Automate Box Method / Area Model for Polynomial Division in LaTeX

Question

I am writing worksheets with polynomial division using area models (often called "box method"):

The process is very slow for making dozens of problems, especially when the problems are bigger, like dividing a 5th degree polynomial by a 3rd degree polynomial. Is there a way to automate the creation of these box divisions in the future?

I would like to be able to type something like \polyboxdiv{6x^3+25x^2-24x+11}{x+5} (similar to \polylongdiv{6x^3+25x^2-24x+11}{x+5}from the polynom package) to automatically generate the box division representation.

Here is a link to the MWE:

\documentclass[addpoints]{exam}
\usepackage[utf8]{inputenc}
\usepackage[margin=.75 in]{geometry}
\usepackage{amsmath,amsfonts,amssymb,amsthm,color,srcltx,enumitem,bm,cancel,thmtools,physics}
\usepackage{multicol} %see http://stackoverflow.com/questions/1398127/breaking-a-list-into-multiple-columns-in-latex
\usepackage{multirow,array} %Used for the "hand-made" payoff matrix
\usepackage{hhline}
\usepackage{float}
\usepackage{polynom} %for polynomial long division and synthetic division, see http://texdoc.net/texmf-dist/doc/latex/polynom/polynom.pdf
\usepackage[table]{xcolor}

%For solution tables, see https://tex.stackexchange.com/questions/133397/printanswer-in-table-in-exam-class
\makeatletter
\newcommand{\st}[1]{\ifprintanswers\begingroup\Solution@Emphasis#1\if@shadedsolutions%
{\cellcolor{white}}
\else
\fi\endgroup\else\phantom{#1}\fi}
\makeatother
\SolutionEmphasis{\color{black}} %color font solutions
%\unframedsolutions  %Not necessary for table environment
%\shadedsolutions    %gives background color to solutions
\printanswers

\begin{document}
\section{Polynomials Division Test}
\begin{questions}

\question[40]
For the division problem
\begin{equation*}
(6x^3+25x^2-24x+11) \divisionsymbol (x+5)
\end{equation*}
\begin{parts}

\begin{multicols}{2}
\part Divide using an area model with a left divisor.
{\renewcommand{\arraystretch}{2}
\begin{table}[H]
\centering
\begin{tabular}{*{1}{>{\centering\arraybackslash}m{1.25cm}}|*{4}{>{\centering\arraybackslash}m{1.25cm}|}}
\multicolumn{1}{c}{} & \multicolumn{1}{c}{$\st{6x^2}$}  & \multicolumn{1}{c}{$\st{-5x}$} & \multicolumn{1}{c}{$\st{+1}$} & \multicolumn{1}{c}{} \\ \hhline{~----}
\multirow{2}*{} $\st{x}$   & $\st{6x^3}$    & $\st{-5x^2}$   & $\st{x}$ & $\st{6}$ \\ \hhline{~----}
                $\st{+5}$  & $\st{30x^2}$   & $\st{-25x}$    & $\st{5}$ & \cellcolor{black!10} \\ \hhline{~----}
\end{tabular}
\end{table}}
\columnbreak
\part Divide using an area model with an upper divisor.
{\renewcommand{\arraystretch}{2}
\begin{table}[H]
\centering
\begin{tabular}{*{1}{>{\centering\arraybackslash}m{1.25cm}}|*{2}{>{\centering\arraybackslash}m{1.25cm}|}}
\multicolumn{1}{c}{} & \multicolumn{1}{c}{$\st{x}$}  & \multicolumn{1}{c}{$\st{+5}$} \\ \hhline{~--}
\multirow{2}*{} $\st{6x^2}$ & $\st{6x^3}$   & $\st{30x^2}$          \\ \hhline{~--}
                $\st{-5x}$  & $\st{-5x^2}$  & $\st{-25x}$           \\ \hhline{~--}
                $\st{+1}$   & $\st{x}$      & $\st{5}$              \\ \hhline{~--}
                            & $\st{6}$      & \cellcolor{black!10}  \\ \hhline{~--}
\end{tabular}
\end{table}}
\end{multicols}
\vspace*{\stretch{1}}

\begin{multicols}{2}
\part Divide using the classic long-division algorithm.
\begin{solution}
\begin{center}
\polylongdiv{6x^3+25x^2-24x+11}{x+5}
\end{center}
\end{solution}
\columnbreak
\part Divide using Horner's method of synthetic division.
\begin{solution}
\begin{center}
\polyhornerscheme[x=-5]{6x^3+25x^2-24x+11}
\end{center}
\end{solution}
\end{multicols}
\vspace*{\stretch{1}}

\end{parts}
\end{questions}
\end{document}

Thank you!

