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Difficulty: HardPolynomials, Factor and Remainder Theorems

The polynomial P(x)=2x3+px2+qx−6P(x) = 2x^3 + px^2 + qx - 6 has (x−2)(x - 2) as a factor. When P(x)P(x) is divided by (x+1)(x + 1), the remainder is −12-12. What is the value of p+qp + q?

  1. −2-2Answer
  2. B
    −8-8
  3. C
    44
  4. D
    −4-4

Answer

The value of p+qp + q is −2-2.
Using the Factor Theorem, P(2)=0P(2) = 0 yields 2p+q=−52p + q = -5. Using the Remainder Theorem, P(−1)=−12P(-1) = -12 yields p−q=−4p - q = -4. Solving these two equations simultaneously gives p=−3p = -3 and q=1q = 1. Therefore, p+q=−3+1=−2p + q = -3 + 1 = -2.

Step-by-Step Solution

1
Apply the Factor Theorem for divisor (x−2)(x - 2)
2p+q=−52p + q = -5
Since (x−2)(x - 2) is a factor of P(x)P(x), P(2)=0P(2) = 0. Substituting x=2x = 2 gives 2(2)3+p(2)2+q(2)−6=0  ⟹  16+4p+2q−6=0  ⟹  4p+2q=−10  ⟹  2p+q=−52(2)^3 + p(2)^2 + q(2) - 6 = 0 \implies 16 + 4p + 2q - 6 = 0 \implies 4p + 2q = -10 \implies 2p + q = -5.
2
Apply the Remainder Theorem for divisor (x+1)(x + 1)
p−q=−4p - q = -4
Dividing P(x)P(x) by (x+1)(x + 1) leaves a remainder of −12-12, so P(−1)=−12P(-1) = -12. Substituting x=−1x = -1 gives 2(−1)3+p(−1)2+q(−1)−6=−12  ⟹  −2+p−q−6=−12  ⟹  p−q=−42(-1)^3 + p(-1)^2 + q(-1) - 6 = -12 \implies -2 + p - q - 6 = -12 \implies p - q = -4.
3
Solve the simultaneous linear equations for pp and qq
p=−3p = -3 and q=1q = 1
Adding the two equations (2p+q)+(p−q)=−5+(−4)(2p + q) + (p - q) = -5 + (-4) yields 3p=−9  ⟹  p=−33p = -9 \implies p = -3. Substituting p=−3p = -3 into p−q=−4p - q = -4 gives −3−q=−4  ⟹  q=1-3 - q = -4 \implies q = 1.
4
Calculate the required expression p+qp + q
p+q=−2p + q = -2
Summing the calculated constants: p+q=−3+1=−2p + q = -3 + 1 = -2.

Key Concept

Factor and Remainder Theorems for Polynomials
Estimated Time:2m 0s
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Topics

Polynomials, Factor and Remainder Theorems

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