Question
Answer
$$(x+3)^3(2x-1)^2=4x^5+32x^4+73x^3+9x^2-81x+27$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3)^3(2x-1)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^3+9x^2+27x+27)(4x^2-4x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^5+32x^4+73x^3+9x^2-81x+27\end{aligned} $$ | |
| ① | Find $ \left(x+3\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = x $ and $ B = 3 $. $$ \left(x+3\right)^3 = x^3+3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2+3^3 = x^3+9x^2+27x+27 $$Find $ \left(2x-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2x-1\right)^2 = \color{blue}{\left( 2x \right)^2} -2 \cdot 2x \cdot 1 + \color{red}{1^2} = 4x^2-4x+1\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^3+9x^2+27x+27}\right) $ by each term in $ \left( 4x^2-4x+1\right) $. $$ \left( \color{blue}{x^3+9x^2+27x+27}\right) \cdot \left( 4x^2-4x+1\right) = \\ = 4x^5-4x^4+x^3+36x^4-36x^3+9x^2+108x^3 -\cancel{108x^2}+27x+ \cancel{108x^2}-108x+27 $$ |
| ③ | Combine like terms: $$ 4x^5 \color{blue}{-4x^4} + \color{red}{x^3} + \color{blue}{36x^4} \color{green}{-36x^3} + \color{orange}{9x^2} + \color{green}{108x^3} \, \color{blue}{ -\cancel{108x^2}} \,+ \color{green}{27x} + \, \color{blue}{ \cancel{108x^2}} \, \color{green}{-108x} +27 = \\ = 4x^5+ \color{blue}{32x^4} + \color{green}{73x^3} + \color{blue}{9x^2} \color{green}{-81x} +27 $$ |
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