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Algebra

Question 2 of 8

Algebraic Division

Question

Use long division to divide \((6x^3 - 11x^2 - 4x + 5)\) by \((2x - 1)\)

Show solution Hide solution Fully worked — 4 steps
  1. Find the first term of the quotient

    Divide \(6x^3\) by \(2x\) to get the first term of the quotient, \(3x^2\). Multiply the divisor \((2x - 1)\) by \(3x^2\) and subtract:

    \[ 6x^3 \div 2x = 3x^2 \]
    \[ (3x^2)(2x - 1) = 6x^3 - 3x^2 \]
    \[ (6x^3 - 11x^2) - (6x^3 - 3x^2) = -8x^2 \]

    Bring down the next term, \(-4x\); the working line is now \(-8x^2 - 4x\).

  2. Find the second term of the quotient

    Divide \(-8x^2\) by \(2x\) to get the second term of the quotient, \(-4x\). Multiply the divisor \((2x - 1)\) by \(-4x\) and subtract:

    \[ -8x^2 \div 2x = -4x \]
    \[ (-4x)(2x - 1) = -8x^2 + 4x \]
    \[ (-8x^2 - 4x) - (-8x^2 + 4x) = -8x \]

    Bring down the next term, \(5\); the working line is now \(-8x + 5\).

  3. Find the third term of the quotient

    Divide \(-8x\) by \(2x\) to get the third term of the quotient, \(-4\). Multiply the divisor \((2x - 1)\) by \(-4\) and subtract:

    \[ -8x \div 2x = -4 \]
    \[ (-4)(2x - 1) = -8x + 4 \]
    \[ (-8x + 5) - (-8x + 4) = 1 \]
  4. State the result

    In this case the remainder is 1. Therefore:

    \[ (6x^3 - 11x^2 - 4x + 5) \div (2x - 1) = 3x^2 - 4x - 4 + \frac{1}{2x - 1} \]

    That is, the quotient is \(3x^2 - 4x - 4\) and the remainder is 1.