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  5. Long Division

Long Division

Long Division is used when dividing polynomials by linear factors. Unlike synthetic division, long division is more versatile as a ploynomial can be divided by another polynomial not just linear expressions.

Steps for Long Division

  1. Setup: In long division format, write the polynomial that is being divided (the dividend) and the polynomial that is dividing into the dividend (the divisor), with the dividend appearing within the division symbol and the divisor outside. Verify that the two polynomials are placed in decreasing order of degree, and with missing terms, use \( 0 \).
  2. Divide: To find the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor. The result should be written below the dividend after multiplying the whole divisor by this term. Subtract the product from the dividend, bringing down the next term. Repeat the process utill there are no more terms in the dividend.
  3. Result: The quotient is the result of the division. A residual expressed as a fraction over the divisor will be present if there is one.

Example

Divide the polynomial \(2x^3 - 5x^2 + 3x + 1\) by \(x-2\) using long division.

First, we can rewrite this expression in the long division format. We can identify the dividend is \(2x^3 - 5x^2 + 3x + 1\) and the divisor is \(x-2\):

$$ \require{enclose} \begin{array}{r} \\[-3pt] x-2 \enclose{longdiv}{2x^3 - 5x^2 + 3x + 1} \\[-3pt] \end{array} $$

Next, we can divide the polynomial using long division:

$$ \require{enclose} \begin{array}{r} 2x^2 - x + 1 \\[-3pt] x - 2 \enclose{longdiv}{2x^3 - 5x^2 + 3x + 1} \\[-3pt] \underline{-2x^3 + 4x^2}\phantom{2} \\[-3pt] -x^2 + 3x + 1 \\[-3pt] \underline{x^2 - 2x}\phantom{2} \\[-3pt] x+ 1 \\[-3pt] \underline{-x + 2}\phantom{2} \\[-3pt] \color{blue}3 \\[-3pt] \end{array} $$

Therefore, we can determine that dividing the polynomial \(2x^3 - 5x^2 + 3x + 1\) by \(x-2\) using long division gives us a remainder of \(\boldsymbol{3}\).


Divide the following polynomials using long division.

\(6x^2 - 5x + 6\) by \(2x + 1\)

First, we can rewrite this expression in the long division format. We can identify the dividend is \( 6x^2 - 5x + 6 \) and the divisor is \(2x + 1 \):

$$ \require{enclose} \begin{array}{r} \\[-3pt] 2x+1 \enclose{longdiv}{6x^2 - 5x + 6 } \\[-3pt] \end{array} $$

Next, we can divide the polynomial using long division:

$$ \require{enclose} \begin{array}{r} 3x-4 \\[-3pt] 2x+1 \enclose{longdiv}{6x^2 - 5x + 6} \\[-3pt] \underline{-6x^2 - 3x}\phantom{2} \\[-3pt] -8x + 6 \\[-3pt] \underline{8x + 4}\phantom{2} \\[-3pt] \color{blue}10 \\[-3pt] \end{array} $$

Therefore, we can determine that dividing the polynomial \(6x^2 - 5x + 6\) by \(2x + 1\) using long division gives us a remainder of \(\boldsymbol{10}\).

\(x^3 + 3x^2 - 8x - 4\) by \(x - 3\)

First, we can rewrite this expression in the long division format. We can identify the dividend is \(x^3 + 3x^2 - 8x - 4\) and the divisor is \(x - 3\):

$$ \require{enclose} \begin{array}{r} \\[-3pt] x-3 \enclose{longdiv}{x^3 + 3x^2 - 8x - 4} \\[-3pt] \end{array} $$

Next, we can divide the polynomial using long division:

$$ \require{enclose} \begin{array}{r} x^2 + 6x + 10 \\[-3pt] x-3 \enclose{longdiv}{x^3 + 3x^2 - 8x - 4} \\[-3pt] \underline{-x^3 + 3x^2}\phantom{2} \\[-3pt] 6x^2 - 8x - 4 \\[-3pt] \underline{-6x^2+18x}\phantom{2} \\[-3pt] 10x-4 \\[-3pt] \underline{-10x + 30}\phantom{2} \\[-3pt] \color{blue}26 \\[-3pt] \end{array} $$

Therefore, we can determine that dividing the polynomial \(x^3 + 3x^2 - 8x - 4\) by \(x - 3\) using long division gives us a remainder of \(\boldsymbol{26}\).

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