Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5(x+h)^2+2(x+h)-(5x^2+2x)}{h}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5(x^2+2hx+h^2)+2(x+h)-(5x^2+2x)}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5x^2+10hx+5h^2+2x+2h-(5x^2+2x)}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5h^2+10hx+5x^2+2h+2x-(5x^2+2x)}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{5h^2+10hx+5x^2+2h+2x-5x^2-2x}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{5h^2+10hx+2h}{h}\end{aligned} $$ | |
| ① | Find $ \left(x+h\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}{ h }$. $$ \begin{aligned}\left(x+h\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot h + \color{red}{h^2} = x^2+2hx+h^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{5} $ by $ \left( x^2+2hx+h^2\right) $ $$ \color{blue}{5} \cdot \left( x^2+2hx+h^2\right) = 5x^2+10hx+5h^2 $$ |
| ③ | Multiply $ \color{blue}{2} $ by $ \left( x+h\right) $ $$ \color{blue}{2} \cdot \left( x+h\right) = 2x+2h $$ |
| ④ | Combine like terms: $$ 5x^2+10hx+5h^2+2x+2h = 5h^2+10hx+5x^2+2h+2x $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 5x^2+2x \right) = -5x^2-2x $$ |
| ⑥ | Simplify numerator $$ 5h^2+10hx+ \, \color{blue}{ \cancel{5x^2}} \,+2h+ \, \color{green}{ \cancel{2x}} \, \, \color{blue}{ -\cancel{5x^2}} \, \, \color{green}{ -\cancel{2x}} \, = 5h^2+10hx+2h $$ |