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Factor \(\displaystyle 18 x^{3} + 54 x^{2} - 14 x - 42\).

Factoring out the GCF \(\displaystyle 2\) from each term gives \(\displaystyle 2(9 x^{3} + 27 x^{2} - 7 x - 21)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 9 x^{2}\), gives \(\displaystyle 9 x^{2}(x + 3)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -7\), gives \(\displaystyle -7(x + 3)\). The polynomial now has a common binomial factor of \(\displaystyle x + 3\). This gives \(\displaystyle 2[9 x^{2} \left(x + 3\right) -7 \cdot \left(x + 3\right)] = 2\left(x + 3\right) \left(9 x^{2} - 7\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $18 x^{3} + 54 x^{2} - 14 x - 42$. 
    \soln{9cm}{Factoring out the GCF $2$ from each term gives $2(9 x^{3} + 27 x^{2} - 7 x - 21)$. Grouping the first two terms and factoring out their GCF, $9 x^{2}$, gives $9 x^{2}(x + 3)$. Grouping the last two terms and factoring out their GCF, $-7$, gives $-7(x + 3)$. The polynomial now has a common binomial factor of $x + 3$. This gives $2[9 x^{2} \left(x + 3\right) -7 \cdot \left(x + 3\right)] = 2\left(x + 3\right) \left(9 x^{2} - 7\right)$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Factor  <img class="equation_image" title=" \displaystyle 18 x^{3} + 54 x^{2} - 14 x - 42 " src="/equation_images/%20%5Cdisplaystyle%2018%20x%5E%7B3%7D%20%2B%2054%20x%5E%7B2%7D%20-%2014%20x%20-%2042%20" alt="LaTeX:  \displaystyle 18 x^{3} + 54 x^{2} - 14 x - 42 " data-equation-content=" \displaystyle 18 x^{3} + 54 x^{2} - 14 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%202%20" alt="LaTeX:  \displaystyle 2 " data-equation-content=" \displaystyle 2 " />  from each term gives  <img class="equation_image" title=" \displaystyle 2(9 x^{3} + 27 x^{2} - 7 x - 21) " src="/equation_images/%20%5Cdisplaystyle%202%289%20x%5E%7B3%7D%20%2B%2027%20x%5E%7B2%7D%20-%207%20x%20-%2021%29%20" alt="LaTeX:  \displaystyle 2(9 x^{3} + 27 x^{2} - 7 x - 21) " data-equation-content=" \displaystyle 2(9 x^{3} + 27 x^{2} - 7 x - 21) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 9 x^{2} " src="/equation_images/%20%5Cdisplaystyle%209%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle 9 x^{2} " data-equation-content=" \displaystyle 9 x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle 9 x^{2}(x + 3) " src="/equation_images/%20%5Cdisplaystyle%209%20x%5E%7B2%7D%28x%20%2B%203%29%20" alt="LaTeX:  \displaystyle 9 x^{2}(x + 3) " data-equation-content=" \displaystyle 9 x^{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(x + 3) " src="/equation_images/%20%5Cdisplaystyle%20-7%28x%20%2B%203%29%20" alt="LaTeX:  \displaystyle -7(x + 3) " data-equation-content=" \displaystyle -7(x + 3) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle x + 3 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%203%20" alt="LaTeX:  \displaystyle x + 3 " data-equation-content=" \displaystyle x + 3 " /> . This gives  <img class="equation_image" title=" \displaystyle 2[9 x^{2} \left(x + 3\right) -7 \cdot \left(x + 3\right)] = 2\left(x + 3\right) \left(9 x^{2} - 7\right) " src="/equation_images/%20%5Cdisplaystyle%202%5B9%20x%5E%7B2%7D%20%5Cleft%28x%20%2B%203%5Cright%29%20-7%20%5Ccdot%20%5Cleft%28x%20%2B%203%5Cright%29%5D%20%3D%202%5Cleft%28x%20%2B%203%5Cright%29%20%5Cleft%289%20x%5E%7B2%7D%20-%207%5Cright%29%20" alt="LaTeX:  \displaystyle 2[9 x^{2} \left(x + 3\right) -7 \cdot \left(x + 3\right)] = 2\left(x + 3\right) \left(9 x^{2} - 7\right) " data-equation-content=" \displaystyle 2[9 x^{2} \left(x + 3\right) -7 \cdot \left(x + 3\right)] = 2\left(x + 3\right) \left(9 x^{2} - 7\right) " /> . </p> </p>