Question
Answer
$$(x-3)((x-2)^2-3(x-2)+3)+(x-1)^2-3(x-1)+3=x^3-9x^2+29x-32$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-3)((x-2)^2-3(x-2)+3)+(x-1)^2-3(x-1)+3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-3)(x^2-4x+4-3(x-2)+3)+x^2-2x+1-3(x-1)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x-3)(x^2-4x+4-(3x-6)+3)+x^2-2x+1-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x-3)(x^2-4x+4-3x+6+3)+x^2-2x+1-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(x-3)(x^2-7x+13)+x^2-2x+1-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^3-7x^2+13x-3x^2+21x-39+x^2-2x+1-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}x^3-10x^2+34x-39+x^2-2x+1-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}x^3-9x^2+32x-38-(3x-3)+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}x^3-9x^2+32x-38-3x+3+3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}x^3-9x^2+29x-32\end{aligned} $$ | |
| ① | Find $ \left(x-2\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}{ 2 }$. $$ \begin{aligned}\left(x-2\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 2 + \color{red}{2^2} = x^2-4x+4\end{aligned} $$Find $ \left(x-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}{ x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$ |
| ② | Multiply $ \color{blue}{3} $ by $ \left( x-2\right) $ $$ \color{blue}{3} \cdot \left( x-2\right) = 3x-6 $$Multiply $ \color{blue}{3} $ by $ \left( x-1\right) $ $$ \color{blue}{3} \cdot \left( x-1\right) = 3x-3 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x-6 \right) = -3x+6 $$ |
| ④ | Combine like terms: $$ x^2 \color{blue}{-4x} + \color{red}{4} \color{blue}{-3x} + \color{green}{6} + \color{green}{3} = x^2 \color{blue}{-7x} + \color{green}{13} $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{x-3}\right) $ by each term in $ \left( x^2-7x+13\right) $. $$ \left( \color{blue}{x-3}\right) \cdot \left( x^2-7x+13\right) = x^3-7x^2+13x-3x^2+21x-39 $$ |
| ⑥ | Combine like terms: $$ x^3 \color{blue}{-7x^2} + \color{red}{13x} \color{blue}{-3x^2} + \color{red}{21x} -39 = x^3 \color{blue}{-10x^2} + \color{red}{34x} -39 $$ |
| ⑦ | Combine like terms: $$ x^3 \color{blue}{-10x^2} + \color{red}{34x} \color{green}{-39} + \color{blue}{x^2} \color{red}{-2x} + \color{green}{1} = x^3 \color{blue}{-9x^2} + \color{red}{32x} \color{green}{-38} $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x-3 \right) = -3x+3 $$ |
| ⑨ | Combine like terms: $$ x^3-9x^2+ \color{blue}{32x} \color{red}{-38} \color{blue}{-3x} + \color{green}{3} + \color{green}{3} = x^3-9x^2+ \color{blue}{29x} \color{green}{-32} $$ |
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