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Algebra

Question 6 of 6

Remainder Theorem

Question

The functions \(x^4 + bx^3 - 3x^2 - 4x - 5\) and \(3x^3 - 8x^2 - 2x - 13\), have the same remainder when divided by \((x - 2)\). What is the value of \(b\)?

Show solution Hide solution Fully worked — 5 steps
  1. Set up the first function and the value to substitute

    Let \(f(x_1) = 3x^3 - 8x^2 - 2x - 13\). The divisor is \(x - 2\), so we evaluate both functions at \(x = 2\). We need to find \(b\).

  2. Substitute \(x = 2\) into \(f(x_1)\) and solve for \(f(2)\)
    \[ f(2) = 3(2)^3 - 8(2)^2 - 2(2) - 13 \]
    \[ f(2) = 3(8) - 8(4) - 4 - 13 \]
    \[ f(2) = 24 - 32 - 4 - 13 \]
    \[ f(2) = -25 \]
  3. Substitute \(x = 2\) into \(f(x_2)\) and solve for \(f(2)\)

    Let \(f(x_2) = x^4 + bx^3 - 3x^2 - 4x - 5\).

    \[ f(2) = (2)^4 + b(2)^3 - 3(2)^2 - 4(2) - 5 \]
    \[ f(2) = (16) + 8b - 3(4) - 8 - 5 \]
    \[ f(2) = 16 + 8b - 12 - 8 - 5 \]
    \[ f(2) = -9 + 8b \]
  4. Equate the two remainders and solve for \(b\)
    \[ -9 + 8b = -25 \]
    \[ 8b = -25 + 9 \]
    \[ 8b = -16 \]

    Divide both sides by 8:

    \[ b = -2 \]
  5. Final answer

    That is, \(b\) is minus 2, and so the function is:

    \[ b = -2 \; \therefore \; f(x) = x^4 - 2x^3 - 3x^2 - 4x - 5 \]