Generate Math Exams and Quizzes in LaTeX and pdf formats

Factoring out the GCF \(\displaystyle -21\) from each term gives \(\displaystyle -21(3 x^{3} + 2 x^{2} - 3 x - 2)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle x^{2}\), gives \(\displaystyle x^{2}(3 x + 2)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -1\), gives \(\displaystyle -1(3 x + 2)\). The polynomial now has a common binomial factor of \(\displaystyle 3 x + 2\). This gives \(\displaystyle -21[x^{2} \left(3 x + 2\right) -1 \cdot \left(3 x + 2\right)] = -21\left(3 x + 2\right) \left(x^{2} - 1\right)\). The quadratic factor can be factored using the difference of squares to give \(\displaystyle -21\left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right). \)

Download \(\LaTeX\)

\begin{question}Factor $- 63 x^{3} - 42 x^{2} + 63 x + 42$. 
    \soln{9cm}{Factoring out the GCF $-21$ from each term gives $-21(3 x^{3} + 2 x^{2} - 3 x - 2)$. Grouping the first two terms and factoring out their GCF, $x^{2}$, gives $x^{2}(3 x + 2)$. Grouping the last two terms and factoring out their GCF, $-1$, gives $-1(3 x + 2)$. The polynomial now has a common binomial factor of $3 x + 2$. This gives $-21[x^{2} \left(3 x + 2\right) -1 \cdot \left(3 x + 2\right)] = -21\left(3 x + 2\right) \left(x^{2} - 1\right)$. The quadratic factor can be factored using the difference of squares to give $-21\left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right). $}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Factor  <img class="equation_image" title=" \displaystyle - 63 x^{3} - 42 x^{2} + 63 x + 42 " src="/equation_images/%20%5Cdisplaystyle%20-%2063%20x%5E%7B3%7D%20-%2042%20x%5E%7B2%7D%20%2B%2063%20x%20%2B%2042%20" alt="LaTeX:  \displaystyle - 63 x^{3} - 42 x^{2} + 63 x + 42 " data-equation-content=" \displaystyle - 63 x^{3} - 42 x^{2} + 63 x + 42 " /> . </p> </p>

HTML for Canvas

<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle -21 " src="/equation_images/%20%5Cdisplaystyle%20-21%20" alt="LaTeX:  \displaystyle -21 " data-equation-content=" \displaystyle -21 " />  from each term gives  <img class="equation_image" title=" \displaystyle -21(3 x^{3} + 2 x^{2} - 3 x - 2) " src="/equation_images/%20%5Cdisplaystyle%20-21%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%203%20x%20-%202%29%20" alt="LaTeX:  \displaystyle -21(3 x^{3} + 2 x^{2} - 3 x - 2) " data-equation-content=" \displaystyle -21(3 x^{3} + 2 x^{2} - 3 x - 2) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle x^{2} " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle x^{2} " data-equation-content=" \displaystyle x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle x^{2}(3 x + 2) " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%283%20x%20%2B%202%29%20" alt="LaTeX:  \displaystyle x^{2}(3 x + 2) " data-equation-content=" \displaystyle x^{2}(3 x + 2) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle -1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20" alt="LaTeX:  \displaystyle -1 " data-equation-content=" \displaystyle -1 " /> , gives  <img class="equation_image" title=" \displaystyle -1(3 x + 2) " src="/equation_images/%20%5Cdisplaystyle%20-1%283%20x%20%2B%202%29%20" alt="LaTeX:  \displaystyle -1(3 x + 2) " data-equation-content=" \displaystyle -1(3 x + 2) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 3 x + 2 " src="/equation_images/%20%5Cdisplaystyle%203%20x%20%2B%202%20" alt="LaTeX:  \displaystyle 3 x + 2 " data-equation-content=" \displaystyle 3 x + 2 " /> . This gives  <img class="equation_image" title=" \displaystyle -21[x^{2} \left(3 x + 2\right) -1 \cdot \left(3 x + 2\right)] = -21\left(3 x + 2\right) \left(x^{2} - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20-21%5Bx%5E%7B2%7D%20%5Cleft%283%20x%20%2B%202%5Cright%29%20-1%20%5Ccdot%20%5Cleft%283%20x%20%2B%202%5Cright%29%5D%20%3D%20-21%5Cleft%283%20x%20%2B%202%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%201%5Cright%29%20" alt="LaTeX:  \displaystyle -21[x^{2} \left(3 x + 2\right) -1 \cdot \left(3 x + 2\right)] = -21\left(3 x + 2\right) \left(x^{2} - 1\right) " data-equation-content=" \displaystyle -21[x^{2} \left(3 x + 2\right) -1 \cdot \left(3 x + 2\right)] = -21\left(3 x + 2\right) \left(x^{2} - 1\right) " /> . The quadratic factor can be factored using the difference of squares to give  <img class="equation_image" title=" \displaystyle -21\left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right).  " src="/equation_images/%20%5Cdisplaystyle%20-21%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%283%20x%20%2B%202%5Cright%29.%20%20" alt="LaTeX:  \displaystyle -21\left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right).  " data-equation-content=" \displaystyle -21\left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right).  " /> </p> </p>