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Difficulty: MediumPolynomials, Factor and Remainder Theorems

When the polynomial P(x)=2x3+3x2−px+qP(x) = 2x^3 + 3x^2 - px + q is divided by (x−1)(x - 1), the remainder is 33. Given that (x+2)(x + 2) is a factor of P(x)P(x), what is the value of p+qp + q?

  1. 2Answer
  2. B
    4
  3. C
    6
  4. D
    -2

Answer

The value of p+qp + q is 22.
According to the Remainder Theorem, P(1)=3P(1) = 3 gives 2(1)3+3(1)2−p(1)+q=32(1)^3 + 3(1)^2 - p(1) + q = 3, which simplifies to p−q=2p - q = 2. According to the Factor Theorem, (x+2)(x + 2) being a factor means P(−2)=0P(-2) = 0, giving 2(−2)3+3(−2)2−p(−2)+q=02(-2)^3 + 3(-2)^2 - p(-2) + q = 0, which simplifies to 2p+q=42p + q = 4. Solving these equations together gives p=2p = 2 and q=0q = 0. Summing them yields p+q=2p + q = 2.

Step-by-Step Solution

1
Apply the Remainder Theorem for divisor (x−1)(x - 1)
P(1)=2(1)3+3(1)2−p(1)+q=3  ⟹  5−p+q=3  ⟹  p−q=2P(1) = 2(1)^3 + 3(1)^2 - p(1) + q = 3 \implies 5 - p + q = 3 \implies p - q = 2
By the Remainder Theorem, dividing P(x)P(x) by (x−a)(x - a) leaves a remainder equal to P(a)P(a).
2
Apply the Factor Theorem for factor (x+2)(x + 2)
P(−2)=2(−2)3+3(−2)2−p(−2)+q=0  ⟹  −16+12+2p+q=0  ⟹  2p+q=4P(-2) = 2(-2)^3 + 3(-2)^2 - p(-2) + q = 0 \implies -16 + 12 + 2p + q = 0 \implies 2p + q = 4
By the Factor Theorem, if (x−a)(x - a) is a factor of P(x)P(x), then P(a)=0P(a) = 0. Here a=−2a = -2.
3
Solve the simultaneous linear equations for pp and qq
Adding p−q=2p - q = 2 and 2p+q=42p + q = 4 yields 3p=6  ⟹  p=23p = 6 \implies p = 2. Substituting p=2p = 2 into p−q=2p - q = 2 gives q=0q = 0.
Eliminating qq allows direct calculation of pp, followed by back-substitution for qq.
4
Calculate the target expression p+qp + q
p+q=2+0=2p + q = 2 + 0 = 2
Combine the values of pp and qq to obtain the required sum.

Key Concept

Factor and Remainder Theorems for Polynomials
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Topics

Polynomials, Factor and Remainder Theorems

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