Multiply polynomial $ \, \color{blue}{ t-2 } \, $ by $ \, \color{orangered}{ 2t+4 }$.

$$ \left( \color{blue}{t-2} \right) \cdot \left( \color{orangered}{2t+4} \right) = 2t^2-8 $$

In this example we are multiplying two binomials so FOIL method can be used.

$$ \begin{aligned} \left( \color{blue}{ t-2}\right) \cdot \left( \color{orangered}{ 2t+4}\right) &= \underbrace{ \color{blue}{t} \cdot \color{orangered}{2t} }_{\text{FIRST}} + \underbrace{ \color{blue}{t} \cdot \color{orangered}{4} }_{\text{OUTER}} + \underbrace{ \left( \color{blue}{-2} \right) \cdot \color{orangered}{2t} }_{\text{INNER}} + \underbrace{ \left( \color{blue}{-2} \right) \cdot \color{orangered}{4} }_{\text{LAST}} = \\ &= 2t^2 + 4t + \left( -4t\right) + \left( -8\right) = \\ &= 2t^2 + 4t + \left( -4t\right) + \left( -8\right) = \\ &= 2t^2-8; \end{aligned} $$

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