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Difficulty: MediumPolynomials, Factor and Remainder Theorems

When the polynomial P(x)=x3−ax2+bx−6P(x) = x^3 - ax^2 + bx - 6 is divided by (x−1)(x - 1), the remainder is −4-4. If (x−2)(x - 2) is a factor of P(x)P(x), what is the value of a+ba + b?

  1. 55Answer
  2. B
    11
  3. C
    −1-1
  4. D
    −3-3

Answer

The value of a+ba + b is 55.
Using the Remainder Theorem with P(1)=−4P(1) = -4 gives the equation −a+b=1-a + b = 1. Using the Factor Theorem with P(2)=0P(2) = 0 gives 2a−b=12a - b = 1. Solving this simultaneous system gives a=2a = 2 and b=3b = 3, leading to a+b=5a + b = 5.

Step-by-Step Solution

1
Apply the Remainder Theorem for division by (x−1)(x - 1).
−a+b=1-a + b = 1
According to the Remainder Theorem, P(1)=−4P(1) = -4. Substituting x=1x = 1 into P(x)P(x) yields 13−a(1)2+b(1)−6=−4  ⟹  −a+b=11^3 - a(1)^2 + b(1) - 6 = -4 \implies -a + b = 1.
2
Apply the Factor Theorem for the factor (x−2)(x - 2).
2a−b=12a - b = 1
Since (x−2)(x - 2) is a factor, P(2)=0P(2) = 0. Substituting x=2x = 2 into P(x)P(x) yields 23−a(2)2+b(2)−6=0  ⟹  8−4a+2b−6=0  ⟹  2a−b=12^3 - a(2)^2 + b(2) - 6 = 0 \implies 8 - 4a + 2b - 6 = 0 \implies 2a - b = 1.
3
Solve the system of linear equations simultaneously.
a=2,b=3a = 2, b = 3
Adding the two equations (−a+b)+(2a−b)=1+1(-a + b) + (2a - b) = 1 + 1 yields a=2a = 2. Substituting a=2a = 2 back into −a+b=1-a + b = 1 gives b=3b = 3.
4
Calculate the required value a+ba + b.
55
a+b=2+3=5a + b = 2 + 3 = 5.

Key Concept

Polynomial Factor and Remainder Theorems
Estimated Time:1m 30s
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Topics

Polynomials, Factor and Remainder Theorems

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