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Algebra

Question 8 of 8

Algebraic Division

Question

Use long division to divide \((2x^4 - 4x^3 + 20x - 50)\) by \((x^2 - 5)\)

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

    Divide \(2x^4\) by \(x^2\) to get the first term of the quotient, \(2x^2\). Multiply the divisor \((x^2 - 5)\) by \(2x^2\) and subtract:

    \[ 2x^4 \div x^2 = 2x^2 \]
    \[ (2x^2)(x^2 - 5) = 2x^4 - 10x^2 \]
    \[ (2x^4 - 4x^3) - (2x^4 - 10x^2) = -4x^3 + 10x^2 \]

    Bring down the next term, \(20x\); the working line is now \(-4x^3 + 10x^2 + 20x\).

  2. Find the second term of the quotient

    Divide \(-4x^3\) by \(x^2\) to get the second term of the quotient, \(-4x\). Multiply the divisor \((x^2 - 5)\) by \(-4x\) and subtract:

    \[ -4x^3 \div x^2 = -4x \]
    \[ (-4x)(x^2 - 5) = -4x^3 + 20x \]
    \[ (-4x^3 + 10x^2 + 20x) - (-4x^3 + 20x) = 10x^2 \]

    Bring down the next term, \(-50\); the working line is now \(10x^2 + 0 - 50\).

  3. Find the third term of the quotient

    Divide \(10x^2\) by \(x^2\) to get the third term of the quotient, \(10\). Multiply the divisor \((x^2 - 5)\) by \(10\) and subtract:

    \[ 10x^2 \div x^2 = 10 \]
    \[ (10)(x^2 - 5) = 10x^2 - 50 \]
    \[ (10x^2 - 50) - (10x^2 - 50) = 0 \]
  4. State the result

    In this case the remainder is 0. This means that \(x^2 - 5\) is a factor of \(2x^4 - 4x^3 + 20x - 50\). Therefore:

    \[ (2x^4 - 4x^3 + 20x - 50) \div (x^2 - 5) = 2x^2 - 4x + 10 \]

    That is, the quotient is \(2x^2 - 4x + 10\) and the remainder is 0.