Question
Answer
$$(x+5)\cdot2x-(x-3)^2=x^2+16x-9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+5)\cdot2x-(x-3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x+5)\cdot2x-(x^2-6x+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(2x+10)x-(x^2-6x+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^2+10x-(x^2-6x+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2x^2+10x-x^2+6x-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^2+16x-9\end{aligned} $$ | |
| ① | Find $ \left(x-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x-3\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 3 + \color{red}{3^2} = x^2-6x+9\end{aligned} $$ |
| ② | $$ \left( \color{blue}{x+5}\right) \cdot 2 = 2x+10 $$ |
| ③ | $$ \left( \color{blue}{2x+10}\right) \cdot x = 2x^2+10x $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2-6x+9 \right) = -x^2+6x-9 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{2x^2} + \color{red}{10x} \color{blue}{-x^2} + \color{red}{6x} -9 = \color{blue}{x^2} + \color{red}{16x} -9 $$ |
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