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

$$ \begin{aligned} \left( \color{blue}{ 2x^2-3x}\right) \cdot \left( \color{orangered}{ x-7}\right) &= \underbrace{ \color{blue}{2x^2} \cdot \color{orangered}{x} }_{\text{FIRST}} + \underbrace{ \color{blue}{2x^2} \cdot \left( \color{orangered}{-7} \right) }_{\text{OUTER}} + \underbrace{ \left( \color{blue}{-3x} \right) \cdot \color{orangered}{x} }_{\text{INNER}} + \underbrace{ \left( \color{blue}{-3x} \right) \cdot \left( \color{orangered}{-7} \right) }_{\text{LAST}} = \\ &= 2x^3 + \left( -14x^2\right) + \left( -3x^2\right) + 21x = \\ &= 2x^3 + \left( -14x^2\right) + \left( -3x^2\right) + 21x = \\ &= 2x^3-17x^2+21x; \end{aligned} $$

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