Answer
$$(2q-3)^4=16q^4-96q^3+216q^2-216q+81$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2q-3)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16q^4-96q^3+216q^2-216q+81\end{aligned} $$ | |
| ① | $$ (2q-3)^4 = (2q-3)^2 \cdot (2q-3)^2 $$ |
| ② | Find $ \left(2q-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}{ 2q } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(2q-3\right)^2 = \color{blue}{\left( 2q \right)^2} -2 \cdot 2q \cdot 3 + \color{red}{3^2} = 4q^2-12q+9\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4q^2-12q+9}\right) $ by each term in $ \left( 4q^2-12q+9\right) $. $$ \left( \color{blue}{4q^2-12q+9}\right) \cdot \left( 4q^2-12q+9\right) = 16q^4-48q^3+36q^2-48q^3+144q^2-108q+36q^2-108q+81 $$ |
| ④ | Combine like terms: $$ 16q^4 \color{blue}{-48q^3} + \color{red}{36q^2} \color{blue}{-48q^3} + \color{green}{144q^2} \color{orange}{-108q} + \color{green}{36q^2} \color{orange}{-108q} +81 = \\ = 16q^4 \color{blue}{-96q^3} + \color{green}{216q^2} \color{orange}{-216q} +81 $$ |
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