Question
Answer
$$nz+mr((f-z)^2+y^2)-fm=f^2mr-2fmrz+mry^2+mrz^2-fm+nz$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}nz+mr((f-z)^2+y^2)-fm& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}nz+mr(1f^2-2fz+z^2+y^2)-fm \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}nz+f^2mr-2fmrz+mrz^2+mry^2-fm \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}f^2mr-2fmrz+mry^2+mrz^2+nz-fm \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}f^2mr-2fmrz+mry^2+mrz^2-fm+nz\end{aligned} $$ | |
| ① | Find $ \left(f-z\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ f } $ and $ B = \color{red}{ z }$. $$ \begin{aligned}\left(f-z\right)^2 = \color{blue}{f^2} -2 \cdot f \cdot z + \color{red}{z^2} = f^2-2fz+z^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{mr} $ by $ \left( f^2-2fz+z^2+y^2\right) $ $$ \color{blue}{mr} \cdot \left( f^2-2fz+z^2+y^2\right) = f^2mr-2fmrz+mrz^2+mry^2 $$ |
| ③ | Combine like terms: $$ nz+f^2mr-2fmrz+mrz^2+mry^2 = f^2mr-2fmrz+mry^2+mrz^2+nz $$ |
| ④ | Combine like terms: $$ f^2mr-2fmrz+mry^2+mrz^2-fm+nz = f^2mr-2fmrz+mry^2+mrz^2-fm+nz $$ |
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