Question
Answer
$$-x^2\cdot(7-9x)(x^2+x+4)=9x^5+2x^4+29x^3-28x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-x^2\cdot(7-9x)(x^2+x+4)& \xlongequal{ }-(7x^2-9x^3)(x^2+x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(7x^4+7x^3+28x^2-9x^5-9x^4-36x^3) \xlongequal{ } \\[1 em] & \xlongequal{ }-(-9x^5-2x^4-29x^3+28x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9x^5+2x^4+29x^3-28x^2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{7x^2-9x^3}\right) $ by each term in $ \left( x^2+x+4\right) $. $$ \left( \color{blue}{7x^2-9x^3}\right) \cdot \left( x^2+x+4\right) = 7x^4+7x^3+28x^2-9x^5-9x^4-36x^3 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(-9x^5-2x^4-29x^3+28x^2 \right) = 9x^5+2x^4+29x^3-28x^2 $$ |
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