Honestly, this is definitely a tough subject. It can be hard to understand, so we will start slow and work up to it.

Dividing a polynomial by a monomial (one term) is a good place to start because it’s not that bad. Let’s say we have:

\(\Large \frac{{ - 12{x^3} + 6{x^2} + 9x}}{{3x}}\)

To be able to divide in this example, we need to divide each piece by 3x and reduce.

\(\Large \frac{{ - 12{x^3}}}{{3x}} + \frac{{6{x^2}}}{{3x}} + \frac{{9x}}{{3x}}\)

Answer: \( - 4{x^2} + 2x + 3\)

It becomes much tougher if we have to divide by a binomial (two terms). If this is the case, we have to do long division. Now, you’ll have to reach back in your memory to try to remember how to do long division. Let’s work though a basic example.

\(245 \div 7\)

Change the form to:

\(7\left){\vphantom{1{245}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{245}}}\)

Seven does not fit into 2, so we have to see how many times it will fit into 24 without going over. Seven times three works!

\(7\mathop{\left){\vphantom{1{245}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{245}}}}
\limits^{\displaystyle\,\,\, {3\_}}\)

Is this ringing a bell at all, yet? Let’s finish.

\(7\mathop{\left){\vphantom{1{245}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{245}}}}
\limits^{\displaystyle\,\,\, {35}}\)

\begin{matrix}
-21 &  \\
\hline
  & 35
  \end{matrix}

\begin{matrix}
-35 &  \\
\hline
  & 0
  \end{matrix}

Answer: 35

Let’s try to apply this to a polynomial.

\((2{x^3} - 8{x^2} + 25) \div (2x - 6)\)

Change to:

\(2x - 6\left){\vphantom{1{2{x^3} - 8{x^2} + 25}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - 8{x^2} + 25}}}\)

You only have to worry about the 2x to start with. What can we multiply 2x by to match 2x³? The answer is x².

Now we want our 2x to match -2x². We can multiply by –x.

6x and 25 cannot be combined since they are not “like terms,” so we leave it as 6x + 25. Almost done! To get 2x to match -6x we multiply by -3.

There is no way to get our 2x to match 7 by multiplying so we leave the last piece of the answer in fraction form or \(\Large \frac{7}{{2x - 6}}\).

So our final answer is: \({x^2} - x - 3 + \Large \frac{7}{{2x - 6}}\).

Phew! That is a tough problem! Let’s try one more.

\((2{k^3} - 15{k^2} + 23k + 11) \div (2k - 5)\)

Change to:

\(2k - 5\left){\vphantom{1{2{k^3} - 15{k^2} + 23k + 11}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{k^3} - 15{k^2} + 23k + 11}}}\)

\(2k - 5\mathop{\left){\vphantom{1{2{k^3} - 15{k^2} + 23k + 11}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{k^3} - 15{k^2} + 23k + 11}}}}
\limits^{\displaystyle \,\,\, {{k^2}\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}\)

\begin{matrix}
- {\text{ 2}}{k^{\text{3}}}-{\text{ 5}}{k^{\text{2}}} & \\
\hline
& - {\text{1}}0{k^{\text{2}}} + {\text{23}}k
\end{matrix}

\(2k - 5\mathop{\left){\vphantom{1{2{k^3} - 15{k^2} + 23k + 11}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{k^3} - 15{k^2} + 23k + 11}}}}
\limits^{\displaystyle \,\,\, {{k^2} - 5k\_\_\_\_\_\_\_\_\_}}\)

\begin{matrix}
- {\text{ 2}}{k^{\text{3}}}--{\text{ 5}}{k^{\text{2}}} & \\
\hline
& - {\text{1}}0{k^{\text{2}}} + {\text{23}}k \\
& - {\text{ }}\left( { - {\text{1}}0{k^2}} \right){\text{ }} + {\text{25}}k \\
\hline
& - {\text{2}}k + {\text{11}}
\end{matrix}

\(2k - 5\mathop{\left){\vphantom{1{2{k^3} - 15{k^2} + 23k + 11}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{k^3} - 15{k^2} + 23k + 11}}}}
\limits^{\displaystyle \,\,\, {{k^2} - 5k - 1\_\_\_\_\_\_\_}}\)

\begin{matrix}
- {\text{ 2}}{k^{\text{3}}}--{\text{ 5}}{k^{\text{2}}} & \\
\hline
& - {\text{1}}0{k^{\text{2}}} + {\text{23}}k \\
& - {\text{ }}\left( { - {\text{1}}0{k^2}} \right){\text{ }} + {\text{25}}k \\
\hline
& - {\text{2}}k + {\text{11}} \\
& - {\text{ }}\left( { - {\text{2}}k} \right) + {\text{ 5}} \\
\hline
& 6
\end{matrix}

Final answer: \({k^2} - 5k - 1 + \Large \frac{6}{{2k - 5}}\)

Below you can download some free math worksheets and practice.