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Difficulty: HardPolynomials, Factor and Remainder Theorems

When the polynomial P(x)=3x3+ax2+bx−10P(x) = 3x^3 + ax^2 + bx - 10 is divided by (x−2)(x - 2), the remainder is 1414, and when it is divided by (x+1)(x + 1), the remainder is −16-16. What is the value of a+ba + b?

Answer: 1

Answer

The value of a+ba + b is 11.
According to the Remainder Theorem, dividing P(x)P(x) by (x−2)(x - 2) and (x+1)(x + 1) gives P(2)=14P(2) = 14 and P(−1)=−16P(-1) = -16 respectively. Expanding these expressions forms two linear equations: 2a+b=02a + b = 0 and a−b=−3a - b = -3. Solving these simultaneously gives a=−1a = -1 and b=2b = 2, so a+b=1a + b = 1.

Step-by-Step Solution

1
Apply the Remainder Theorem for the first divisor (x−2)(x - 2)
2a+b=02a + b = 0
By the Remainder Theorem, P(2)=14P(2) = 14. Substituting x=2x = 2 into P(x)P(x) yields 3(8)+4a+2b−10=143(8) + 4a + 2b - 10 = 14, which simplifies to 2a+b=02a + b = 0.
2
Apply the Remainder Theorem for the second divisor (x+1)(x + 1)
a−b=−3a - b = -3
By the Remainder Theorem, P(−1)=−16P(-1) = -16. Substituting x=−1x = -1 into P(x)P(x) yields 3(−1)+a−b−10=−163(-1) + a - b - 10 = -16, which simplifies to a−b=−3a - b = -3.
3
Solve the simultaneous equations for aa and bb
a=−1a = -1 and b=2b = 2
Adding 2a+b=02a + b = 0 and a−b=−3a - b = -3 yields 3a=−33a = -3, giving a=−1a = -1. Substituting a=−1a = -1 into 2a+b=02a + b = 0 gives b=2b = 2.
4
Calculate the value of a+ba + b
1
Adding the computed values yields a+b=−1+2=1a + b = -1 + 2 = 1.

Key Concept

Polynomial Remainder Theorem and Systems of Linear Equations
Estimated Time:2m 0s
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Topics

Polynomials, Factor and Remainder Theorems

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