Answer
$$(3t^2-1.2t+1)(3t^2-1.2t+x)=9t^4-6t^3+3t^2x+4t^2-tx-t+x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3t^2-1.2t+1)(3t^2-1.2t+x)& \xlongequal{ }(3t^2-t+1)(3t^2-t+x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9t^4-6t^3+3t^2x+4t^2-tx-t+x\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3t^2-t+1}\right) $ by each term in $ \left( 3t^2-t+x\right) $. $$ \left( \color{blue}{3t^2-t+1}\right) \cdot \left( 3t^2-t+x\right) = 9t^4-3t^3+3t^2x-3t^3+t^2-tx+3t^2-t+x $$ |
| ② | Combine like terms: $$ 9t^4 \color{blue}{-3t^3} +3t^2x \color{blue}{-3t^3} + \color{red}{t^2} -tx+ \color{red}{3t^2} -t+x = 9t^4 \color{blue}{-6t^3} +3t^2x+ \color{red}{4t^2} -tx-t+x $$ |
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