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Algebra

Question 6 of 8

Algebraic Division

Question

Use long division to divide \((x^4 + 4x^3 - 8x^2 + 56x - 32)\) by \((x^2 - 2x + 8)\)

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

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

    \[ x^4 \div x^2 = x^2 \]
    \[ (x^2)(x^2 - 2x + 8) = x^4 - 2x^3 + 8x^2 \]
    \[ (x^4 + 4x^3 - 8x^2) - (x^4 - 2x^3 + 8x^2) = 6x^3 - 16x^2 \]

    Bring down the next term, \(56x\); the working line is now \(6x^3 - 16x^2 + 56x\).

  2. Find the second term of the quotient

    Divide \(6x^3\) by \(x^2\) to get the second term of the quotient, \(6x\). Multiply the divisor \((x^2 - 2x + 8)\) by \(6x\) and subtract:

    \[ 6x^3 \div x^2 = 6x \]
    \[ (6x)(x^2 - 2x + 8) = 6x^3 - 12x^2 + 48x \]
    \[ (6x^3 - 16x^2 + 56x) - (6x^3 - 12x^2 + 48x) = -4x^2 + 8x \]

    Bring down the next term, \(-32\); the working line is now \(-4x^2 + 8x - 32\).

  3. Find the third term of the quotient

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

    \[ -4x^2 \div x^2 = -4 \]
    \[ (-4)(x^2 - 2x + 8) = -4x^2 + 8x - 32 \]
    \[ (-4x^2 + 8x - 32) - (-4x^2 + 8x - 32) = 0 \]
  4. State the result

    In this case the remainder is 0. This means that \(x^2 - 2x + 8\) is a factor of \(x^4 + 4x^3 - 8x^2 + 56x - 32\). Therefore:

    \[ (x^4 + 4x^3 - 8x^2 + 56x - 32) \div (x^2 - 2x + 8) = x^2 + 6x - 4 \]

    That is, the quotient is \(x^2 + 6x - 4\) and the remainder is 0.