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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−2)(x - 2), the remainder is 00. When P(x)P(x) is divided by (x+1)(x + 1), the remainder is 1212. What is the value of a+ba + b?

  1. −7-7Answer
  2. B
    1717
  3. C
    77
  4. D
    1919

Answer

The value of a+ba + b is −7-7.
According to the Remainder Theorem, dividing P(x)P(x) by (x−2)(x - 2) with a remainder of 00 means P(2)=0P(2) = 0, giving the equation 2a+b=−12a + b = -1. Dividing P(x)P(x) by (x+1)(x + 1) with a remainder of 1212 means P(−1)=12P(-1) = 12, giving a−b=19a - b = 19. Solving these two linear equations simultaneously yields a=6a = 6 and b=−13b = -13. Adding these values together gives a+b=−7a + b = -7.

Step-by-Step Solution

1
Apply the Factor/Remainder Theorem for divisor (x−2)(x - 2)
P(2)=23+a(2)2+b(2)−6=0  ⟹  4a+2b+2=0  ⟹  2a+b=−1P(2) = 2^3 + a(2)^2 + b(2) - 6 = 0 \implies 4a + 2b + 2 = 0 \implies 2a + b = -1
Since dividing P(x)P(x) by (x−2)(x - 2) leaves a remainder of 00, P(2)=0P(2) = 0.
2
Apply the Remainder Theorem for divisor (x+1)(x + 1)
P(−1)=(−1)3+a(−1)2+b(−1)−6=12  ⟹  a−b−7=12  ⟹  a−b=19P(-1) = (-1)^3 + a(-1)^2 + b(-1) - 6 = 12 \implies a - b - 7 = 12 \implies a - b = 19
Setting the linear divisor x+1=0x + 1 = 0 gives x=−1x = -1, so P(−1)=12P(-1) = 12.
3
Solve the system of simultaneous linear equations for aa and bb
Adding (2a+b=−1)(2a + b = -1) and (a−b=19)(a - b = 19) yields 3a=18  ⟹  a=63a = 18 \implies a = 6. Substituting a=6a = 6 into a−b=19a - b = 19 gives 6−b=19  ⟹  b=−136 - b = 19 \implies b = -13.
Eliminating bb allows finding the values of constants aa and bb.
4
Calculate a+ba + b
a+b=6+(−13)=−7a + b = 6 + (-13) = -7
Summing the determined constants aa and bb gives the target expression.

Key Concept

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

Polynomials, Factor and Remainder Theorems

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