Question
Answer
$$-(x+c)(x+a)(a-c)=-a^2c-a^2x+ac^2-ax^2+c^2x+cx^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(x+c)(x+a)(a-c)& \xlongequal{ }-(x^2+ax+cx+ac)(a-c) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-(1a^2c+a^2x-ac^2+ax^2-c^2x-cx^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-a^2c-a^2x+ac^2-ax^2+c^2x+cx^2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2+ax+cx+ac}\right) $ by each term in $ \left( a-c\right) $. $$ \left( \color{blue}{x^2+ax+cx+ac}\right) \cdot \left( a-c\right) = \\ = ax^2-cx^2+a^2x -\cancel{acx}+ \cancel{acx}-c^2x+a^2c-ac^2 $$ |
| ② | Combine like terms: $$ ax^2-cx^2+a^2x \, \color{blue}{ -\cancel{acx}} \,+ \, \color{blue}{ \cancel{acx}} \,-c^2x+a^2c-ac^2 = a^2c+a^2x-ac^2+ax^2-c^2x-cx^2 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left(a^2c+a^2x-ac^2+ax^2-c^2x-cx^2 \right) = -a^2c-a^2x+ac^2-ax^2+c^2x+cx^2 $$ |
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