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Factoring out the GCF \(\displaystyle -6\) from each term gives \(\displaystyle -6(5 x^{3} - 2 x^{2} - 5 x + 2)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle x^{2}\), gives \(\displaystyle x^{2}(5 x - 2)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -1\), gives \(\displaystyle -1(5 x - 2)\). The polynomial now has a common binomial factor of \(\displaystyle 5 x - 2\). This gives \(\displaystyle -6[x^{2} \left(5 x - 2\right) -1 \cdot \left(5 x - 2\right)] = -6\left(5 x - 2\right) \left(x^{2} - 1\right)\). The quadratic factor can be factored using the difference of squares to give \(\displaystyle -6\left(x - 1\right) \left(x + 1\right) \left(5 x - 2\right). \)

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\begin{question}Factor $- 30 x^{3} + 12 x^{2} + 30 x - 12$. 
    \soln{9cm}{Factoring out the GCF $-6$ from each term gives $-6(5 x^{3} - 2 x^{2} - 5 x + 2)$. Grouping the first two terms and factoring out their GCF, $x^{2}$, gives $x^{2}(5 x - 2)$. Grouping the last two terms and factoring out their GCF, $-1$, gives $-1(5 x - 2)$. The polynomial now has a common binomial factor of $5 x - 2$. This gives $-6[x^{2} \left(5 x - 2\right) -1 \cdot \left(5 x - 2\right)] = -6\left(5 x - 2\right) \left(x^{2} - 1\right)$. The quadratic factor can be factored using the difference of squares to give $-6\left(x - 1\right) \left(x + 1\right) \left(5 x - 2\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 - 30 x^{3} + 12 x^{2} + 30 x - 12 " src="/equation_images/%20%5Cdisplaystyle%20-%2030%20x%5E%7B3%7D%20%2B%2012%20x%5E%7B2%7D%20%2B%2030%20x%20-%2012%20" alt="LaTeX:  \displaystyle - 30 x^{3} + 12 x^{2} + 30 x - 12 " data-equation-content=" \displaystyle - 30 x^{3} + 12 x^{2} + 30 x - 12 " /> . </p> </p>

HTML for Canvas

<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle -6 " src="/equation_images/%20%5Cdisplaystyle%20-6%20" alt="LaTeX:  \displaystyle -6 " data-equation-content=" \displaystyle -6 " />  from each term gives  <img class="equation_image" title=" \displaystyle -6(5 x^{3} - 2 x^{2} - 5 x + 2) " src="/equation_images/%20%5Cdisplaystyle%20-6%285%20x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20%2B%202%29%20" alt="LaTeX:  \displaystyle -6(5 x^{3} - 2 x^{2} - 5 x + 2) " data-equation-content=" \displaystyle -6(5 x^{3} - 2 x^{2} - 5 x + 2) " /> . 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 - 2) " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%285%20x%20-%202%29%20" alt="LaTeX:  \displaystyle x^{2}(5 x - 2) " data-equation-content=" \displaystyle x^{2}(5 x - 2) " /> . 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(5 x - 2) " src="/equation_images/%20%5Cdisplaystyle%20-1%285%20x%20-%202%29%20" alt="LaTeX:  \displaystyle -1(5 x - 2) " data-equation-content=" \displaystyle -1(5 x - 2) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 5 x - 2 " src="/equation_images/%20%5Cdisplaystyle%205%20x%20-%202%20" alt="LaTeX:  \displaystyle 5 x - 2 " data-equation-content=" \displaystyle 5 x - 2 " /> . This gives  <img class="equation_image" title=" \displaystyle -6[x^{2} \left(5 x - 2\right) -1 \cdot \left(5 x - 2\right)] = -6\left(5 x - 2\right) \left(x^{2} - 1\right) " src="/equation_images/%20%5Cdisplaystyle%20-6%5Bx%5E%7B2%7D%20%5Cleft%285%20x%20-%202%5Cright%29%20-1%20%5Ccdot%20%5Cleft%285%20x%20-%202%5Cright%29%5D%20%3D%20-6%5Cleft%285%20x%20-%202%5Cright%29%20%5Cleft%28x%5E%7B2%7D%20-%201%5Cright%29%20" alt="LaTeX:  \displaystyle -6[x^{2} \left(5 x - 2\right) -1 \cdot \left(5 x - 2\right)] = -6\left(5 x - 2\right) \left(x^{2} - 1\right) " data-equation-content=" \displaystyle -6[x^{2} \left(5 x - 2\right) -1 \cdot \left(5 x - 2\right)] = -6\left(5 x - 2\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 -6\left(x - 1\right) \left(x + 1\right) \left(5 x - 2\right).  " src="/equation_images/%20%5Cdisplaystyle%20-6%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%285%20x%20-%202%5Cright%29.%20%20" alt="LaTeX:  \displaystyle -6\left(x - 1\right) \left(x + 1\right) \left(5 x - 2\right).  " data-equation-content=" \displaystyle -6\left(x - 1\right) \left(x + 1\right) \left(5 x - 2\right).  " /> </p> </p>