\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Algebra
Quadratics
New Random

Factor \(\displaystyle 10 x^{3} + 16 x^{2} - 20 x - 32\).

Factoring out the GCF \(\displaystyle 2\) from each term gives \(\displaystyle 2(5 x^{3} + 8 x^{2} - 10 x - 16)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle x^{2}\), gives \(\displaystyle x^{2}(5 x + 8)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -2\), gives \(\displaystyle -2(5 x + 8)\). The polynomial now has a common binomial factor of \(\displaystyle 5 x + 8\). This gives \(\displaystyle 2[x^{2} \left(5 x + 8\right) -2 \cdot \left(5 x + 8\right)] = 2\left(5 x + 8\right) \left(x^{2} - 2\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $10 x^{3} + 16 x^{2} - 20 x - 32$. 
    \soln{9cm}{Factoring out the GCF $2$ from each term gives $2(5 x^{3} + 8 x^{2} - 10 x - 16)$. Grouping the first two terms and factoring out their GCF, $x^{2}$, gives $x^{2}(5 x + 8)$. Grouping the last two terms and factoring out their GCF, $-2$, gives $-2(5 x + 8)$. The polynomial now has a common binomial factor of $5 x + 8$. This gives $2[x^{2} \left(5 x + 8\right) -2 \cdot \left(5 x + 8\right)] = 2\left(5 x + 8\right) \left(x^{2} - 2\right)$. }

\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 10 x^{3} + 16 x^{2} - 20 x - 32 " src="/equation_images/%20%5Cdisplaystyle%2010%20x%5E%7B3%7D%20%2B%2016%20x%5E%7B2%7D%20-%2020%20x%20-%2032%20" alt="LaTeX:  \displaystyle 10 x^{3} + 16 x^{2} - 20 x - 32 " data-equation-content=" \displaystyle 10 x^{3} + 16 x^{2} - 20 x - 32 " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle 2 " src="/equation_images/%20%5Cdisplaystyle%202%20" alt="LaTeX:  \displaystyle 2 " data-equation-content=" \displaystyle 2 " />  from each term gives  <img class="equation_image" title=" \displaystyle 2(5 x^{3} + 8 x^{2} - 10 x - 16) " src="/equation_images/%20%5Cdisplaystyle%202%285%20x%5E%7B3%7D%20%2B%208%20x%5E%7B2%7D%20-%2010%20x%20-%2016%29%20" alt="LaTeX:  \displaystyle 2(5 x^{3} + 8 x^{2} - 10 x - 16) " data-equation-content=" \displaystyle 2(5 x^{3} + 8 x^{2} - 10 x - 16) " /> . 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}(5 x + 8) " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%285%20x%20%2B%208%29%20" alt="LaTeX:  \displaystyle x^{2}(5 x + 8) " data-equation-content=" \displaystyle x^{2}(5 x + 8) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle -2 " src="/equation_images/%20%5Cdisplaystyle%20-2%20" alt="LaTeX:  \displaystyle -2 " data-equation-content=" \displaystyle -2 " /> , gives  <img class="equation_image" title=" \displaystyle -2(5 x + 8) " src="/equation_images/%20%5Cdisplaystyle%20-2%285%20x%20%2B%208%29%20" alt="LaTeX:  \displaystyle -2(5 x + 8) " data-equation-content=" \displaystyle -2(5 x + 8) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 5 x + 8 " src="/equation_images/%20%5Cdisplaystyle%205%20x%20%2B%208%20" alt="LaTeX:  \displaystyle 5 x + 8 " data-equation-content=" \displaystyle 5 x + 8 " /> . This gives  <img class="equation_image" title=" \displaystyle 2[x^{2} \left(5 x + 8\right) -2 \cdot \left(5 x + 8\right)] = 2\left(5 x + 8\right) \left(x^{2} - 2\right) " src="/equation_images/%20%5Cdisplaystyle%202%5Bx%5E%7B2%7D%20%5Cleft%285%20x%20%2B%208%5Cright%29%20-2%20%5Ccdot%20%5Cleft%285%20x%20%2B%208%5Cright%29%5D%20%3D%202%5Cleft%285%20x%20%2B%208%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%202%5Cright%29%20" alt="LaTeX:  \displaystyle 2[x^{2} \left(5 x + 8\right) -2 \cdot \left(5 x + 8\right)] = 2\left(5 x + 8\right) \left(x^{2} - 2\right) " data-equation-content=" \displaystyle 2[x^{2} \left(5 x + 8\right) -2 \cdot \left(5 x + 8\right)] = 2\left(5 x + 8\right) \left(x^{2} - 2\right) " /> . </p> </p>