Multiply polynomial $ \, \color{blue}{ x^2-5 } \, $ by $ \, \color{orangered}{ x+4 }$.

$$ \left( \color{blue}{x^2-5} \right) \cdot \left( \color{orangered}{x+4} \right) = x^3+4x^2-5x-20 $$

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

$$ \begin{aligned} \left( \color{blue}{ x^2-5}\right) \cdot \left( \color{orangered}{ x+4}\right) &= \underbrace{ \color{blue}{x^2} \cdot \color{orangered}{x} }_{\text{FIRST}} + \underbrace{ \color{blue}{x^2} \cdot \color{orangered}{4} }_{\text{OUTER}} + \underbrace{ \left( \color{blue}{-5} \right) \cdot \color{orangered}{x} }_{\text{INNER}} + \underbrace{ \left( \color{blue}{-5} \right) \cdot \color{orangered}{4} }_{\text{LAST}} = \\ &= x^3 + 4x^2 + \left( -5x\right) + \left( -20\right) = \\ &= x^3 + 4x^2 + \left( -5x\right) + \left( -20\right) = \\ &= x^3+4x^2-5x-20; \end{aligned} $$

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