Answer
$$3(x^2+3x+5)-(x-5)(x-3)=2x^2+17x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3(x^2+3x+5)-(x-5)(x-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^2+9x+15-(x^2-3x-5x+15) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^2+9x+15-(x^2-8x+15) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^2+9x+15-x^2+8x-15 \xlongequal{ } \\[1 em] & \xlongequal{ }3x^2+9x+ \cancel{15}-x^2+8x -\cancel{15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2x^2+17x\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{3} $ by $ \left( x^2+3x+5\right) $ $$ \color{blue}{3} \cdot \left( x^2+3x+5\right) = 3x^2+9x+15 $$ Multiply each term of $ \left( \color{blue}{x-5}\right) $ by each term in $ \left( x-3\right) $. $$ \left( \color{blue}{x-5}\right) \cdot \left( x-3\right) = x^2-3x-5x+15 $$ |
| ② | Combine like terms: $$ x^2 \color{blue}{-3x} \color{blue}{-5x} +15 = x^2 \color{blue}{-8x} +15 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2-8x+15 \right) = -x^2+8x-15 $$ |
| ④ | Combine like terms: $$ \color{blue}{3x^2} + \color{red}{9x} + \, \color{green}{ \cancel{15}} \, \color{blue}{-x^2} + \color{red}{8x} \, \color{green}{ -\cancel{15}} \, = \color{blue}{2x^2} + \color{red}{17x} $$ |
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