☰

如何创建因子“菱形”和因子“盒子”以逐步解决代数问题

tag-icon • tag-icon formatting
如何创建因子“菱形”和因子“盒子”以逐步解决代数问题

请帮助我格式化代码：

  1. 将数值放置在因子菱形的 4 个象限中
  2. 中心号码之内因子框表和将数字置于中心位于顶行 2 个单元格上方
  3. 将数字放到第一列中每个单元格的左侧

这显示了我需要在解决方案中放置“钻石”和盒子的位置

注意：显示的数字对于因数来说并不正确。我只是粘贴了图像以供说明。

当然，数字和变量会因问题而异。正如你从我的 MWE 中看到的，解决方案集采用 2 列格式。

非常感谢！如果没有您的慷慨帮助，我会不知所措。

最大误差.....

\documentclass[12pt]{exam}

%\printanswers % un-comment to print solutions.
\printanswers
\renewcommand{\solutiontitle}{}

\usepackage{multirow, tabularx}
\newcolumntype{C}{>{\centering\arraybackslash}X}

\usepackage{array}
% p centers horizontally in table cell
% m centers vertically in table cell
\newcolumntype{P}[1]{>{\centering\arraybackslash}p{#1}}
\newcolumntype{M}[1]{>{\centering\arraybackslash}m{#1}}

\usepackage{amsmath}
\usepackage{cancel}
\usepackage{graphicx}
\usepackage{pgfplots}
\usepackage{amssymb}
\usepackage{mathrsfs}
\usepackage{framed}
\usepackage{tkz-euclide}
\usepackage{multicol}
\usepackage{tasks}
\pgfplotsset{compat=1.10}
\usepackage[a4paper,margin=0.5in,include head]{geometry}

\everymath{\displaystyle}

\setlength\parindent{1em}

\pagestyle{head}
\header{Algebra II Review Ch 3.2: Operations Rational Expressions and Equations: K E Y}
       {}
       {01/13-14/21} 

\newcommand{\pagetop}{%
\noindent 
  \fbox{\fbox{\parbox{\dimexpr\textwidth-4\fboxsep-4\fboxrule}{
    \textbf {Obj. 3.2.a: I can simplify factored rational expressions and find their restrictions.
    \bigskip
    \bigskipSimplify expression and state the excluded values (+1 pt numerator, +1 pt denominator, +1 pt restrictions.) each equation. Show all work/steps on this page.}
  }}}
  \bigskip
  \vspace{0.5mm}
}

\settasks{after-item-skip=1em,
          after-skip=2cm,
          label-width=2em,
          item-indent=3em,
          label=(\arabic*),
          column-sep=2em
          }
 
% ------------ DOCUMENT STARTS HERE---------------- 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

%definition for bigskip = 1 line to replace all \bigskip
\def\bigskip{\vskip\bigskipamount}

%macro \diamondscheme. The lines are drawn too, you need not tikz. The size is controlled by the \diaw parameter. It is set to 1.5cm in this example.

% definition
\newdimen\diaw \diaw=1.5cm
\def\diamondscheme#1#2#3#4{%
    \vbox to\diaw{
       \kern-1ex
       \hbox to\diaw{\hss$#1$\hss}
       \vss
       \hbox to\diaw{\hbox to0pt{\hss$#2$\hss}\hss
                     \raise.7ex\hbox{\diacross}\hss
                     \hbox to0pt{\hss$#3$\hss}}
       \vss
       \hbox to\diaw{\hss$#4$\hss}
       \kern-1ex
    }
}
\newcount\tmpnum 
\tmpnum=\diaw 
\divide\tmpnum by65781  % tmpnum is now equal to \diaw in bp units
\divide\tmpnum by2
\def\tmp{\the\tmpnum\space}
\edef\diacross{\pdfliteral
   {q 0.4 w -\tmp \tmp m \tmp -\tmp l S -\tmp -\tmp m \tmp \tmp l S Q}}

\begin{tasks}
[style=enumerate](2)
% Prob #1
\task $\dfrac{10k^2+32k+24}{1k5+18}$
\begin{solutionorbox}[5cm]
Factor 2 out of the numerator.\bigskip

$\dfrac{2(5k^2+16k+12)} {15k+18}$\bigskip

Now factor the numerator using diamond method. 
\bigskip
%call diamondscheme macro
\hspace{2cm}
\diamondscheme{60}{10}{6}{16}
\bigskip
The 5 in front of $5k^2$ means this is a non-monic quadratic trinomial. So you must proceed to the box method to complete the factorization.

