Question
Answer
$$(x+5)(x^2-3x-7)=x^3+2x^2-22x-35$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+5)(x^2-3x-7)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^3-3x^2-7x+5x^2-15x-35 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3+2x^2-22x-35\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x+5}\right) $ by each term in $ \left( x^2-3x-7\right) $. $$ \left( \color{blue}{x+5}\right) \cdot \left( x^2-3x-7\right) = x^3-3x^2-7x+5x^2-15x-35 $$ |
| ② | Combine like terms: $$ x^3 \color{blue}{-3x^2} \color{red}{-7x} + \color{blue}{5x^2} \color{red}{-15x} -35 = x^3+ \color{blue}{2x^2} \color{red}{-22x} -35 $$ |
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