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Difficulty: HardPolynomials, Factor and Remainder Theorems

The polynomial P(x)=2x3+ax2+bx+6P(x) = 2x^3 + ax^2 + bx + 6 leaves a remainder of 1212 when divided by (x−1)(x - 1) and has (x+3)(x + 3) as a factor. What is the value of a−ba - b?

  1. 66Answer
  2. B
    88
  3. C
    44
  4. D
    −28-28

Answer

The value of a−ba - b is 66.
Applying the Remainder Theorem with P(1)=12P(1) = 12 gives a+b=4a + b = 4. Applying the Factor Theorem with P(−3)=0P(-3) = 0 gives 3a−b=163a - b = 16. Solving these simultaneous linear equations gives a=5a = 5 and b=−1b = -1. Evaluating a−b=5−(−1)a - b = 5 - (-1) gives 66.

Step-by-Step Solution

1
Apply the Remainder Theorem for the divisor (x−1)(x - 1)
a+b=4a + b = 4
According to the Remainder Theorem, P(1)=12P(1) = 12. Substituting x=1x = 1 gives 2(1)3+a(1)2+b(1)+6=12  ⟹  a+b+8=12  ⟹  a+b=42(1)^3 + a(1)^2 + b(1) + 6 = 12 \implies a + b + 8 = 12 \implies a + b = 4.
2
Apply the Factor Theorem for the factor (x+3)(x + 3)
3a−b=163a - b = 16
According to the Factor Theorem, P(−3)=0P(-3) = 0. Substituting x=−3x = -3 gives 2(−3)3+a(−3)2+b(−3)+6=0  ⟹  −54+9a−3b+6=0  ⟹  9a−3b=48  ⟹  3a−b=162(-3)^3 + a(-3)^2 + b(-3) + 6 = 0 \implies -54 + 9a - 3b + 6 = 0 \implies 9a - 3b = 48 \implies 3a - b = 16.
3
Solve the system of linear equations for aa and bb
a=5a = 5 and b=−1b = -1
Adding (a+b=4)(a + b = 4) and (3a−b=16)(3a - b = 16) yields 4a=20  ⟹  a=54a = 20 \implies a = 5. Substituting a=5a = 5 into a+b=4a + b = 4 gives b=−1b = -1.
4
Calculate the target value a−ba - b
a−b=6a - b = 6
Subtracting bb from aa gives 5−(−1)=5+1=65 - (-1) = 5 + 1 = 6.

Key Concept

Polynomial Remainder and Factor Theorems
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Topics

Polynomials, Factor and Remainder Theorems

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