Standard High contrast

Algebra

Question 5 of 6

Remainder Theorem

Question

The expression, \(2x^3 + ax^2 + b\), has a remainder of 9 when divided by \((x - 1)\) and a remainder of \(-3\) when divided by \((x + 2)\). Find the values of \(a\) and \(b\)

Show solution Hide solution Fully worked — 8 steps
  1. Set up the function and the values to substitute

    Let \(f(x) = 2x^3 + ax^2 + b\). The first divisor is \(x - 1\), so \(x = 1\) and the remainder gives \(f(1) = 9\). The second divisor is \(x + 2\), so \(x = -2\) and the remainder gives \(f(-2) = -3\). We need to find \(a\) and \(b\).

  2. Substitute \(x = 1\) into \(f(x)\) and solve for \(f(1)\)
    \[ f(1) = 2(1)^3 + a(1)^2 + b \]
    \[ f(1) = 2 + a + b \]
  3. Equate \(f(1) = 2 + a + b\) and \(f(1) = 9\), then solve for \(b\)
    \[ 2 + a + b = 9 \]
    \[ a + b = 7 \]

    This gives equation (1):

    \[ b = 7 - a \tag{1} \]
  4. Substitute \(x = -2\) into \(f(x)\) and solve for \(f(-2)\)
    \[ f(-2) = 2(-2)^3 + a(-2)^2 + b \]
    \[ f(-2) = 2(-8) + 4a + b \]
    \[ f(-2) = -16 + 4a + b \]
  5. Equate \(f(-2) = -16 + 4a + b\) and \(f(-2) = -3\)
    \[ -16 + 4a + b = -3 \]

    This gives equation (2):

    \[ 4a + b = 13 \tag{2} \]
  6. Substitute the \(b\)-value from equation (1) into \(4a + b = 13\)
    \[ 4a + (7 - a) = 13 \]

    Combine similar terms:

    \[ 3a + 7 = 13 \]

    Subtract 7 from both sides:

    \[ 3a + 7 - 7 = 13 - 7 \]

    Combine similar terms, then divide both sides by 3:

    \[ 3a = 6 \]
    \[ a = 2 \]
  7. Substitute the \(a\)-value into \(b = 7 - a\) to get the \(b\)-value
    \[ b = 7 - (2) \]
    \[ b = 5 \]
  8. Final answer

    That is, \(a\) is 2 and \(b\) is 5.

    \[ a = 2 \quad \text{and} \quad b = 5 \]