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

Download \(\LaTeX\)

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

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\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} + 5 x^{2} + 10 x - 5 " src="/equation_images/%20%5Cdisplaystyle%20-%2010%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%2010%20x%20-%205%20" alt="LaTeX:  \displaystyle - 10 x^{3} + 5 x^{2} + 10 x - 5 " data-equation-content=" \displaystyle - 10 x^{3} + 5 x^{2} + 10 x - 5 " /> . </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(2 x^{3} - x^{2} - 2 x + 1) " src="/equation_images/%20%5Cdisplaystyle%20-5%282%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%202%20x%20%2B%201%29%20" alt="LaTeX:  \displaystyle -5(2 x^{3} - x^{2} - 2 x + 1) " data-equation-content=" \displaystyle -5(2 x^{3} - x^{2} - 2 x + 1) " /> . 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}(2 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%282%20x%20-%201%29%20" alt="LaTeX:  \displaystyle x^{2}(2 x - 1) " data-equation-content=" \displaystyle x^{2}(2 x - 1) " /> . 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(2 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20-1%282%20x%20-%201%29%20" alt="LaTeX:  \displaystyle -1(2 x - 1) " data-equation-content=" \displaystyle -1(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[x^{2} \left(2 x - 1\right) -1 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(x^{2} - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20-5%5Bx%5E%7B2%7D%20%5Cleft%282%20x%20-%201%5Cright%29%20-1%20%5Ccdot%20%5Cleft%282%20x%20-%201%5Cright%29%5D%20%3D%20-5%5Cleft%282%20x%20-%201%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%201%5Cright%29%20" alt="LaTeX:  \displaystyle -5[x^{2} \left(2 x - 1\right) -1 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\right) \left(x^{2} - 1\right) " data-equation-content=" \displaystyle -5[x^{2} \left(2 x - 1\right) -1 \cdot \left(2 x - 1\right)] = -5\left(2 x - 1\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 -5\left(x - 1\right) \left(x + 1\right) \left(2 x - 1\right).  " src="/equation_images/%20%5Cdisplaystyle%20-5%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%282%20x%20-%201%5Cright%29.%20%20" alt="LaTeX:  \displaystyle -5\left(x - 1\right) \left(x + 1\right) \left(2 x - 1\right).  " data-equation-content=" \displaystyle -5\left(x - 1\right) \left(x + 1\right) \left(2 x - 1\right).  " /> </p> </p>