Question
Answer
$$(4t^3-5)^2=16t^6-40t^3+25$$
Explanation
| $$ \begin{aligned}(4t^3-5)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16t^6-40t^3+25\end{aligned} $$ | |
| ① | Find $ \left(4t^3-5\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4t^3 } $ and $ B = \color{red}{ 5 }$. $$ \begin{aligned}\left(4t^3-5\right)^2 = \color{blue}{\left( 4t^3 \right)^2} -2 \cdot 4t^3 \cdot 5 + \color{red}{5^2} = 16t^6-40t^3+25\end{aligned} $$ |
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