\vspace{0.75cm}
%Example of fixed column width and text long (array package required)
% for the vertical padding / row height
\hspace{1cm}
{\renewcommand{\arraystretch}{1.5}
\begin{tabular}{ | m{5em} | m{2cm}| m{2cm} | } 

  \hline
  $x^2$ & $6x$  \\ 
  \hline
  $2x$ & $12$  \\ 
  \hline
 
\end{tabular}
}
\vspace{0.5cm}
\bigskip
simplified form: $7x$

Zeros: $x=0$

Holes: $none$

VA: $none$
\end{solutionorbox}
% Prob #2
\task $\dfrac{3x^2-13x-10}{x-5}$
\begin{solutionorbox}[5cm]
$\dfrac{3x^2-13x-10}{x-5}$\bigskip

Nothing can be factored out of the numerator and denominator.

Now factor the numerator using diamond method. 
\bigskip
%call diamondscheme macro
\hspace{2cm}
\diamondscheme{-30}{-15}{2}{-13}
\bigskip

The 3 in front of $3k^2$ means this is a non-monic quadratic trinomial. So you must proceed to the box method to complete the factorization.

\vspace{0.75cm}
%Example of fixed column width and text long (array package required)
% for the vertical padding / row height
\hspace{1cm}
{\renewcommand{\arraystretch}{1.5}
\begin{tabular}{ | m{5em} | m{2cm}| m{2cm} | } 

  \hline
  $x^2$ & $6x$  \\ 
  \hline
  $2x$ & $12$  \\ 
  \hline
 
\end{tabular}
}
\vspace{0.5cm}

simplified form: $?$

Zeros: $x=0$

Holes: $none$

VA: $none$
\end{solutionorbox}
\end{tasks}
\end{document}

答案1

你的问题 1 应该可以通过以下宏来解决\diamondscheme。线条也会被绘制，你不需要 tikz。尺寸由参数控制\diaw。在这个例子中，它设置为 1.5cm。

% definition
\newdimen\diaw \diaw=1.5cm
\def\diamondscheme#1#2#3#4{%
    \vbox to\diaw{
       \kern-1ex
       \hbox to\diaw{\hss$#1$\hss}
       \vss
       \hbox to\diaw{\hbox to0pt{\hss$#2$\hss}\hss
                     \raise.7ex\hbox{\diacross}\hss
                     \hbox to0pt{\hss$#3$\hss}}
       \vss
       \hbox to\diaw{\hss$#4$\hss}
       \kern-1ex
    }
}
\newcount\tmpnum 
\tmpnum=\diaw 
\divide\tmpnum by65781  % tmpnum is now equal to \diaw in bp units
\divide\tmpnum by2
\def\tmp{\the\tmpnum\space}
\edef\diacross{\pdfliteral
   {q 0.4 w -\tmp \tmp m \tmp -\tmp l S -\tmp -\tmp m \tmp \tmp l S Q}}

%test:
\diamondscheme{20}{-4}{-5}{-9}

根据\\文档中的内容。仔细阅读日志文件。Underfull \hbox这里的每个警告都是\\不正确使用的。如果您想在段落后添加一行空格，请\bigskip通过以下方式重新定义 LaTeX：

\def\bigskip{\vskip\bigskipamount}

并将所有内容替换\\为\bigskip。当然，不要在tabular环境中执行此替换，因为环境\\使用正确。注意：原始 LaTeX\bigskip是\vadjust{\vskip\bigskipamount}，它不会完成当前段落，只会带来问题。

请注意，您的日志中也有一个Overfull \hbox。这是由于第二个错误的 LaTeX 概念：它插入\leavemode到由 LaTeX 创建的每个框之前。因此，您\fbox的定义中的向右\pagetop移动了。在此之前插入\indent\noindent\fbox

相关内容