Question
Answer
$$(x^2+4x)(x^2+4x-1)-20=x^4+8x^3+15x^2-4x-20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2+4x)(x^2+4x-1)-20& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^4+4x^3-x^2+4x^3+16x^2-4x-20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^4+8x^3+15x^2-4x-20\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2+4x}\right) $ by each term in $ \left( x^2+4x-1\right) $. $$ \left( \color{blue}{x^2+4x}\right) \cdot \left( x^2+4x-1\right) = x^4+4x^3-x^2+4x^3+16x^2-4x $$ |
| ② | Combine like terms: $$ x^4+ \color{blue}{4x^3} \color{red}{-x^2} + \color{blue}{4x^3} + \color{red}{16x^2} -4x-20 = x^4+ \color{blue}{8x^3} + \color{red}{15x^2} -4x-20 $$ |
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