Question

Difficulty: HardPolynomials, Factor and Remainder Theorems

If the polynomial P(x)=x4+ax3−7x2+bx+12P(x) = x^4 + ax^3 - 7x^2 + bx + 12 is completely divisible by x2−2x−3x^2 - 2x - 3, what is the value of a−ba - b?

  1. −10-10Answer
  2. B
    1010
  3. C
    −5-5
  4. D
    55

Answer

The value of a−ba - b is −10-10.
Factoring x2−2x−3x^2 - 2x - 3 gives (x−3)(x+1)(x - 3)(x + 1). By the Factor Theorem, P(3)=0P(3) = 0 and P(−1)=0P(-1) = 0. Substituting these into P(x)P(x) produces the linear system 9a+b=−109a + b = -10 and a+b=6a + b = 6. Solving this system yields a=−2a = -2 and b=8b = 8. Subtracting gives a−b=−2−8=−10a - b = -2 - 8 = -10.

Step-by-Step Solution

1
Factor the quadratic divisor to find the roots.
x2−2x−3=(x−3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1), so the roots are x=3x = 3 and x=−1x = -1.
By the Factor Theorem, if a polynomial is divisible by a quadratic expression, P(x)P(x) must evaluate to zero at each root of the divisor.
2
Set up equations by evaluating P(3)=0P(3) = 0 and P(−1)=0P(-1) = 0.
For x=3x = 3: 34+a(3)3−7(3)2+b(3)+12=0  ⟹  81+27a−63+3b+12=0  ⟹  9a+b=−103^4 + a(3)^3 - 7(3)^2 + b(3) + 12 = 0 \implies 81 + 27a - 63 + 3b + 12 = 0 \implies 9a + b = -10.
For x=−1x = -1: (−1)4+a(−1)3−7(−1)2+b(−1)+12=0  ⟹  1−a−7−b+12=0  ⟹  a+b=6(-1)^4 + a(-1)^3 - 7(-1)^2 + b(-1) + 12 = 0 \implies 1 - a - 7 - b + 12 = 0 \implies a + b = 6.
Evaluating the polynomial at each root yields a system of two linear equations in variables aa and bb.
3
Solve the simultaneous equations for aa and bb.
Subtracting (a+b=6)(a + b = 6) from (9a+b=−10)(9a + b = -10) gives 8a=−16  ⟹  a=−28a = -16 \implies a = -2.
Substituting a=−2a = -2 into a+b=6a + b = 6 gives −2+b=6  ⟹  b=8-2 + b = 6 \implies b = 8.
Elimination isolates aa, allowing both aa and bb to be uniquely determined.
4
Calculate a−ba - b.
a−b=−2−8=−10a - b = -2 - 8 = -10.
This computes the required expression value.

Key Concept

Factor Theorem for quadratic divisors
Estimated Time:2m 0s
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