Editing Polynomial long division

{{Short description|Algorithm for division of polynomials}}
{{For|a shorthand version of this method|synthetic division}}
In [[algebra]], '''polynomial long division''' is an [[algorithm]] for dividing a [[polynomial]] by another polynomial of the same or lower [[Degree of a polynomial|degree]], a generalized version of the arithmetic technique called [[long division]]. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones. Sometimes using a shorthand version called [[synthetic division]] is faster, with less writing and fewer calculations. Another abbreviated method is polynomial [[short division]] (Blomqvist's method).

Polynomial long division is an algorithm that implements the [[Euclidean division of polynomials]], which starting from two polynomials ''A'' (the ''dividend'') and ''B'' (the ''divisor'') produces, if ''B'' is not zero, a ''[[quotient]]'' ''Q'' and a ''remainder'' ''R'' such that
:''A'' = ''BQ'' + ''R'',
and either ''R'' = 0 or the degree of ''R'' is lower than the degree of ''B''. These conditions uniquely define  ''Q'' and ''R'', which means that ''Q'' and ''R'' do not depend on the method used to compute them.

The result ''R'' = 0 occurs [[if and only if]] the polynomial ''A'' has ''B'' as a [[polynomial factorization|factor]]. Thus long division is a means for testing whether one polynomial has another as a factor, and, if it does, for factoring it out. For example, if a [[root of a polynomial|root]] ''r'' of ''A'' is known, it can be factored out by dividing ''A'' by (''x''&nbsp;–&nbsp;''r'').

==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'').

==Pseudocode==

The algorithm can be represented in [[pseudocode]] as follows, where +, −, and × represent polynomial arithmetic, and lead(r) / lead(d) represents the polynomial obtained by dividing the two leading terms:

 '''function''' n / d '''is'''
     require d ≠ 0
     q ← 0
     r ← n             // At each step n = d × q + r
 
     '''while''' r ≠ 0 '''and''' degree(r) ≥ degree(d) '''do'''
         t ← lead(r) / lead(d)       // Divide the leading terms
         q ← q + t
         r ← r − t × d
 
     '''return''' (q, r)

This works equally well when degree(''n'') < degree(''d''); in that case the result is just the trivial (0, ''n'').

This algorithm describes exactly the above paper and pencil method: {{var|d}} is written on the left of the ")"; {{var|q}} is written, term after term, above the horizontal line, the last term being the value of {{var|t}}; the region under the horizontal line is used to compute and write down the successive values of {{var|r}}.

== Euclidean division ==
{{anchor|Division transformation}}
{{main|Euclidean division of polynomials}}
For every pair of polynomials (''A'', ''B'') such that ''B'' ≠ 0, polynomial division provides a ''quotient'' ''Q'' and a ''remainder'' ''R'' such that 
:<math>A=BQ+R,</math>
and either ''R''=0 or degree(''R'') < degree(''B''). Moreover (''Q'', ''R'') is the unique pair of polynomials having this property.

The process of getting the uniquely defined polynomials ''Q'' and ''R'' from ''A'' and ''B'' is called ''Euclidean division'' (sometimes ''division transformation''). Polynomial long division is thus an [[algorithm]] for Euclidean division.<ref>{{cite book|author=S. Barnard|title=Higher Algebra|year=2008|publisher=READ BOOKS|isbn=978-1-4437-3086-0|page=24}}</ref>

==Applications==

===Factoring polynomials===

Sometimes one or more roots of a polynomial are known, perhaps having been found using the [[rational root theorem]]. If one root ''r'' of a polynomial ''P''(''x'') of degree ''n'' is known then polynomial long division can be used to factor ''P''(''x'') into the form {{nowrap|(''x'' − ''r'')''Q''(''x'')}} where ''Q''(''x'') is a polynomial of degree ''n'' − 1.  ''Q''(''x'') is simply the quotient obtained from the division process; since ''r'' is known to be a root of ''P''(''x''), it is known that the remainder must be zero.

Likewise, if several roots ''r'', ''s'', . . .  of ''P''(''x'') are known, a linear factor {{nowrap|(''x'' − ''r'')}} can be divided out to obtain ''Q''(''x''), and then {{nowrap|(''x'' − ''s'')}} can be divided out of ''Q''(''x''), etc. Alternatively, the quadratic factor <math>(x-r)(x-s)=x^2-(r{+}s)x+rs</math> can be divided out of ''P''(''x'') to obtain a quotient of degree {{nowrap|''n'' − 2.}}

This method is especially useful for cubic polynomials, and sometimes all the roots of a higher-degree polynomial can be obtained. For example, if the rational root theorem produces a single (rational) root of a [[quintic function|quintic polynomial]], it can be factored out to obtain a quartic (fourth degree) quotient; the explicit formula for the roots of a [[quartic function|quartic polynomial]] can then be used to find the other four roots of the quintic. There is, however, no general way to solve a quintic by purely algebraic methods, see [[Abel–Ruffini theorem]].

===Finding tangents to polynomial functions===

Polynomial long division can be used to find the equation of the line that is [[tangent]] to the [[graph of a function|graph of the function]] defined by the polynomial ''P''(''x'') at a particular point {{nowrap|''x'' {{=}} ''r''.}}<ref>Strickland-Constable, Charles, "A simple method for finding tangents to polynomial graphs", ''[[Mathematical Gazette]]'' 89, November 2005: 466-467.</ref> If ''R''(''x'') is the remainder of the division of ''P''(''x'') by {{nowrap|(''x'' − ''r'')<sup>2</sup>,}} then the equation of the tangent line at {{nowrap|''x'' {{=}} ''r''}} to the graph of the function {{nowrap|''y'' {{=}} ''P''(''x'')}} is {{nowrap|''y'' {{=}} ''R''(''x''),}} regardless of whether or not ''r''  is a root of the polynomial.

====Example====
Find the equation of the line that is tangent to the following curve
<math>y = (x^3 - 12x^2 - 42)  </math>
:at: <math>x = 1 </math>
Begin by dividing the polynomial by:
<math> (x-1)^2=(x^2-2x+1)</math>

: <math>
\begin{array}{r}
 x - 10\\
 x^2-2x+1\ \overline{)\ x^3 - 12x^2 + 0x - 42}\\
 \underline{x^3 - {\color{White}0}2x^2 + {\color{White}1}x} {\color{White} {} - 42}\\
 -10x^2 - {\color{White}01}x - 42\\
 \underline{-10x^2 + 20x - 10}\\
 -21x - 32
\end{array}
</math>
The tangent line is 
<math> y=(-21x-32)</math>

===Cyclic redundancy check===

A [[cyclic redundancy check]] uses the remainder of polynomial division to detect errors in transmitted messages.

==See also==
*[[Polynomial remainder theorem]]
*[[Synthetic division]], a more concise method of performing Euclidean polynomial division
*[[Ruffini's rule]]
*[[Euclidean domain]]
*[[Gröbner basis]]
*[[Greatest common divisor of two polynomials]]

==References==
{{reflist}}

{{Polynomials}}

[[Category:Polynomials]]
[[Category:Computer algebra]]
[[Category:Division (mathematics)]]