Question
Answer
$$\dfrac{(a-2)(2a-3)^4}{(a-2)^3(2a-3)^2}=\dfrac{16a^5-128a^4+408a^3-648a^2+513a-162}{4a^5-36a^4+129a^3-230a^2+204a-72}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(a-2)(2a-3)^4}{(a-2)^3(2a-3)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{(a-2)(16a^4-96a^3+216a^2-216a+81)}{(1a^3-6a^2+12a-8)(4a^2-12a+9)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{16a^5-128a^4+408a^3-648a^2+513a-162}{4a^5-36a^4+129a^3-230a^2+204a-72}\end{aligned} $$ | |
| ① | $$ (2a-3)^4 = (2a-3)^2 \cdot (2a-3)^2 $$ |
| ② | Find $ \left(2a-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2a } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(2a-3\right)^2 = \color{blue}{\left( 2a \right)^2} -2 \cdot 2a \cdot 3 + \color{red}{3^2} = 4a^2-12a+9\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4a^2-12a+9}\right) $ by each term in $ \left( 4a^2-12a+9\right) $. $$ \left( \color{blue}{4a^2-12a+9}\right) \cdot \left( 4a^2-12a+9\right) = 16a^4-48a^3+36a^2-48a^3+144a^2-108a+36a^2-108a+81 $$ |
| ④ | Combine like terms: $$ 16a^4 \color{blue}{-48a^3} + \color{red}{36a^2} \color{blue}{-48a^3} + \color{green}{144a^2} \color{orange}{-108a} + \color{green}{36a^2} \color{orange}{-108a} +81 = \\ = 16a^4 \color{blue}{-96a^3} + \color{green}{216a^2} \color{orange}{-216a} +81 $$ |
| ⑤ | Find $ \left(a-2\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = a $ and $ B = 2 $. $$ \left(a-2\right)^3 = a^3-3 \cdot a^2 \cdot 2 + 3 \cdot a \cdot 2^2-2^3 = a^3-6a^2+12a-8 $$ |
| ⑥ | Find $ \left(2a-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2a } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(2a-3\right)^2 = \color{blue}{\left( 2a \right)^2} -2 \cdot 2a \cdot 3 + \color{red}{3^2} = 4a^2-12a+9\end{aligned} $$ |
| ⑦ | Multiply each term of $ \left( \color{blue}{a-2}\right) $ by each term in $ \left( 16a^4-96a^3+216a^2-216a+81\right) $. $$ \left( \color{blue}{a-2}\right) \cdot \left( 16a^4-96a^3+216a^2-216a+81\right) = \\ = 16a^5-96a^4+216a^3-216a^2+81a-32a^4+192a^3-432a^2+432a-162 $$ |
| ⑧ | Combine like terms: $$ 16a^5 \color{blue}{-96a^4} + \color{red}{216a^3} \color{green}{-216a^2} + \color{orange}{81a} \color{blue}{-32a^4} + \color{red}{192a^3} \color{green}{-432a^2} + \color{orange}{432a} -162 = \\ = 16a^5 \color{blue}{-128a^4} + \color{red}{408a^3} \color{green}{-648a^2} + \color{orange}{513a} -162 $$ |
| ⑨ | Multiply each term of $ \left( \color{blue}{a^3-6a^2+12a-8}\right) $ by each term in $ \left( 4a^2-12a+9\right) $. $$ \left( \color{blue}{a^3-6a^2+12a-8}\right) \cdot \left( 4a^2-12a+9\right) = \\ = 4a^5-12a^4+9a^3-24a^4+72a^3-54a^2+48a^3-144a^2+108a-32a^2+96a-72 $$ |
| ⑩ | Combine like terms: $$ 4a^5 \color{blue}{-12a^4} + \color{red}{9a^3} \color{blue}{-24a^4} + \color{green}{72a^3} \color{orange}{-54a^2} + \color{green}{48a^3} \color{blue}{-144a^2} + \color{red}{108a} \color{blue}{-32a^2} + \color{red}{96a} -72 = \\ = 4a^5 \color{blue}{-36a^4} + \color{green}{129a^3} \color{blue}{-230a^2} + \color{red}{204a} -72 $$ |
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