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Algebra

Question 7 of 8

Algebraic Division

Question

Use long division to divide \((x^4 + 5x^3 - 3x^2 - 10x + 12)\) by \((x^2 + x - 1)\)

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  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 \((x^2 + x - 1)\) by \(x^2\) and subtract:

    \[ x^4 \div x^2 = x^2 \]
    \[ (x^2)(x^2 + x - 1) = x^4 + x^3 - x^2 \]
    \[ (x^4 + 5x^3 - 3x^2) - (x^4 + x^3 - x^2) = 4x^3 - 2x^2 \]

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

  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 \((x^2 + x - 1)\) by \(4x\) and subtract:

    \[ 4x^3 \div x^2 = 4x \]
    \[ (4x)(x^2 + x - 1) = 4x^3 + 4x^2 - 4x \]
    \[ (4x^3 - 2x^2 - 10x) - (4x^3 + 4x^2 - 4x) = -6x^2 - 6x \]

    Bring down the next term, \(12\); the working line is now \(-6x^2 - 6x + 12\).

  3. Find the third term of the quotient

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

    \[ -6x^2 \div x^2 = -6 \]
    \[ (-6)(x^2 + x - 1) = -6x^2 - 6x + 6 \]
    \[ (-6x^2 - 6x + 12) - (-6x^2 - 6x + 6) = 6 \]
  4. State the result

    In this case the remainder is 6. Therefore:

    \[ (x^4 + 5x^3 - 3x^2 - 10x + 12) \div (x^2 + x - 1) = x^2 + 4x - 6 + \frac{6}{x^2 + x - 1} \]

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