# Dividing Polynomials Using Long Division

1. Polynomial Long Division Steps
2. Examples

The sections on the box method and on synthetic division showed how both approaches could be used to divide polynomials in certain circumstances.

But we can also go about dividing polynomials using long division, and sometimes this can be a preferred approach to some of the other methods available.

This page will show a range of examples of long division involving polynomials, the likes of which can appear often in Math.

## Division of Polynomials using Long DivisionSteps:

We’ll look at the following division sum,    ( x^{\tt{2}} \space {\text{--}} \space 3x \space {\text{--}} \space 10 ) \space \div \space ( x + 2 ).

Set it up as a standard division sum with a half box like we do with numbers, and proceed from there.

\begin{array}{r} x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ \end{array}

1)  We divide the leading term in the dividend by the leading term in the divisor, and place the result above.

( \frac{x^{\tiny{2}}}{x}  =  x )
\begin{array}{r} x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ \end{array}

2)  Now multiply this result by the divisor, and place the result below the dividend.

( x \times ( x + 2 )   =   x^2 + 2x )
\begin{array}{r} x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ \space\space x^{\tiny{2}} + 2x \space\space\space\space\space\space\space\space \\ \end{array}

3)  Then subtract this result from the dividend, writing the result underneath.
\begin{array}{r} x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ {\text{--}} \space\space \underline{x^{\tiny{2}} + 2x \space\space\space\space\space\space\space\space} \\ {\text{-}}5x \space {\text{--}} \space 10\\ \end{array}

After this point, it’s a repeat of the same steps but with the new polynomial we have underneath,  {\text{-}}5x \space {\text{--}} \space 10.
We actually repeat the same steps with each new polynomial we obtain until the result is either zero, or a polynomial of a degree less than the original divisor polynomial.
When dividing polynomials using long division.

4)     ( \frac{{\text{-}}5x}{x}  =  {\text{-}}5 )

\begin{array}{r} x \space \space \space {\text{-}}5 \space \space \space \space \space\space \space\space\space\space\space\\ x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ {\text{--}} \space\space \underline{x^{\tiny{2}} + 2x \space\space\space\space\space\space\space\space} \\ {\text{-}}5x \space {\text{--}} \space 10\\ \end{array}

5)     ( {\text{-}}5 \times ( x + 2 )  =  {\text{-}}5x \space {\text{--}} \space 10 )

\begin{array}{r} x \space \space \space {\text{-}}5 \space \space \space \space \space\space \space\space\space\space\space\\ x + 2 \space\space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x \space {\text{--}} \space 10}\\ {\text{--}} \space\space \underline{x^{\tiny{2}} + 2x \space\space\space\space\space\space\space\space} \\ {\text{-}}5x \space {\text{--}} \space 10\\ {\text{--}} \space\space \underline{{\text{-}}5x \space {\text{--}} \space 10}\\ 0\\ \end{array}

After carrying out the calculations a 2nd time, the result is  0.
Thus there is no remainder, and the answer is the result from above we placed on top.

( x^{\tt{2}} \space {\text{--}} \space 3x \space {\text{--}} \space 10 ) \space \div \space ( x + 2 ) \space = \space x \space {\text{--}} \space 5

Examples

(1.1)

( x^{\tt{2}} \space {\text{--}} \space 3x + 6 ) \space \div \space ( x + 1 )

Solution

1)       [ \frac{x^{\tiny{2}}}{x}  =  x ]
\begin{array}{r} x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x + 1 \space|\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x + 6}\\ \end{array}

2)     [ x \times ( x + 1 )   =   x^2 + x ]
\begin{array}{r} x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x + 1 \space |\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x + 6}\\ {\text{--}} \space \underline{x^{\tiny{2}} \space + \space x \space\space\space\space\space\space\space} \\ {\text{-}}4x + 6\\ \end{array}

