Question
Multiply polynomial $ \, \color{blue}{ x^5-x } \, $ by $ \, \color{orangered}{ x^8-x }$.
Answer
$$ \left( \color{blue}{x^5-x} \right) \cdot \left( \color{orangered}{x^8-x} \right) = x^{13}-x^9-x^6+x^2 $$
Explanation
In this example we are multiplying two binomials so FOIL method can be used.
$$ \begin{aligned} \left( \color{blue}{ x^5-x}\right) \cdot \left( \color{orangered}{ x^8-x}\right) &= \underbrace{ \color{blue}{x^5} \cdot \color{orangered}{x^8} }_{\text{FIRST}} + \underbrace{ \color{blue}{x^5} \cdot \left( \color{orangered}{-x} \right) }_{\text{OUTER}} + \underbrace{ \left( \color{blue}{-x} \right) \cdot \color{orangered}{x^8} }_{\text{INNER}} + \underbrace{ \left( \color{blue}{-x} \right) \cdot \left( \color{orangered}{-x} \right) }_{\text{LAST}} = \\ &= x^{13} + \left( -x^6\right) + \left( -x^9\right) + x^2 = \\ &= x^{13} + \left( -x^6\right) + \left( -x^9\right) + x^2 = \\ &= x^{13}-x^9-x^6+x^2; \end{aligned} $$This page was created using
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