Question
Answer
$$(x+3)(3x^2+7x-5)=3x^3+16x^2+16x-15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3)(3x^2+7x-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^3+7x^2-5x+9x^2+21x-15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^3+16x^2+16x-15\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x+3}\right) $ by each term in $ \left( 3x^2+7x-5\right) $. $$ \left( \color{blue}{x+3}\right) \cdot \left( 3x^2+7x-5\right) = 3x^3+7x^2-5x+9x^2+21x-15 $$ |
| ② | Combine like terms: $$ 3x^3+ \color{blue}{7x^2} \color{red}{-5x} + \color{blue}{9x^2} + \color{red}{21x} -15 = 3x^3+ \color{blue}{16x^2} + \color{red}{16x} -15 $$ |
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