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Factor \(\displaystyle - 32 x^{3} - 48 x^{2} + 28 x + 42\).

Factoring out the GCF \(\displaystyle -2\) from each term gives \(\displaystyle -2(16 x^{3} + 24 x^{2} - 14 x - 21)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 8 x^{2}\), gives \(\displaystyle 8 x^{2}(2 x + 3)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -7\), gives \(\displaystyle -7(2 x + 3)\). The polynomial now has a common binomial factor of \(\displaystyle 2 x + 3\). This gives \(\displaystyle -2[8 x^{2} \left(2 x + 3\right) -7 \cdot \left(2 x + 3\right)] = -2\left(2 x + 3\right) \left(8 x^{2} - 7\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $- 32 x^{3} - 48 x^{2} + 28 x + 42$. 
    \soln{9cm}{Factoring out the GCF $-2$ from each term gives $-2(16 x^{3} + 24 x^{2} - 14 x - 21)$. Grouping the first two terms and factoring out their GCF, $8 x^{2}$, gives $8 x^{2}(2 x + 3)$. Grouping the last two terms and factoring out their GCF, $-7$, gives $-7(2 x + 3)$. The polynomial now has a common binomial factor of $2 x + 3$. This gives $-2[8 x^{2} \left(2 x + 3\right) -7 \cdot \left(2 x + 3\right)] = -2\left(2 x + 3\right) \left(8 x^{2} - 7\right)$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Factor  <img class="equation_image" title=" \displaystyle - 32 x^{3} - 48 x^{2} + 28 x + 42 " src="/equation_images/%20%5Cdisplaystyle%20-%2032%20x%5E%7B3%7D%20-%2048%20x%5E%7B2%7D%20%2B%2028%20x%20%2B%2042%20" alt="LaTeX:  \displaystyle - 32 x^{3} - 48 x^{2} + 28 x + 42 " data-equation-content=" \displaystyle - 32 x^{3} - 48 x^{2} + 28 x + 42 " /> . </p> </p>
HTML for Canvas
<p> <p>Factoring out the GCF  <img class="equation_image" title=" \displaystyle -2 " src="/equation_images/%20%5Cdisplaystyle%20-2%20" alt="LaTeX:  \displaystyle -2 " data-equation-content=" \displaystyle -2 " />  from each term gives  <img class="equation_image" title=" \displaystyle -2(16 x^{3} + 24 x^{2} - 14 x - 21) " src="/equation_images/%20%5Cdisplaystyle%20-2%2816%20x%5E%7B3%7D%20%2B%2024%20x%5E%7B2%7D%20-%2014%20x%20-%2021%29%20" alt="LaTeX:  \displaystyle -2(16 x^{3} + 24 x^{2} - 14 x - 21) " data-equation-content=" \displaystyle -2(16 x^{3} + 24 x^{2} - 14 x - 21) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 8 x^{2} " src="/equation_images/%20%5Cdisplaystyle%208%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle 8 x^{2} " data-equation-content=" \displaystyle 8 x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle 8 x^{2}(2 x + 3) " src="/equation_images/%20%5Cdisplaystyle%208%20x%5E%7B2%7D%282%20x%20%2B%203%29%20" alt="LaTeX:  \displaystyle 8 x^{2}(2 x + 3) " data-equation-content=" \displaystyle 8 x^{2}(2 x + 3) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle -7 " src="/equation_images/%20%5Cdisplaystyle%20-7%20" alt="LaTeX:  \displaystyle -7 " data-equation-content=" \displaystyle -7 " /> , gives  <img class="equation_image" title=" \displaystyle -7(2 x + 3) " src="/equation_images/%20%5Cdisplaystyle%20-7%282%20x%20%2B%203%29%20" alt="LaTeX:  \displaystyle -7(2 x + 3) " data-equation-content=" \displaystyle -7(2 x + 3) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle 2 x + 3 " src="/equation_images/%20%5Cdisplaystyle%202%20x%20%2B%203%20" alt="LaTeX:  \displaystyle 2 x + 3 " data-equation-content=" \displaystyle 2 x + 3 " /> . This gives  <img class="equation_image" title=" \displaystyle -2[8 x^{2} \left(2 x + 3\right) -7 \cdot \left(2 x + 3\right)] = -2\left(2 x + 3\right) \left(8 x^{2} - 7\right) " src="/equation_images/%20%5Cdisplaystyle%20-2%5B8%20x%5E%7B2%7D%20%5Cleft%282%20x%20%2B%203%5Cright%29%20-7%20%5Ccdot%20%5Cleft%282%20x%20%2B%203%5Cright%29%5D%20%3D%20-2%5Cleft%282%20x%20%2B%203%5Cright%29%20%5Cleft%288%20x%5E%7B2%7D%20-%207%5Cright%29%20" alt="LaTeX:  \displaystyle -2[8 x^{2} \left(2 x + 3\right) -7 \cdot \left(2 x + 3\right)] = -2\left(2 x + 3\right) \left(8 x^{2} - 7\right) " data-equation-content=" \displaystyle -2[8 x^{2} \left(2 x + 3\right) -7 \cdot \left(2 x + 3\right)] = -2\left(2 x + 3\right) \left(8 x^{2} - 7\right) " /> . </p> </p>