Answer 1

I can help with a partial answer: how to automate the problem. The formatting, however, is quite tedious. Your top row is the quotient and the box in the rightmost column is your remainder. The computer algebra system SAGE, which is Python based, can easily tell us the quotient and remainder with q,r = numerator.quo_rem(denominator). The function FormatTerm(a,deg) will help format the individual terms and the function PolyBoxDivL(f,g) will format the table. Not as nicely as you wanted, though.

\documentclass[addpoints]{exam}
\usepackage[utf8]{inputenc}
\usepackage[margin=.75 in]{geometry}
\usepackage{amsmath,amsfonts,amssymb,amsthm,color,srcltx,enumitem,bm,cancel,thmtools,physics}
\usepackage{multicol}
\usepackage{multirow,array}
\usepackage[table]{xcolor}
\usepackage{sagetex}          %gives us access to SAGE
%%%%%%%%%%%%%%%
\begin{document}
\section{Polynomials Division Test}
\begin{sagesilent}
R.<x> = PolynomialRing(ZZ)    #### Ring of polynomials with integer coefficients
def FormatTerm(a,deg):
    if deg == 0:
        return "$%s$"%(a)

    if deg == 1:
        if a == 1:
            return "$x$"
        elif a == -1:
            return "$-x$"
        else:
            return "$%s x$"%(a)

    if deg >1:
        if a == 1:
            return "$x^{%s}$"%(deg)
        else:
            return "$%s x^{%s}$"%(a,deg)

def PolyBoxDivL(f,g):
    numerator = f
    denominator = g
    q,r = numerator.quo_rem(denominator)
    degreeq = q.degree()
    length = q.number_of_terms()+2   ##### +1 to include the remainder and vertical poly
    width = g.number_of_terms()
########### Table header    
    output = r""
    output += r"\begin{tabular}{c|"+"c|"*(length)+"}"
    output += r"\cline{2-%s}"%(length)
    output += r" & "
    for i in range(degreeq,-1,-1):
        if q[i] != 0:
            output += r"%s & "%(FormatTerm(q[i],i))
    output += r"\rule{0pt}{13pt} \\[6pt]\cline{2-%s}"%(length)
#### rows which aren't the header 

    for i in range(g.degree(),-1,-1):
        if denominator[i] != 0:
            output += r"\rule{0pt}{18pt} %s &"%(FormatTerm(denominator[i],i))
            for j in range(q.degree(),-1,-1):
                if q[j] != 0:
                    output += r"%s & "%(FormatTerm(denominator[i]*q[j],i+j))
            if i == g.degree():
                output += r"$%s$ \\[8pt]\cline{2-%s}"%(latex(r),length)
            else:
                output += r"\cellcolor{black!10} \\[8pt]\cline{2-%s}"%(length)
    output += r"\end{tabular}"

    return output
\end{sagesilent}

Use the area model with a left divisor to find $(6x^3+25x^2-24x+11)   \divisionsymbol (x+5)$:\\\\
\begin{sagesilent}
Q1 = PolyBoxDivL(6*x^3+25*x^2-24*x+11,x+5)
\end{sagesilent}
\sagestr{Q1}

\vspace{.5cm}
Use the area model with a left divisor to find $(5x^5-3x^2+x-1) \divisionsymbol (x^2+x+1)$:\\\\
\begin{sagesilent}
Q2 = PolyBoxDivL(5*x^5-3*x^2+x-1,x^2+x+1)
\end{sagesilent}
\sagestr{Q2}

\vspace{.5cm}
Use the area model with a left divisor to find $(x^4-1) \divisionsymbol (x^2-1)$:\\\\
\begin{sagesilent}
Q3 = PolyBoxDivL(x^4-1,x^2-1)
\end{sagesilent}
\sagestr{Q3}

\vspace{.5cm}
Use the area model with a left divisor to find $(x^9-7x^4) \divisionsymbol (x^3-x+4)$:\\\\
\begin{sagesilent}
Q4 = PolyBoxDivL(x^9-7*x^4,x^3-x+4)
\end{sagesilent}
\sagestr{Q4}
\end{document}

