Question
Answer
$$(x^4-2)(x-4)^2(x^3-1)=x^9-8x^8+16x^7-x^6+6x^5-32x^3+2x^2-16x+32$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^4-2)(x-4)^2(x^3-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^4-2)(x^2-8x+16)(x^3-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^6-8x^5+16x^4-2x^2+16x-32)(x^3-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^9-8x^8+16x^7-x^6+6x^5-32x^3+2x^2-16x+32\end{aligned} $$ | |
| ① | Find $ \left(x-4\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(x-4\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 4 + \color{red}{4^2} = x^2-8x+16\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^4-2}\right) $ by each term in $ \left( x^2-8x+16\right) $. $$ \left( \color{blue}{x^4-2}\right) \cdot \left( x^2-8x+16\right) = x^6-8x^5+16x^4-2x^2+16x-32 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^6-8x^5+16x^4-2x^2+16x-32}\right) $ by each term in $ \left( x^3-1\right) $. $$ \left( \color{blue}{x^6-8x^5+16x^4-2x^2+16x-32}\right) \cdot \left( x^3-1\right) = \\ = x^9-x^6-8x^8+8x^5+16x^7 -\cancel{16x^4}-2x^5+2x^2+ \cancel{16x^4}-16x-32x^3+32 $$ |
| ④ | Combine like terms: $$ x^9-x^6-8x^8+ \color{blue}{8x^5} +16x^7 \, \color{red}{ -\cancel{16x^4}} \, \color{blue}{-2x^5} +2x^2+ \, \color{red}{ \cancel{16x^4}} \,-16x-32x^3+32 = x^9-8x^8+16x^7-x^6+ \color{blue}{6x^5} -32x^3+2x^2-16x+32 $$ |
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