Standard High contrast

Algebra

Question 4 of 6

Remainder Theorem

Question

The functions \(x^3 - 7x - 4\) and \(3x^3 - 3x^2 + bx + 14\), have the same remainder when divided by \((x - 3)\). 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 \(f(x_1) = x^3 - 7x - 4\). The divisor is \(x - 3\), so we evaluate both functions at \(x = 3\). We need to find \(b\).

  2. Find \(f(3)\) of \(f(x_1)\): substitute \(x = 3\) into \(f(x_1)\) and solve for \(f(3)\)
    \[ f(3) = (3)^3 - 7(3) - 4 \]
    \[ f(3) = 27 - 21 - 4 \]
    \[ f(3) = 2 \]
  3. Find \(f(3)\) of \(f(x_2)\): substitute \(x = 3\) into \(f(x_2)\) and solve for \(f(3)\)

    Let \(f(x_2) = 3x^3 - 3x^2 + bx + 14\).

    \[ f(3) = 3(3)^3 - 3(3)^2 + b(3) + 14 \]
    \[ f(3) = 3(27) - 3(9) + 3b + 14 \]
    \[ f(3) = 81 - 27 + 3b + 14 \]
    \[ f(3) = 3b + 68 \]
  4. To find \(b\): equate \(f(3) = 3b + 68\) and \(f(3) = 2\), then solve for \(b\)
    \[ 3b + 68 = 2 \]
    \[ 3b = 2 - 68 \]
    \[ 3b = -66 \]
  5. Final answer

    That is, \(b\) is minus 22.

    \[ b = -22 \]