SAGE is not part of LaTeX. It needs to be downloaded to your machine or, even easier, accessed through a free Cocalc account. The code running in Cocalc looks like this:

A very important line to notice is output += r"\begin{tabular}{c|"+"c|"*(length)+"}". This allows the table to vary the number of columns because SAGE can calculate the number of nonzero terms needed to calculate the number of columns. The area model with an upper(?) divisor is handled in similar fashion. The problem is calculated by \begin{sagesilent} Q4 = PolyBoxDivL(x^9-7*x^4,x^3-x+4) \end{sagesilent} and then placing the resulting string into the document with \sagestr{Q4}.

EDIT: In response to comments below, I've added 0 coefficient terms so that the exponents are the same on diagonals. I don't know how to get the table formatting to handle vertical lines for some rows. I've chosen to color the cells to resolve the problem

\documentclass[addpoints]{exam}
\usepackage[utf8]{inputenc}
\usepackage[margin=.75 in]{geometry}
     \usepackage{amsmath,amsfonts,amssymb,amsthm,color,srcltx,enumitem,bm,cancel,thmtools,physics}
\usepackage{multicol}
\usepackage{multirow,array}
\usepackage[table]{xcolor}
\usepackage{sagetex}          %gives us access to SAGE
%%%%%%%%%%%%%%%
\begin{document}
\section{Polynomials Division Test}
\begin{sagesilent}
R.<x> = PolynomialRing(ZZ)    #### Ring of polynomials with integer   coefficients
def FormatTerm(a,deg):
    if deg == 0:
        return "$%s$"%(a)

    if deg == 1:
        if a == 1:
            return "$x$"
        elif a == -1:
            return "$-x$"
        else:
            return "$%s x$"%(a)

    if deg >1:
        if a == 1:
            return "$x^{%s}$"%(deg)
        else:
            return "$%s x^{%s}$"%(a,deg)

def PolyBoxDivL(f,g):
    numerator = f
    denominator = g
    q,r = numerator.quo_rem(denominator)
    length = q.degree()+3
    width = g.degree()+1
########### Table header    
    output = r""
    output += r"\begin{tabular}{"+"c"*(length)+"}"
    output += r" & "
    for i in range(q.degree(),-1,-1):
        output += r" \cellcolor{black!10} %s & "%(FormatTerm(q[i],i))
    output += r"\rule{0pt}{13pt} \\[6pt]"
#### rows which aren't the header

    for i in range(g.degree(),-1,-1):
        output += r"\rule{0pt}{18pt} \cellcolor{black!10} %s &"%(FormatTerm(denominator[i],i))
        for j in range(q.degree(),-1,-1):
            output += r"\cellcolor{orange!10} %s & "%(FormatTerm(denominator[i]*q[j],i+j))
        if i == g.degree():
            output += r"\cellcolor{blue!10} $%s$ \\[8pt]"%(latex(r))
        else:
            output += r"\cellcolor{blue!10} \\[8pt]"
    output += r"\end{tabular}"

    return output
\end{sagesilent}

Use the area model with a left divisor to find $(6x^3+25x^2-24x+11)     \divisionsymbol (x+5)$:\\\\
\begin{sagesilent}
Q1 = PolyBoxDivL(6*x^3+25*x^2-24*x+11,x+5)
\end{sagesilent}
\sagestr{Q1}

\vspace{.5cm}
Use the area model with a left divisor to find $(5x^5-3x^2+x-1) \divisionsymbol (x^2+x+1)$:\\\\
\begin{sagesilent}
Q2 = PolyBoxDivL(5*x^5-3*x^2+x-1,x^2+x+1)
\end{sagesilent}
\sagestr{Q2}

\vspace{.5cm}
Use the area model with a left divisor to find $(x^4-1) \divisionsymbol (x^2-1)$:\\\\
\begin{sagesilent}
Q3 = PolyBoxDivL(x^4-1,x^2-1)
\end{sagesilent}
\sagestr{Q3}

\vspace{.5cm}
Use the area model with a left divisor to find $(x^9-7x^4) \divisionsymbol (x^3-x+4)$:\\\\
\begin{sagesilent}
Q4 = PolyBoxDivL(x^9-7*x^4,x^3-x+4)
\end{sagesilent}
\sagestr{Q4}
\end{document}

The output is

Answer 2

What I am describing is not an answer but something to point you in the right direction

Since it will not fit in the comments column I am posting it here

Hope the mods will not mind!

