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

$$ \begin{aligned} \left( \color{blue}{ x+3}\right) \cdot \left( \color{orangered}{ 2x-1}\right) &= \underbrace{ \color{blue}{x} \cdot \color{orangered}{2x} }_{\text{FIRST}} + \underbrace{ \color{blue}{x} \cdot \left( \color{orangered}{-1} \right) }_{\text{OUTER}} + \underbrace{ \color{blue}{3} \cdot \color{orangered}{2x} }_{\text{INNER}} + \underbrace{ \color{blue}{3} \cdot \left( \color{orangered}{-1} \right) }_{\text{LAST}} = \\ &= 2x^2 + \left( -x\right) + 6x + \left( -3\right) = \\ &= 2x^2 + \left( -x\right) + 6x + \left( -3\right) = \\ &= 2x^2+5x-3; \end{aligned} $$

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