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Algebraic Division

Question

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

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  1. Find the first term of the quotient

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

    \[ 3x^4 \div x = 3x^3 \]
    \[ (3x^3)(x + 1) = 3x^4 + 3x^3 \]
    \[ (3x^4 - 5x^3) - (3x^4 + 3x^3) = -8x^3 \]

    Bring down the next term, \(-5x^2\); the working line is now \(-8x^3 - 5x^2\).

  2. Find the second term of the quotient

    Divide \(-8x^3\) by \(x\) to get the second term of the quotient, \(-8x^2\). Multiply the divisor \((x + 1)\) by \(-8x^2\) and subtract:

    \[ -8x^3 \div x = -8x^2 \]
    \[ (-8x^2)(x + 1) = -8x^3 - 8x^2 \]
    \[ (-8x^3 - 5x^2) - (-8x^3 - 8x^2) = 3x^2 \]

    Bring down the next term, \(5x\); the working line is now \(3x^2 + 5x\).

  3. Find the third term of the quotient

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

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

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

  4. Find the fourth term of the quotient

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

    \[ 2x \div x = 2 \]
    \[ (2)(x + 1) = 2x + 2 \]
    \[ (2x + 2) - (2x + 2) = 0 \]
  5. State the result

    In this case the remainder is 0. This means \(x + 1\) is a factor of \(3x^4 - 5x^3 - 5x^2 + 5x + 2\). Therefore:

    \[ (3x^4 - 5x^3 - 5x^2 + 5x + 2) \div (x + 1) = 3x^3 - 8x^2 + 3x + 2 \]

    That is, the quotient is \(3x^3 - 8x^2 + 3x + 2\) and the remainder is 0.