Question
Answer
$$(3x-1)(3x^2+4x-5)=9x^3+9x^2-19x+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x-1)(3x^2+4x-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9x^3+12x^2-15x-3x^2-4x+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9x^3+9x^2-19x+5\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x-1}\right) $ by each term in $ \left( 3x^2+4x-5\right) $. $$ \left( \color{blue}{3x-1}\right) \cdot \left( 3x^2+4x-5\right) = 9x^3+12x^2-15x-3x^2-4x+5 $$ |
| ② | Combine like terms: $$ 9x^3+ \color{blue}{12x^2} \color{red}{-15x} \color{blue}{-3x^2} \color{red}{-4x} +5 = 9x^3+ \color{blue}{9x^2} \color{red}{-19x} +5 $$ |
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