Answer
$$(x-1)(x^2+3x-5)=x^3+2x^2-8x+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-1)(x^2+3x-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^3+3x^2-5x-x^2-3x+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3+2x^2-8x+5\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x^2+3x-5\right) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x^2+3x-5\right) = x^3+3x^2-5x-x^2-3x+5 $$ |
| ② | Combine like terms: $$ x^3+ \color{blue}{3x^2} \color{red}{-5x} \color{blue}{-x^2} \color{red}{-3x} +5 = x^3+ \color{blue}{2x^2} \color{red}{-8x} +5 $$ |
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