Question
Answer
$$(x-1)^3(x^2+3x+4)-1=x^5-2x^3-4x^2+9x-5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-1)^3(x^2+3x+4)-1& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^3-3x^2+3x-1)(x^2+3x+4)-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^5-2x^3-4x^2+9x-4-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^5-2x^3-4x^2+9x-5\end{aligned} $$ | |
| ① | Find $ \left(x-1\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = x $ and $ B = 1 $. $$ \left(x-1\right)^3 = x^3-3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2-1^3 = x^3-3x^2+3x-1 $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^3-3x^2+3x-1}\right) $ by each term in $ \left( x^2+3x+4\right) $. $$ \left( \color{blue}{x^3-3x^2+3x-1}\right) \cdot \left( x^2+3x+4\right) = \\ = x^5+ \cancel{3x^4}+4x^3 -\cancel{3x^4}-9x^3-12x^2+3x^3+9x^2+12x-x^2-3x-4 $$ |
| ③ | Combine like terms: $$ x^5+ \, \color{blue}{ \cancel{3x^4}} \,+ \color{green}{4x^3} \, \color{blue}{ -\cancel{3x^4}} \, \color{orange}{-9x^3} \color{blue}{-12x^2} + \color{orange}{3x^3} + \color{red}{9x^2} + \color{green}{12x} \color{red}{-x^2} \color{green}{-3x} -4 = x^5 \color{orange}{-2x^3} \color{red}{-4x^2} + \color{green}{9x} -4 $$ |
| ④ | Combine like terms: $$ x^5-2x^3-4x^2+9x \color{blue}{-4} \color{blue}{-1} = x^5-2x^3-4x^2+9x \color{blue}{-5} $$ |
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