Editing Polynomial long division (section)

==Example==

=== Polynomial long division ===
Find the quotient and the remainder of the division of  <math>(x^3 - 2x^2 - 4)</math>, the ''dividend'', by <math>(x-3) </math>, the ''divisor''.

The dividend is first rewritten like this:

:<math>x^3 - 2x^2 + 0x - 4.</math>

The quotient and remainder can then be determined as follows:

<ol>
<li>
Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of ''x'', which in this case is ''x''). Place the result above the bar (''x''<sup>3</sup> ÷ ''x'' = ''x''<sup>2</sup>).
:<math>
\begin{array}{l}
{\color{White} x-3\ )\ x^3 - 2}x^2\\
x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}
\end{array}
</math>
</li>
<li>
Multiply the divisor by the result just obtained (the first term of the eventual quotient). Write the result under the first two terms of the dividend ({{math|1=''x''<sup>2</sup> · (''x'' − 3) = ''x''<sup>3</sup> − 3''x''<sup>2</sup>}}).
:<math>
\begin{array}{l}
{\color{White} x-3\ )\ x^3 - 2}x^2\\
x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\
{\color{White} x-3\ )\ } x^3 - 3x^2
\end{array}
</math>
</li>
<li>
Subtract the product just obtained from the appropriate terms of the original dividend (being careful that subtracting something having a minus sign is equivalent to adding something having a plus sign), and write the result underneath {{math|({{math|1=''x''<sup>3</sup> − 2''x''<sup>2</sup>) − (''x''<sup>3</sup> − 3''x''<sup>2</sup>) = −2''x''<sup>2</sup> + 3''x''<sup>2</sup> = ''x''<sup>2</sup>}}}}
Then, "bring down" the next term from the dividend.

:<math>
\begin{array}{l}
{\color{White} x-3\ )\ x^3 - 2}x^2\\
x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\
{\color{White} x-3\ )\ } \underline{x^3 - 3x^2}\\
{\color{White} x-3\ )\ 0x^3} + {\color{White}}x^2 + 0x
\end{array}
</math>
</li>
<li>
Repeat the previous three steps, except this time use the two terms that have just been written as the dividend.
:<math>
\begin{array}{r}
 x^2 + {\color{White}1}x {\color{White} {} + 3}\\
 x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\
 \underline{x^3 - 3x^2 {\color{White} {} + 0x - 4}}\\
 +x^2 + 0x {\color{White} {} - 4}\\
 \underline{+x^2 - 3x {\color{White} {} - 4}}\\
 +3x - 4\\
\end{array}
</math>
</li>
<li>
Repeat step 4. This time, there is nothing to "bring down".
:<math>
\begin{array}{r}
 x^2 + {\color{White}1}x + 3\\
 x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\
 \underline{x^3 - 3x^2 {\color{White} {} + 0x - 4}}\\
 +x^2 + 0x {\color{White} {} - 4}\\
 \underline{+x^2 - 3x {\color{White} {} - 4}}\\
 +3x - 4\\
 \underline{+3x - 9}\\
 +5
\end{array}
</math>
</li>
</ol>

The polynomial above the bar is the quotient ''q''(''x''), and the number left over (5) is the remainder ''r''(''x'').

:<math>{x^3 - 2x^2 - 4} = (x-3)\,\underbrace{(x^2 + x + 3)}_{q(x)}  +\underbrace{5}_{r(x)}</math>

The [[long division]] algorithm for arithmetic is very similar to the above algorithm, in which the variable ''x'' is replaced (in base 10) by the specific number 10.

=== Polynomial short division===
Blomqvist's method<ref>Archived at [https://ghostarchive.org/varchive/youtube/20211211/Ad16hxs809I Ghostarchive]{{cbignore}} and the [https://web.archive.org/web/20200401062354/https://www.youtube.com/watch?v=Ad16hxs809I&gl=US&hl=en Wayback Machine]{{cbignore}}: {{Citation|title=Blomqvist's division: the simplest method for solving divisions?| date=7 December 2019 |url=https://www.youtube.com/watch?v=Ad16hxs809I|language=en|access-date=2019-12-10}}{{cbignore}}</ref> is an abbreviated version of the long division above. This pen-and-paper method uses the same algorithm as polynomial long division, but [[mental calculation]] is used to determine remainders. This requires less writing, and can therefore be a faster method once mastered.

The division is at first written in a similar way as [[Multiplication algorithm|long multiplication]] with the dividend at the top, and the divisor below it. The quotient is to be written below the bar from left to right.

:<math>\begin{matrix} \qquad \qquad x^3-2x^2+{0x}-4 \\ \underline{ \div \quad \qquad \qquad \qquad \qquad  x-3 }\end{matrix}</math>

Divide the first term of the dividend by the highest term of the divisor (''x''<sup>3</sup> ÷ ''x'' = ''x''<sup>2</sup>). Place the result below the bar. ''x''<sup>3</sup> has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result ''x''<sup>2</sup> is then multiplied by the second term in the divisor −3 = −3''x''<sup>2</sup>. Determine the partial remainder by subtracting −2''x''<sup>2</sup> − (−3''x''<sup>2</sup>) = ''x''<sup>2</sup>. Mark −2''x''<sup>2</sup> as used and place the new remainder ''x''<sup>2</sup> above it.

:<math>\begin{matrix} \qquad x^2 \\ \qquad \quad \bcancel{x^3}+\bcancel{-2x^2}+{0x}-4 \\ \underline{ \div \qquad \qquad \qquad \qquad \qquad  x-3 }\\x^2 \qquad \qquad \end{matrix}
</math>

Divide the highest term of the remainder by the highest term of the divisor (''x''<sup>2</sup> ÷ ''x'' = ''x''). Place the result (+x) below the bar. ''x''<sup>2</sup> has been divided leaving no remainder, and can therefore be marked as used. The result ''x'' is then multiplied by the second term in the divisor −3 = −3''x''. Determine the partial remainder by subtracting 0''x'' − (−3''x'') = 3''x''. Mark 0''x'' as used and place the new remainder 3''x'' above it.

:<math>\begin{matrix} \qquad \qquad \quad\bcancel{x^2} \quad3x\\ \qquad \quad \bcancel{x^3}+\bcancel{-2x^2}+\bcancel{0x}-4 \\ \underline{ \div \qquad \qquad \qquad \qquad \qquad  x-3 }\\x^2 +x \qquad \end{matrix}
</math>

Divide the highest term of the remainder by the highest term of the divisor (3x ÷ ''x'' = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (−9) = 5. Mark −4 as used and place the new remainder 5 above it.

:<math>\begin{matrix} \quad \qquad \qquad \qquad\bcancel{x^2} \quad \bcancel{3x} \quad5\\
\qquad \quad \bcancel{x^3}+\bcancel{-2x^2}+\bcancel{0x}\bcancel{-4} \\
\underline{ \div \qquad \qquad \qquad \qquad \qquad  x-3 }\\
x^2 +x +3\qquad \end{matrix}
</math>

The polynomial below the bar is the quotient ''q''(''x''), and the number left over (5) is the remainder ''r''(''x'').