Question
Answer
$$b(x-3)(x+2)(x+5)=bx^3+4bx^2-11bx-30b$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}b(x-3)(x+2)(x+5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1bx-3b)(x+2)(x+5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1bx^2+2bx-3bx-6b)(x+5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1bx^2-bx-6b)(x+5) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}bx^3+5bx^2-bx^2-5bx-6bx-30b \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}bx^3+4bx^2-11bx-30b\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{b} $ by $ \left( x-3\right) $ $$ \color{blue}{b} \cdot \left( x-3\right) = bx-3b $$ |
| ② | Multiply each term of $ \left( \color{blue}{bx-3b}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{bx-3b}\right) \cdot \left( x+2\right) = bx^2+2bx-3bx-6b $$ |
| ③ | Combine like terms: $$ bx^2+ \color{blue}{2bx} \color{blue}{-3bx} -6b = bx^2 \color{blue}{-bx} -6b $$ |
| ④ | Multiply each term of $ \left( \color{blue}{bx^2-bx-6b}\right) $ by each term in $ \left( x+5\right) $. $$ \left( \color{blue}{bx^2-bx-6b}\right) \cdot \left( x+5\right) = bx^3+5bx^2-bx^2-5bx-6bx-30b $$ |
| ⑤ | Combine like terms: $$ bx^3+ \color{blue}{5bx^2} \color{blue}{-bx^2} \color{red}{-5bx} \color{red}{-6bx} -30b = bx^3+ \color{blue}{4bx^2} \color{red}{-11bx} -30b $$ |
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