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Difficulty: Very hardPolynomials, Factor and Remainder Theorems

When the cubic polynomial P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 is divided by x2−1x^2 - 1, the remainder is 3x+23x + 2. What is the remainder when P(x)P(x) is divided by 2x−32x - 3?

  1. 214\frac{21}{4}Answer
  2. B
    −454-\frac{45}{4}
  3. C
    −154-\frac{15}{4}
  4. D
    2727

Answer

The remainder when P(x)P(x) is divided by 2x−32x - 3 is 214\frac{21}{4}.
Using the Remainder Theorem on the quadratic divisor (x2−1)=(x−1)(x+1)(x^2 - 1) = (x - 1)(x + 1), we find P(1)=5P(1) = 5 and P(−1)=−1P(-1) = -1. Substituting these values into P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 produces the simultaneous linear equations p+q=−3p + q = -3 and p−q=−5p - q = -5. Solving gives p=−4p = -4 and q=1q = 1, leading to P(x)=2x3−4x2+x+6P(x) = 2x^3 - 4x^2 + x + 6. Dividing by (2x−3)(2x - 3) requires evaluating P(32)P\left(\frac{3}{2}\right), which equals 214\frac{21}{4}.

Step-by-Step Solution

1
Express the Division Algorithm for quadratic divisor x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1).
P(x)=(x2−1)Q(x)+(3x+2)P(x) = (x^2 - 1)Q(x) + (3x + 2)
By the Remainder Theorem, evaluating at roots of x2−1=0x^2 - 1 = 0 (x=1x = 1 and x=−1x = -1) yields the values of P(1)P(1) and P(−1)P(-1).
2
Calculate P(1)P(1) and P(−1)P(-1) from the remainder expression.
P(1)=3(1)+2=5P(1) = 3(1) + 2 = 5 and P(−1)=3(−1)+2=−1P(-1) = 3(-1) + 2 = -1
The quotient term (x2−1)Q(x)(x^2 - 1)Q(x) vanishes at x=1x = 1 and x=−1x = -1.
3
Substitute x=1x = 1 and x=−1x = -1 into P(x)=2x3+px2+qx+6P(x) = 2x^3 + px^2 + qx + 6 to build a system of linear equations.
P(1)=2+p+q+6=p+q+8=5  ⟹  p+q=−3P(1) = 2 + p + q + 6 = p + q + 8 = 5 \implies p + q = -3, and P(−1)=−2+p−q+6=p−q+4=−1  ⟹  p−q=−5P(-1) = -2 + p - q + 6 = p - q + 4 = -1 \implies p - q = -5
This sets up two simultaneous linear equations in terms of pp and qq.
4
Solve the system of equations for pp and qq.
Adding the equations gives 2p=−8  ⟹  p=−42p = -8 \implies p = -4, and substituting into p+q=−3p + q = -3 gives q=1q = 1. Thus, P(x)=2x3−4x2+x+6P(x) = 2x^3 - 4x^2 + x + 6.
Determining pp and qq gives the explicit formula for the polynomial.
5
Apply the Remainder Theorem to find the remainder when P(x)P(x) is divided by 2x−32x - 3.
Set 2x−3=0  ⟹  x=322x - 3 = 0 \implies x = \frac{3}{2}. Evaluate P(32)=2(32)3−4(32)2+32+6=2(278)−4(94)+32+6=274−9+64+6=334−3=214P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 4\left(\frac{3}{2}\right)^2 + \frac{3}{2} + 6 = 2\left(\frac{27}{8}\right) - 4\left(\frac{9}{4}\right) + \frac{3}{2} + 6 = \frac{27}{4} - 9 + \frac{6}{4} + 6 = \frac{33}{4} - 3 = \frac{21}{4}.
The remainder of a polynomial P(x)P(x) divided by (ax−b)(ax - b) is P(ba)P\left(\frac{b}{a}\right).

Key Concept

Polynomial Division Algorithm and Remainder Theorem for Linear and Quadratic Divisors
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Topics

Polynomials, Factor and Remainder Theorems

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