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.