Multiply polynomial $ \, \color{blue}{ 4n^2-5+4 } \, $ by $ \, \color{orangered}{ 8n-1 }$.

$$ \left( \color{blue}{4n^2-5+4} \right) \cdot \left( \color{orangered}{8n-1} \right) = 32n^3-4n^2-8n+1 $$

Step 1: First we have to write polynomials in descending order.

$$ \begin{aligned} P(x) &= 4n^2-1 \\ Q(x) &= 8n-1 \\ \end{aligned} $$

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

$$ \begin{aligned} \left( \color{blue}{ 4n^2-1}\right) \cdot \left( \color{orangered}{ 8n-1}\right) &= \underbrace{ \color{blue}{4n^2} \cdot \color{orangered}{8n} }_{\text{FIRST}} + \underbrace{ \color{blue}{4n^2} \cdot \left( \color{orangered}{-1} \right) }_{\text{OUTER}} + \underbrace{ \left( \color{blue}{-1} \right) \cdot \color{orangered}{8n} }_{\text{INNER}} + \underbrace{ \left( \color{blue}{-1} \right) \cdot \left( \color{orangered}{-1} \right) }_{\text{LAST}} = \\ &= 32n^3 + \left( -4n^2\right) + \left( -8n\right) + 1 = \\ &= 32n^3 + \left( -4n^2\right) + \left( -8n\right) + 1 = \\ &= 32n^3-4n^2-8n+1; \end{aligned} $$

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