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Algebra

Question 4 of 8

Algebraic Division

Question

Use long division to divide \((3x^4 - 7x^3 - 43x^2 + 123x - 36)\) by \((3x - 1)\)

Show solution Hide solution Fully worked — 5 steps
  1. Find the first term of the quotient

    Divide \(3x^4\) by \(3x\) to get the first term of the quotient, \(x^3\). Multiply the divisor \((3x - 1)\) by \(x^3\) and subtract:

    \[ 3x^4 \div 3x = x^3 \]
    \[ (x^3)(3x - 1) = 3x^4 - x^3 \]
    \[ (3x^4 - 7x^3) - (3x^4 - x^3) = -6x^3 \]

    Bring down the next term, \(-43x^2\); the working line is now \(-6x^3 - 43x^2\).

  2. Find the second term of the quotient

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

    \[ -6x^3 \div 3x = -2x^2 \]
    \[ (-2x^2)(3x - 1) = -6x^3 + 2x^2 \]
    \[ (-6x^3 - 43x^2) - (-6x^3 + 2x^2) = -45x^2 \]

    Bring down the next term, \(123x\); the working line is now \(-45x^2 + 123x\).

  3. Find the third term of the quotient

    Divide \(-45x^2\) by \(3x\) to get the third term of the quotient, \(-15x\). Multiply the divisor \((3x - 1)\) by \(-15x\) and subtract:

    \[ -45x^2 \div 3x = -15x \]
    \[ (-15x)(3x - 1) = -45x^2 + 15x \]
    \[ (-45x^2 + 123x) - (-45x^2 + 15x) = 108x \]

    Bring down the next term, \(-36\); the working line is now \(108x - 36\).

  4. Find the fourth term of the quotient

    Divide \(108x\) by \(3x\) to get the fourth term of the quotient, \(36\). Multiply the divisor \((3x - 1)\) by \(36\) and subtract:

    \[ 108x \div 3x = 36 \]
    \[ (36)(3x - 1) = 108x - 36 \]
    \[ (108x - 36) - (108x - 36) = 0 \]
  5. State the result

    In this case the remainder is 0. This means that \(3x - 1\) is a factor of \(3x^4 - 7x^3 - 43x^2 + 123x - 36\). Therefore:

    \[ (3x^4 - 7x^3 - 43x^2 + 123x - 36) \div (3x - 1) = x^3 - 2x^2 - 15x + 36 \]

    That is, the quotient is \(x^3 - 2x^2 - 15x + 36\) and the remainder is 0.