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Factor \(\displaystyle - 6 x^{3} - 60 x^{2} - 3 x - 30\).

Factoring out the GCF \(\displaystyle -3\) from each term gives \(\displaystyle -3(2 x^{3} + 20 x^{2} + x + 10)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 2 x^{2}\), gives \(\displaystyle 2 x^{2}(x + 10)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle 1\), gives \(\displaystyle 1(x + 10)\). The polynomial now has a common binomial factor of \(\displaystyle x + 10\). This gives \(\displaystyle -3[2 x^{2} \left(x + 10\right) +1 \cdot \left(x + 10\right)] = -3\left(x + 10\right) \left(2 x^{2} + 1\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $- 6 x^{3} - 60 x^{2} - 3 x - 30$. 
    \soln{9cm}{Factoring out the GCF $-3$ from each term gives $-3(2 x^{3} + 20 x^{2} + x + 10)$. Grouping the first two terms and factoring out their GCF, $2 x^{2}$, gives $2 x^{2}(x + 10)$. Grouping the last two terms and factoring out their GCF, $1$, gives $1(x + 10)$. The polynomial now has a common binomial factor of $x + 10$. This gives $-3[2 x^{2} \left(x + 10\right) +1 \cdot \left(x + 10\right)] = -3\left(x + 10\right) \left(2 x^{2} + 1\right)$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Factor  <img class="equation_image" title=" \displaystyle - 6 x^{3} - 60 x^{2} - 3 x - 30 " src="/equation_images/%20%5Cdisplaystyle%20-%206%20x%5E%7B3%7D%20-%2060%20x%5E%7B2%7D%20-%203%20x%20-%2030%20" alt="LaTeX:  \displaystyle - 6 x^{3} - 60 x^{2} - 3 x - 30 " data-equation-content=" \displaystyle - 6 x^{3} - 60 x^{2} - 3 x - 30 " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle -3 " src="/equation_images/%20%5Cdisplaystyle%20-3%20" alt="LaTeX:  \displaystyle -3 " data-equation-content=" \displaystyle -3 " />  from each term gives  <img class="equation_image" title=" \displaystyle -3(2 x^{3} + 20 x^{2} + x + 10) " src="/equation_images/%20%5Cdisplaystyle%20-3%282%20x%5E%7B3%7D%20%2B%2020%20x%5E%7B2%7D%20%2B%20x%20%2B%2010%29%20" alt="LaTeX:  \displaystyle -3(2 x^{3} + 20 x^{2} + x + 10) " data-equation-content=" \displaystyle -3(2 x^{3} + 20 x^{2} + x + 10) " /> . 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}(x + 10) " src="/equation_images/%20%5Cdisplaystyle%202%20x%5E%7B2%7D%28x%20%2B%2010%29%20" alt="LaTeX:  \displaystyle 2 x^{2}(x + 10) " data-equation-content=" \displaystyle 2 x^{2}(x + 10) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 1 " src="/equation_images/%20%5Cdisplaystyle%201%20" alt="LaTeX:  \displaystyle 1 " data-equation-content=" \displaystyle 1 " /> , gives  <img class="equation_image" title=" \displaystyle 1(x + 10) " src="/equation_images/%20%5Cdisplaystyle%201%28x%20%2B%2010%29%20" alt="LaTeX:  \displaystyle 1(x + 10) " data-equation-content=" \displaystyle 1(x + 10) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle x + 10 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2010%20" alt="LaTeX:  \displaystyle x + 10 " data-equation-content=" \displaystyle x + 10 " /> . This gives  <img class="equation_image" title=" \displaystyle -3[2 x^{2} \left(x + 10\right) +1 \cdot \left(x + 10\right)] = -3\left(x + 10\right) \left(2 x^{2} + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20-3%5B2%20x%5E%7B2%7D%20%5Cleft%28x%20%2B%2010%5Cright%29%20%2B1%20%5Ccdot%20%5Cleft%28x%20%2B%2010%5Cright%29%5D%20%3D%20-3%5Cleft%28x%20%2B%2010%5Cright%29%20%5Cleft%282%20x%5E%7B2%7D%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle -3[2 x^{2} \left(x + 10\right) +1 \cdot \left(x + 10\right)] = -3\left(x + 10\right) \left(2 x^{2} + 1\right) " data-equation-content=" \displaystyle -3[2 x^{2} \left(x + 10\right) +1 \cdot \left(x + 10\right)] = -3\left(x + 10\right) \left(2 x^{2} + 1\right) " /> . </p> </p>