Question
Answer
$$\dfrac{(x+h)^2-3-(x^2-3)}{h}=\dfrac{h^2+2hx}{h}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(x+h)^2-3-(x^2-3)}{h}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+2hx+h^2-3-(x^2-3)}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2+2hx+h^2-3-x^2+3}{h} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{h^2+2hx}{h}\end{aligned} $$ | |
| 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 | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2-3 \right) = -x^2+3 $$ |
| ③ | Simplify numerator $$ \, \color{blue}{ \cancel{x^2}} \,+2hx+h^2 \, \color{green}{ -\cancel{3}} \, \, \color{blue}{ -\cancel{x^2}} \,+ \, \color{green}{ \cancel{3}} \, = h^2+2hx $$ |
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