implies 4A + 2 - 2A = 4 \implies 2A + 2 = 4 \implies 2A = 2 \implies A = 1 \]
Then:
\[ B = 1 - 1 = 0 \]

Step 8: So,
\[ P(n) = n^2 + 1 \]
Check for \( n=3 \): \( 3^2 + 1 = 9 + 1 = 10 \) correct.
Check for \( n=4 \): \( 4^2 + 1 = 16 + 1 = 17 \) correct.

Step 9: But the given polynomial is degree 5, so this contradicts the assumption that \( P(n) \) is quadratic. The problem states a 5th-degree polynomial with leading coefficient 1.

Step 10: Consider the polynomial \( Q(x) = P(x) - x^2 - 1 \). Then \( Q(n) = 0 \) for \( n = 0,1,2,3,4 \).
So \( Q(x) \) has roots at 0,1,2,3,4 and degree 5. Hence,
\[ Q(x) = k x (x-1)(x-2)(x-3)(x-4) \]

Step 11: Since \( P(x) = x^2 + 1 + Q(x) \) and \( P(x) \) is degree 5 with leading coefficient 1:
The leading term of \( Q(x) \) is \( k x^5 \).
The leading term of \( P(x) \) is \( x^5 \).
So \( k = 1 \).

Step 12: Therefore:
\[ P(x) = x^2 + 1 + x (x-1)(x-2)(x-3)(x-4) \]

Step 13: Find \( P(5) \):
Calculate \( Q(5) = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Calculate \( 5^2 + 1 = 25 + 1 = 26 \).

So:
\[ P(5) = 26 + 120 = 146 \]

Hence,
\[ \boxed{146} \]