Question
Answer
$$x^2+14x+(mx+14)^2-6(mx+14)=m^2x^2+22mx+x^2+14x+112$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}x^2+14x+(mx+14)^2-6(mx+14)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+14x+m^2x^2+28mx+196-6(mx+14) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}m^2x^2+28mx+x^2+14x+196-6(mx+14) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}m^2x^2+28mx+x^2+14x+196-(6mx+84) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}m^2x^2+28mx+x^2+14x+196-6mx-84 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}m^2x^2+22mx+x^2+14x+112\end{aligned} $$ | |
| ① | Find $ \left(mx+14\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ mx } $ and $ B = \color{red}{ 14 }$. $$ \begin{aligned}\left(mx+14\right)^2 = \color{blue}{\left( mx \right)^2} +2 \cdot mx \cdot 14 + \color{red}{14^2} = m^2x^2+28mx+196\end{aligned} $$ |
| ② | Combine like terms: $$ x^2+14x+m^2x^2+28mx+196 = m^2x^2+28mx+x^2+14x+196 $$ |
| ③ | Multiply $ \color{blue}{6} $ by $ \left( mx+14\right) $ $$ \color{blue}{6} \cdot \left( mx+14\right) = 6mx+84 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 6mx+84 \right) = -6mx-84 $$ |
| ⑤ | Combine like terms: $$ m^2x^2+ \color{blue}{28mx} +x^2+14x+ \color{red}{196} \color{blue}{-6mx} \color{red}{-84} = m^2x^2+ \color{blue}{22mx} +x^2+14x+ \color{red}{112} $$ |
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