3)     [ \frac{{\text{-}}4x}{x}  =  {\text{-}}4 ]

\begin{array}{r} x \space \space \space \space {\text{-}}4 \space \space \space \space\space \space\space\space\space\space\\ x + 1 \space |\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x + 6}\\ {\text{--}} \space\space \underline{x^{\tiny{2}} \space + \space x \space\space\space\space\space\space\space} \\ {\text{-}}4x + 6\\ \end{array}

4)     [ {\text{-}}4 \times ( x + 1 )  =  {\text{-}}4x \space {\text{--}} \space 4 ]

\begin{array}{r} x \space \space \space {\text{-}}4 \space \space \space \space \space\space \space\space\space\space\space\\ x + 1 \space |\overline{\space x^{\tiny{\tt{2}}} \space\space {\text{--}} \space 3x + 6}\\ {\text{--}} \space\space \underline{x^{\tiny{2}} \space + \space x \space\space\space\space\space\space\space} \\ {\text{-}}4x + 6\\ {\text{--}} \space\space \underline{{\text{-}}4x \space {\text{--}} \space\space 4}\\ 10\\ \end{array}

( x^{\tt{2}} \space {\text{--}} \space 3x + 6 ) \space \div \space ( x + 1 ) \space = \space x \space {\text{--}} \space 4 + \frac{10}{x + 1}

(1.2)

( 4x^{\tt{2}} + 5x \space {\text{--}} \space 3 ) \space \div \space ( x \space {\text{--}} \space 1 )

Solution

1)       [ \frac{4x^{\tiny{2}}}{x}  =  4x ]
\begin{array}{r} 4x \space \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x \space {\text{--}} \space 1 \space\space|\overline{\space 4x^{\tiny{\tt{2}}} + 5x \space\space {\text{--}} \space 3}\\ \end{array}

2)   [ 4x \times ( x \space {\text{--}} \space 1 )  =  4x^2 \space {\text{--}} \space 4x ]
\begin{array}{r} 4x \space \space \space \space \space \space \space\space\space\space \space\space \space\space\space\space\space\\ x \space {\text{--}} \space 1 \space\space|\overline{\space 4x^{\tiny{\tt{2}}} + 5x \space {\text{--}} \space 3}\\ {\text{--}} \space \underline{4x^{\tiny{2}} \space\space {\text{--}} \space 4x \space\space\space\space\space\space} \\ 9x \space {\text{--}} \space 3\\ \end{array}

3)     [ \frac{9x}{x}  =  9 ]

\begin{array}{r} 4x \space \space\space\space 9 \space \space\space\space \space\space \space\space\space\space\space\\ x \space {\text{--}} \space 1 \space\space|\overline{\space 4x^{\tiny{\tt{2}}} + 5x \space {\text{--}} \space 3}\\ {\text{--}} \space \underline{4x^{\tiny{2}} \space\space {\text{--}} \space 4x \space\space\space\space\space\space} \\ 9x \space {\text{--}} \space 3\\ \end{array}

4)     [ 9 \times ( x \space {\text{--}} \space 1 )  =  9x \space {\text{--}} \space 9 ]

\begin{array}{r} 4x \space \space\space\space 9 \space \space\space\space \space\space \space\space\space\space\space\\ x \space {\text{--}} \space 1 \space\space|\overline{\space 4x^{\tiny{\tt{2}}} + 5x \space {\text{--}} \space 3}\\ {\text{--}} \space \underline{4x^{\tiny{2}} \space\space {\text{--}} \space 4x \space\space\space\space\space\space} \\ 9x \space {\text{--}} \space 3\\ {\text{--}} \space\space \underline{9x \space {\text{--}} \space 9}\\ 6\\ \end{array}

( 4x^{\tt{2}} + 5x \space {\text{--}} \space 3 ) \space \div \space ( x \space {\text{--}} \space 1 ) \space = \space 4x + 9 + \frac{6}{x \space {\text{--}} \space 1}

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