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

Please login to create an exam or a quiz.

Algebra
Quadratics
New Random

Factor \(\displaystyle - 20 x^{3} + 10 x^{2} - 90 x + 45\).

Factoring out the GCF \(\displaystyle -5\) from each term gives \(\displaystyle -5(4 x^{3} - 2 x^{2} + 18 x - 9)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 2 x^{2}\), gives \(\displaystyle 2 x^{2}(2 x - 1)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle 9\), gives \(\displaystyle 9(2 x - 1)\). The polynomial now has a common binomial factor of \(\displaystyle 2 x - 1\). This gives \(\displaystyle -5[2 x^{2} \left(2 x - 1\right) +9 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(2 x^{2} + 9\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $- 20 x^{3} + 10 x^{2} - 90 x + 45$. 
    \soln{9cm}{Factoring out the GCF $-5$ from each term gives $-5(4 x^{3} - 2 x^{2} + 18 x - 9)$. Grouping the first two terms and factoring out their GCF, $2 x^{2}$, gives $2 x^{2}(2 x - 1)$. Grouping the last two terms and factoring out their GCF, $9$, gives $9(2 x - 1)$. The polynomial now has a common binomial factor of $2 x - 1$. This gives $-5[2 x^{2} \left(2 x - 1\right) +9 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(2 x^{2} + 9\right)$. }

\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 - 20 x^{3} + 10 x^{2} - 90 x + 45 " src="/equation_images/%20%5Cdisplaystyle%20-%2020%20x%5E%7B3%7D%20%2B%2010%20x%5E%7B2%7D%20-%2090%20x%20%2B%2045%20" alt="LaTeX:  \displaystyle - 20 x^{3} + 10 x^{2} - 90 x + 45 " data-equation-content=" \displaystyle - 20 x^{3} + 10 x^{2} - 90 x + 45 " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle -5 " src="/equation_images/%20%5Cdisplaystyle%20-5%20" alt="LaTeX:  \displaystyle -5 " data-equation-content=" \displaystyle -5 " />  from each term gives  <img class="equation_image" title=" \displaystyle -5(4 x^{3} - 2 x^{2} + 18 x - 9) " src="/equation_images/%20%5Cdisplaystyle%20-5%284%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20%2B%2018%20x%20-%209%29%20" alt="LaTeX:  \displaystyle -5(4 x^{3} - 2 x^{2} + 18 x - 9) " data-equation-content=" \displaystyle -5(4 x^{3} - 2 x^{2} + 18 x - 9) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 2 x^{2} " src="/equation_images/%20%5Cdisplaystyle%202%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle 2 x^{2} " data-equation-content=" \displaystyle 2 x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle 2 x^{2}(2 x - 1) " src="/equation_images/%20%5Cdisplaystyle%202%20x%5E%7B2%7D%282%20x%20-%201%29%20" alt="LaTeX:  \displaystyle 2 x^{2}(2 x - 1) " data-equation-content=" \displaystyle 2 x^{2}(2 x - 1) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 9 " src="/equation_images/%20%5Cdisplaystyle%209%20" alt="LaTeX:  \displaystyle 9 " data-equation-content=" \displaystyle 9 " /> , gives  <img class="equation_image" title=" \displaystyle 9(2 x - 1) " src="/equation_images/%20%5Cdisplaystyle%209%282%20x%20-%201%29%20" alt="LaTeX:  \displaystyle 9(2 x - 1) " data-equation-content=" \displaystyle 9(2 x - 1) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 2 x - 1 " src="/equation_images/%20%5Cdisplaystyle%202%20x%20-%201%20" alt="LaTeX:  \displaystyle 2 x - 1 " data-equation-content=" \displaystyle 2 x - 1 " /> . This gives  <img class="equation_image" title=" \displaystyle -5[2 x^{2} \left(2 x - 1\right) +9 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(2 x^{2} + 9\right) " src="/equation_images/%20%5Cdisplaystyle%20-5%5B2%20x%5E%7B2%7D%20%5Cleft%282%20x%20-%201%5Cright%29%20%2B9%20%5Ccdot%20%5Cleft%282%20x%20-%201%5Cright%29%5D%20%3D%20-5%5Cleft%282%20x%20-%201%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%209%5Cright%29%20" alt="LaTeX:  \displaystyle -5[2 x^{2} \left(2 x - 1\right) +9 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(2 x^{2} + 9\right) " data-equation-content=" \displaystyle -5[2 x^{2} \left(2 x - 1\right) +9 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(2 x^{2} + 9\right) " /> . </p> </p>