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Questions: Algebra BusinessCalculus
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Factor \(\displaystyle 18 x^{3} + 90 x^{2} - 14 x - 70\).
Factoring out the GCF \(\displaystyle 2\) from each term gives \(\displaystyle 2(9 x^{3} + 45 x^{2} - 7 x - 35)\). Grouping the first two terms and factoring out their GCF, \(\displaystyle 9 x^{2}\), gives \(\displaystyle 9 x^{2}(x + 5)\). Grouping the last two terms and factoring out their GCF, \(\displaystyle -7\), gives \(\displaystyle -7(x + 5)\). The polynomial now has a common binomial factor of \(\displaystyle x + 5\). This gives \(\displaystyle 2[9 x^{2} \left(x + 5\right) -7 \cdot \left(x + 5\right)] = 2\left(x + 5\right) \left(9 x^{2} - 7\right)\).
\begin{question}Factor $18 x^{3} + 90 x^{2} - 14 x - 70$.
\soln{9cm}{Factoring out the GCF $2$ from each term gives $2(9 x^{3} + 45 x^{2} - 7 x - 35)$. Grouping the first two terms and factoring out their GCF, $9 x^{2}$, gives $9 x^{2}(x + 5)$. Grouping the last two terms and factoring out their GCF, $-7$, gives $-7(x + 5)$. The polynomial now has a common binomial factor of $x + 5$. This gives $2[9 x^{2} \left(x + 5\right) -7 \cdot \left(x + 5\right)] = 2\left(x + 5\right) \left(9 x^{2} - 7\right)$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Factor <img class="equation_image" title=" \displaystyle 18 x^{3} + 90 x^{2} - 14 x - 70 " src="/equation_images/%20%5Cdisplaystyle%2018%20x%5E%7B3%7D%20%2B%2090%20x%5E%7B2%7D%20-%2014%20x%20-%2070%20" alt="LaTeX: \displaystyle 18 x^{3} + 90 x^{2} - 14 x - 70 " data-equation-content=" \displaystyle 18 x^{3} + 90 x^{2} - 14 x - 70 " /> . </p> </p><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} + 45 x^{2} - 7 x - 35) " src="/equation_images/%20%5Cdisplaystyle%202%289%20x%5E%7B3%7D%20%2B%2045%20x%5E%7B2%7D%20-%207%20x%20-%2035%29%20" alt="LaTeX: \displaystyle 2(9 x^{3} + 45 x^{2} - 7 x - 35) " data-equation-content=" \displaystyle 2(9 x^{3} + 45 x^{2} - 7 x - 35) " /> . 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 + 5) " src="/equation_images/%20%5Cdisplaystyle%209%20x%5E%7B2%7D%28x%20%2B%205%29%20" alt="LaTeX: \displaystyle 9 x^{2}(x + 5) " data-equation-content=" \displaystyle 9 x^{2}(x + 5) " /> . 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 + 5) " src="/equation_images/%20%5Cdisplaystyle%20-7%28x%20%2B%205%29%20" alt="LaTeX: \displaystyle -7(x + 5) " data-equation-content=" \displaystyle -7(x + 5) " /> . The polynomial now has a common binomial factor of <img class="equation_image" title=" \displaystyle x + 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%205%20" alt="LaTeX: \displaystyle x + 5 " data-equation-content=" \displaystyle x + 5 " /> . This gives <img class="equation_image" title=" \displaystyle 2[9 x^{2} \left(x + 5\right) -7 \cdot \left(x + 5\right)] = 2\left(x + 5\right) \left(9 x^{2} - 7\right) " src="/equation_images/%20%5Cdisplaystyle%202%5B9%20x%5E%7B2%7D%20%5Cleft%28x%20%2B%205%5Cright%29%20-7%20%5Ccdot%20%5Cleft%28x%20%2B%205%5Cright%29%5D%20%3D%202%5Cleft%28x%20%2B%205%5Cright%29%20%5Cleft%289%20x%5E%7B2%7D%20-%207%5Cright%29%20" alt="LaTeX: \displaystyle 2[9 x^{2} \left(x + 5\right) -7 \cdot \left(x + 5\right)] = 2\left(x + 5\right) \left(9 x^{2} - 7\right) " data-equation-content=" \displaystyle 2[9 x^{2} \left(x + 5\right) -7 \cdot \left(x + 5\right)] = 2\left(x + 5\right) \left(9 x^{2} - 7\right) " /> . </p> </p>