What you are asking is to

• Parse the value of the polynomial into a tabular environment

• The table design varies to four different designs

• The process is automated from the time you feed in the polynomial in the numerator and denominator portion

Please see the answers on this website at

Automatically creating a table from datatool using references in the text

Formatting complex table from CSV using datatool

\begin{filecontents*}{\jobname.dat}
  Hammer001,   Hammer,    1 ,  0 , 1 , 10 , 1 , light (add some words here to wrap around)
  Hammer002,   Hammer,    2 ,  0 , 1 , 10 , 1 , heavy
  Hammer003,   Hammer,    3 ,  0 , 1 , 10 , 1 , really heavy
  Longsword001,Longsword, 1 , -1 , 2 , 75 , 2 , one-handed
  Longsword002,Longsword, 2 , -1 , 2 , 75 , 2 , two-handed
  Longsword003,Longsword, 3 , -1 , 2 , 75 , 2 , three-handed
\end{filecontents*}

\documentclass{article}
\usepackage{array}
\usepackage{datatool}
\usepackage{longtable}
\usepackage{etoolbox}

\newcommand{\colhead}[1]{\multicolumn{1}{>{\bfseries}l}{#1}}
\newcounter{tabenum}\setcounter{tabenum}{0}
\newcommand{\nextnuml}[1]{%
  \refstepcounter{tabenum}\thetabenum.\label{#1}%
}

\newcommand{\PrintDTLTable}[2][]{%
 % #1 = list of rowIDs
 % #2 = database to search
  \begin{longtable}{l l l l l l l m{2in}}
  & \colhead{Label} & \colhead{Cost} & \colhead{Weight} &
    \colhead{PropA} & \colhead{PropB} & \colhead{PropC} & \colhead{Description}\\
  \hline
  \DTLforeach
    [%
     \(\equal{#1}{}\AND\DTLisSubString{\ReferencedIDs}{\RowID}\)
     \OR
     \(\DTLisSubString{#1}{\RowID}\AND\DTLisSubString{\ReferencedIDs}{\RowID}\)%
    ]
    {#2}{%
      \RowID=RowID,%
      \Label=Label,%
      \Cost=Cost,%
      \Weight=Weight,%
      \PropA=PropA,%
      \PropB=PropB,%
      \PropC=PropC,%
      \Description=Description%
     }{%
       \nextnuml{\RowID} & \Label &\Cost & \Weight & \PropA & \PropB & \PropC & \Description \\
    }%
    \end{longtable}
}

\makeatletter
\let\oldref\ref
\def\ref#1{%
  \immediate\write\@auxout{%
    \string\gappto\string\ReferencedIDs{#1,}%
  }%
  \oldref{#1}%
}
\def\ReferencedIDs{}
\makeatother


\begin{document}

% \DTLsetseparator{&}% Define separator of the data
\DTLloaddb[noheader,keys={RowID,Label,Cost,Weight,PropA,PropB,PropC,Description}]{myDB}{\jobname.dat}

% \DTLdisplaydb{myDB}% Useful for debugging.

\PrintDTLTable[Hammer001,Hammer003,Longsword003]{myDB}

\PrintDTLTable[Hammer002,Longsword002]{myDB}

This is a reference to ~\ref{Hammer003}.

This is a reference to ~\ref{Longsword002}.

\end{document}

Output in tabular form is as below

The first part of the code

\begin{filecontents*}{\jobname.dat} Hammer001, Hammer, 1 , 0 , 1 , 10 , 1 , light (add some words here to wrap around) Hammer002, Hammer, 2 , 0 , 1 , 10 , 1 , heavy Hammer003, Hammer, 3 , 0 , 1 , 10 , 1 , really heavy Longsword001,Longsword, 1 , -1 , 2 , 75 , 2 , one-handed Longsword002,Longsword, 2 , -1 , 2 , 75 , 2 , two-handed Longsword003,Longsword, 3 , -1 , 2 , 75 , 2 , three-handed \end{filecontents*}

is similar to advertising the polynomial which is then parsed into a table form

There are similar queries on the webpage which you may like to explore

I am sure one of the experts on this site will shortly get back to you