Question
Answer
$$(3-x)\cdot(3-x)\cdot(4-x)-8=-x^3+10x^2-33x+28$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-x)\cdot(3-x)\cdot(4-x)-8& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(9-3x-3x+x^2)\cdot(4-x)-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^2-6x+9)\cdot(4-x)-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^2-x^3-24x+6x^2+36-9x-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-x^3+10x^2-33x+28\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-x}\right) $ by each term in $ \left( 3-x\right) $. $$ \left( \color{blue}{3-x}\right) \cdot \left( 3-x\right) = 9-3x-3x+x^2 $$ |
| ② | Combine like terms: $$ 9 \color{blue}{-3x} \color{blue}{-3x} +x^2 = x^2 \color{blue}{-6x} +9 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^2-6x+9}\right) $ by each term in $ \left( 4-x\right) $. $$ \left( \color{blue}{x^2-6x+9}\right) \cdot \left( 4-x\right) = 4x^2-x^3-24x+6x^2+36-9x $$ |
| ④ | Combine like terms: $$ \color{blue}{4x^2} -x^3 \color{red}{-24x} + \color{blue}{6x^2} + \color{green}{36} \color{red}{-9x} \color{green}{-8} = -x^3+ \color{blue}{10x^2} \color{red}{-33x} + \color{green}{28} $$ |
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