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Factor \(\displaystyle 14 x^{3} + 28 x^{2} + 20 x + 40\).

Factoring out the GCF \(\displaystyle 2\) from each term gives \(\displaystyle 2(7 x^{3} + 14 x^{2} + 10 x + 20)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 7 x^{2}\), gives \(\displaystyle 7 x^{2}(x + 2)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle 10\), gives \(\displaystyle 10(x + 2)\). The polynomial now has a common binomial factor of \(\displaystyle x + 2\). This gives \(\displaystyle 2[7 x^{2} \left(x + 2\right) +10 \cdot \left(x + 2\right)] = 2\left(x + 2\right) \left(7 x^{2} + 10\right)\).

Download \(\LaTeX\) 

\begin{question}Factor $14 x^{3} + 28 x^{2} + 20 x + 40$. 
    \soln{9cm}{Factoring out the GCF $2$ from each term gives $2(7 x^{3} + 14 x^{2} + 10 x + 20)$. Grouping the first two terms and factoring out their GCF, $7 x^{2}$, gives $7 x^{2}(x + 2)$. Grouping the last two terms and factoring out their GCF, $10$, gives $10(x + 2)$. The polynomial now has a common binomial factor of $x + 2$. This gives $2[7 x^{2} \left(x + 2\right) +10 \cdot \left(x + 2\right)] = 2\left(x + 2\right) \left(7 x^{2} + 10\right)$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Factor  <img class="equation_image" title=" \displaystyle 14 x^{3} + 28 x^{2} + 20 x + 40 " src="/equation_images/%20%5Cdisplaystyle%2014%20x%5E%7B3%7D%20%2B%2028%20x%5E%7B2%7D%20%2B%2020%20x%20%2B%2040%20" alt="LaTeX:  \displaystyle 14 x^{3} + 28 x^{2} + 20 x + 40 " data-equation-content=" \displaystyle 14 x^{3} + 28 x^{2} + 20 x + 40 " /> . </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(7 x^{3} + 14 x^{2} + 10 x + 20) " src="/equation_images/%20%5Cdisplaystyle%202%287%20x%5E%7B3%7D%20%2B%2014%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%2020%29%20" alt="LaTeX:  \displaystyle 2(7 x^{3} + 14 x^{2} + 10 x + 20) " data-equation-content=" \displaystyle 2(7 x^{3} + 14 x^{2} + 10 x + 20) " /> . Grouping the first two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 7 x^{2} " src="/equation_images/%20%5Cdisplaystyle%207%20x%5E%7B2%7D%20" alt="LaTeX:  \displaystyle 7 x^{2} " data-equation-content=" \displaystyle 7 x^{2} " /> , gives  <img class="equation_image" title=" \displaystyle 7 x^{2}(x + 2) " src="/equation_images/%20%5Cdisplaystyle%207%20x%5E%7B2%7D%28x%20%2B%202%29%20" alt="LaTeX:  \displaystyle 7 x^{2}(x + 2) " data-equation-content=" \displaystyle 7 x^{2}(x + 2) " /> . Grouping the last two terms and factoring out their GCF,  <img class="equation_image" title=" \displaystyle 10 " src="/equation_images/%20%5Cdisplaystyle%2010%20" alt="LaTeX:  \displaystyle 10 " data-equation-content=" \displaystyle 10 " /> , gives  <img class="equation_image" title=" \displaystyle 10(x + 2) " src="/equation_images/%20%5Cdisplaystyle%2010%28x%20%2B%202%29%20" alt="LaTeX:  \displaystyle 10(x + 2) " data-equation-content=" \displaystyle 10(x + 2) " /> . The polynomial now has a common binomial factor of  <img class="equation_image" title=" \displaystyle x + 2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%202%20" alt="LaTeX:  \displaystyle x + 2 " data-equation-content=" \displaystyle x + 2 " /> . This gives  <img class="equation_image" title=" \displaystyle 2[7 x^{2} \left(x + 2\right) +10 \cdot \left(x + 2\right)] = 2\left(x + 2\right) \left(7 x^{2} + 10\right) " src="/equation_images/%20%5Cdisplaystyle%202%5B7%20x%5E%7B2%7D%20%5Cleft%28x%20%2B%202%5Cright%29%20%2B10%20%5Ccdot%20%5Cleft%28x%20%2B%202%5Cright%29%5D%20%3D%202%5Cleft%28x%20%2B%202%5Cright%29%20%5Cleft%287%20x%5E%7B2%7D%20%2B%2010%5Cright%29%20" alt="LaTeX:  \displaystyle 2[7 x^{2} \left(x + 2\right) +10 \cdot \left(x + 2\right)] = 2\left(x + 2\right) \left(7 x^{2} + 10\right) " data-equation-content=" \displaystyle 2[7 x^{2} \left(x + 2\right) +10 \cdot \left(x + 2\right)] = 2\left(x + 2\right) \left(7 x^{2} + 10\right) " /> . </p> </p>