Question
Answer
$$(y-7)(y^2+y-4)=y^3-6y^2-11y+28$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(y-7)(y^2+y-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}y^3+y^2-4y-7y^2-7y+28 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}y^3-6y^2-11y+28\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{y-7}\right) $ by each term in $ \left( y^2+y-4\right) $. $$ \left( \color{blue}{y-7}\right) \cdot \left( y^2+y-4\right) = y^3+y^2-4y-7y^2-7y+28 $$ |
| ② | Combine like terms: $$ y^3+ \color{blue}{y^2} \color{red}{-4y} \color{blue}{-7y^2} \color{red}{-7y} +28 = y^3 \color{blue}{-6y^2} \color{red}{-11y} +28 $$ |
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