Question
Answer
$$-\dfrac{3}{5}(x-3)^2(x+3)=\dfrac{3x^3-9x^2-27x+81}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{3}{5}(x-3)^2(x+3)& \xlongequal{ }-\frac{3}{5}(x^2-6x+9)(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3x^2-18x+27}{5}(x+3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3x^3-9x^2-27x+81}{5}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{3}{5} $ by $ x^2-6x+9 $ to get $ \dfrac{ 3x^2-18x+27 }{ 5 } $. Step 1: Write $ x^2-6x+9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{3}{5} \cdot x^2-6x+9 & \xlongequal{\text{Step 1}} \frac{3}{5} \cdot \frac{x^2-6x+9}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 3 \cdot \left( x^2-6x+9 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x^2-18x+27 }{ 5 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{3x^2-18x+27}{5} $ by $ x+3 $ to get $ \dfrac{3x^3-9x^2-27x+81}{5} $. Step 1: Write $ x+3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{3x^2-18x+27}{5} \cdot x+3 & \xlongequal{\text{Step 1}} \frac{3x^2-18x+27}{5} \cdot \frac{x+3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 3x^2-18x+27 \right) \cdot \left( x+3 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x^3+9x^2-18x^2-54x+27x+81 }{ 5 } = \frac{3x^3-9x^2-27x+81}{5} \end{aligned} $$ |
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