Answer
$$(t^2-16)(t-4)(t+4)=t^4-32t^2+256$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(t^2-16)(t-4)(t+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1t^3-4t^2-16t+64)(t+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}t^4-32t^2+256\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{t^2-16}\right) $ by each term in $ \left( t-4\right) $. $$ \left( \color{blue}{t^2-16}\right) \cdot \left( t-4\right) = t^3-4t^2-16t+64 $$ |
| ② | Multiply each term of $ \left( \color{blue}{t^3-4t^2-16t+64}\right) $ by each term in $ \left( t+4\right) $. $$ \left( \color{blue}{t^3-4t^2-16t+64}\right) \cdot \left( t+4\right) = \\ = t^4+ \cancel{4t^3} -\cancel{4t^3}-16t^2-16t^2 -\cancel{64t}+ \cancel{64t}+256 $$ |
| ③ | Combine like terms: $$ t^4+ \, \color{blue}{ \cancel{4t^3}} \, \, \color{blue}{ -\cancel{4t^3}} \, \color{green}{-16t^2} \color{green}{-16t^2} \, \color{orange}{ -\cancel{64t}} \,+ \, \color{orange}{ \cancel{64t}} \,+256 = t^4 \color{green}{-32t^2} +256 $$ |
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