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

Please login to create an exam or a quiz.

Algebra
Quadratics
New Random

Factor \(\displaystyle - 10 x^{3} - 45 x^{2} - 12 x - 54\).

Factoring out the GCF \(\displaystyle -1\) from each term gives \(\displaystyle -(10 x^{3} + 45 x^{2} + 12 x + 54)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 5 x^{2}\), gives \(\displaystyle 5 x^{2}(2 x + 9)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle 6\), gives \(\displaystyle 6(2 x + 9)\). The polynomial now has a common binomial factor of \(\displaystyle 2 x + 9\). This gives \(\displaystyle -1[5 x^{2} \left(2 x + 9\right) +6 \cdot \left(2 x + 9\right)] = -\left(2 x + 9\right) \left(5 x^{2} + 6\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $- 10 x^{3} - 45 x^{2} - 12 x - 54$. 
    \soln{9cm}{Factoring out the GCF $-1$ from each term gives $-(10 x^{3} + 45 x^{2} + 12 x + 54)$. Grouping the first two terms and factoring out their GCF, $5 x^{2}$, gives $5 x^{2}(2 x + 9)$. Grouping the last two terms and factoring out their GCF, $6$, gives $6(2 x + 9)$. The polynomial now has a common binomial factor of $2 x + 9$. This gives $-1[5 x^{2} \left(2 x + 9\right) +6 \cdot \left(2 x + 9\right)] = -\left(2 x + 9\right) \left(5 x^{2} + 6\right)$. }

\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} - 45 x^{2} - 12 x - 54 " src="/equation_images/%20%5Cdisplaystyle%20-%2010%20x%5E%7B3%7D%20-%2045%20x%5E%7B2%7D%20-%2012%20x%20-%2054%20" alt="LaTeX:  \displaystyle - 10 x^{3} - 45 x^{2} - 12 x - 54 " data-equation-content=" \displaystyle - 10 x^{3} - 45 x^{2} - 12 x - 54 " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring out the 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 " />  from each term gives  <img class="equation_image" title=" \displaystyle -(10 x^{3} + 45 x^{2} + 12 x + 54) " src="/equation_images/%20%5Cdisplaystyle%20-%2810%20x%5E%7B3%7D%20%2B%2045%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%2054%29%20" alt="LaTeX:  \displaystyle -(10 x^{3} + 45 x^{2} + 12 x + 54) " data-equation-content=" \displaystyle -(10 x^{3} + 45 x^{2} + 12 x + 54) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 5 x^{2} " src="/equation_images/%20%5Cdisplaystyle%205%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle 5 x^{2} " data-equation-content=" \displaystyle 5 x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle 5 x^{2}(2 x + 9) " src="/equation_images/%20%5Cdisplaystyle%205%20x%5E%7B2%7D%282%20x%20%2B%209%29%20" alt="LaTeX:  \displaystyle 5 x^{2}(2 x + 9) " data-equation-content=" \displaystyle 5 x^{2}(2 x + 9) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 6 " src="/equation_images/%20%5Cdisplaystyle%206%20" alt="LaTeX:  \displaystyle 6 " data-equation-content=" \displaystyle 6 " /> , gives  <img class="equation_image" title=" \displaystyle 6(2 x + 9) " src="/equation_images/%20%5Cdisplaystyle%206%282%20x%20%2B%209%29%20" alt="LaTeX:  \displaystyle 6(2 x + 9) " data-equation-content=" \displaystyle 6(2 x + 9) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 2 x + 9 " src="/equation_images/%20%5Cdisplaystyle%202%20x%20%2B%209%20" alt="LaTeX:  \displaystyle 2 x + 9 " data-equation-content=" \displaystyle 2 x + 9 " /> . This gives  <img class="equation_image" title=" \displaystyle -1[5 x^{2} \left(2 x + 9\right) +6 \cdot \left(2 x + 9\right)] = -\left(2 x + 9\right) \left(5 x^{2} + 6\right) " src="/equation_images/%20%5Cdisplaystyle%20-1%5B5%20x%5E%7B2%7D%20%5Cleft%282%20x%20%2B%209%5Cright%29%20%2B6%20%5Ccdot%20%5Cleft%282%20x%20%2B%209%5Cright%29%5D%20%3D%20-%5Cleft%282%20x%20%2B%209%5Cright%29%20%5Cleft%285%20x%5E%7B2%7D%20%2B%206%5Cright%29%20" alt="LaTeX:  \displaystyle -1[5 x^{2} \left(2 x + 9\right) +6 \cdot \left(2 x + 9\right)] = -\left(2 x + 9\right) \left(5 x^{2} + 6\right) " data-equation-content=" \displaystyle -1[5 x^{2} \left(2 x + 9\right) +6 \cdot \left(2 x + 9\right)] = -\left(2 x + 9\right) \left(5 x^{2} + 6\right) " /> . </p> </p>