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

$$ \begin{aligned} \left( \color{blue}{ 3x^2+4}\right) \cdot \left( \color{orangered}{ x^5-3}\right) &= \underbrace{ \color{blue}{3x^2} \cdot \color{orangered}{x^5} }_{\text{FIRST}} + \underbrace{ \color{blue}{3x^2} \cdot \left( \color{orangered}{-3} \right) }_{\text{OUTER}} + \underbrace{ \color{blue}{4} \cdot \color{orangered}{x^5} }_{\text{INNER}} + \underbrace{ \color{blue}{4} \cdot \left( \color{orangered}{-3} \right) }_{\text{LAST}} = \\ &= 3x^7 + \left( -9x^2\right) + 4x^5 + \left( -12\right) = \\ &= 3x^7 + \left( -9x^2\right) + 4x^5 + \left( -12\right) = \\ &= 3x^7+4x^5-9x^2-12; \end{aligned} $$

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