Please login to create an exam or a quiz.
Factor \(\displaystyle 10 x^{3} - 60 x^{2} - 10 x + 60\).
Factoring out the GCF \(\displaystyle 10\) from each term gives \(\displaystyle 10(x^{3} - 6 x^{2} - x + 6)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle x^{2}\), gives \(\displaystyle x^{2}(x - 6)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -1\), gives \(\displaystyle -1(x - 6)\). The polynomial now has a common binomial factor of \(\displaystyle x - 6\). This gives \(\displaystyle 10[x^{2} \left(x - 6\right) -1 \cdot \left(x - 6\right)] = 10\left(x - 6\right) \left(x^{2} - 1\right)\). The quadratic factor can be factored using the difference of squares to give \(\displaystyle 10\left(x - 6\right) \left(x - 1\right) \left(x + 1\right). \)
\begin{question}Factor $10 x^{3} - 60 x^{2} - 10 x + 60$.
\soln{9cm}{Factoring out the GCF $10$ from each term gives $10(x^{3} - 6 x^{2} - x + 6)$. Grouping the first two terms and factoring out their GCF, $x^{2}$, gives $x^{2}(x - 6)$. Grouping the last two terms and factoring out their GCF, $-1$, gives $-1(x - 6)$. The polynomial now has a common binomial factor of $x - 6$. This gives $10[x^{2} \left(x - 6\right) -1 \cdot \left(x - 6\right)] = 10\left(x - 6\right) \left(x^{2} - 1\right)$. The quadratic factor can be factored using the difference of squares to give $10\left(x - 6\right) \left(x - 1\right) \left(x + 1\right). $}
\end{question}
\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}<p> <p>Factor <img class="equation_image" title=" \displaystyle 10 x^{3} - 60 x^{2} - 10 x + 60 " src="/equation_images/%20%5Cdisplaystyle%2010%20x%5E%7B3%7D%20-%2060%20x%5E%7B2%7D%20-%2010%20x%20%2B%2060%20" alt="LaTeX: \displaystyle 10 x^{3} - 60 x^{2} - 10 x + 60 " data-equation-content=" \displaystyle 10 x^{3} - 60 x^{2} - 10 x + 60 " /> . </p> </p><p> <p>Factoring out the GCF <img class="equation_image" title=" \displaystyle 10 " src="/equation_images/%20%5Cdisplaystyle%2010%20" alt="LaTeX: \displaystyle 10 " data-equation-content=" \displaystyle 10 " /> from each term gives <img class="equation_image" title=" \displaystyle 10(x^{3} - 6 x^{2} - x + 6) " src="/equation_images/%20%5Cdisplaystyle%2010%28x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20-%20x%20%2B%206%29%20" alt="LaTeX: \displaystyle 10(x^{3} - 6 x^{2} - x + 6) " data-equation-content=" \displaystyle 10(x^{3} - 6 x^{2} - x + 6) " /> . 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}(x - 6) " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%28x%20-%206%29%20" alt="LaTeX: \displaystyle x^{2}(x - 6) " data-equation-content=" \displaystyle x^{2}(x - 6) " /> . 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(x - 6) " src="/equation_images/%20%5Cdisplaystyle%20-1%28x%20-%206%29%20" alt="LaTeX: \displaystyle -1(x - 6) " data-equation-content=" \displaystyle -1(x - 6) " /> . The polynomial now has a common binomial factor of <img class="equation_image" title=" \displaystyle x - 6 " src="/equation_images/%20%5Cdisplaystyle%20x%20-%206%20" alt="LaTeX: \displaystyle x - 6 " data-equation-content=" \displaystyle x - 6 " /> . This gives <img class="equation_image" title=" \displaystyle 10[x^{2} \left(x - 6\right) -1 \cdot \left(x - 6\right)] = 10\left(x - 6\right) \left(x^{2} - 1\right) " src="/equation_images/%20%5Cdisplaystyle%2010%5Bx%5E%7B2%7D%20%5Cleft%28x%20-%206%5Cright%29%20-1%20%5Ccdot%20%5Cleft%28x%20-%206%5Cright%29%5D%20%3D%2010%5Cleft%28x%20-%206%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%201%5Cright%29%20" alt="LaTeX: \displaystyle 10[x^{2} \left(x - 6\right) -1 \cdot \left(x - 6\right)] = 10\left(x - 6\right) \left(x^{2} - 1\right) " data-equation-content=" \displaystyle 10[x^{2} \left(x - 6\right) -1 \cdot \left(x - 6\right)] = 10\left(x - 6\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 10\left(x - 6\right) \left(x - 1\right) \left(x + 1\right). " src="/equation_images/%20%5Cdisplaystyle%2010%5Cleft%28x%20-%206%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29.%20%20" alt="LaTeX: \displaystyle 10\left(x - 6\right) \left(x - 1\right) \left(x + 1\right). " data-equation-content=" \displaystyle 10\left(x - 6\right) \left(x - 1\right) \left(x + 1\right). " /> </p> </p>