Tap the blue circles to see an explanation.
| $$ \begin{aligned}4(x-3)(x-1)^2(x+1)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4(x-3)(x^2-2x+1)(x^3+3x^2+3x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(4x-12)(x^2-2x+1)(x^3+3x^2+3x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(4x^3-8x^2+4x-12x^2+24x-12)(x^3+3x^2+3x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(4x^3-20x^2+28x-12)(x^3+3x^2+3x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}4x^6-8x^5-20x^4+16x^3+28x^2-8x-12\end{aligned} $$ | |
| ① | 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 }$. $$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$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 $ \color{blue}{4} $ by $ \left( x-3\right) $ $$ \color{blue}{4} \cdot \left( x-3\right) = 4x-12 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4x-12}\right) $ by each term in $ \left( x^2-2x+1\right) $. $$ \left( \color{blue}{4x-12}\right) \cdot \left( x^2-2x+1\right) = 4x^3-8x^2+4x-12x^2+24x-12 $$ |
| ④ | Combine like terms: $$ 4x^3 \color{blue}{-8x^2} + \color{red}{4x} \color{blue}{-12x^2} + \color{red}{24x} -12 = 4x^3 \color{blue}{-20x^2} + \color{red}{28x} -12 $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{4x^3-20x^2+28x-12}\right) $ by each term in $ \left( x^3+3x^2+3x+1\right) $. $$ \left( \color{blue}{4x^3-20x^2+28x-12}\right) \cdot \left( x^3+3x^2+3x+1\right) = \\ = 4x^6+12x^5+12x^4+4x^3-20x^5-60x^4-60x^3-20x^2+28x^4+84x^3+84x^2+28x-12x^3-36x^2-36x-12 $$ |
| ⑥ | Combine like terms: $$ 4x^6+ \color{blue}{12x^5} + \color{red}{12x^4} + \color{green}{4x^3} \color{blue}{-20x^5} \color{orange}{-60x^4} \color{blue}{-60x^3} \color{red}{-20x^2} + \color{orange}{28x^4} + \color{green}{84x^3} + \color{orange}{84x^2} + \color{blue}{28x} \color{green}{-12x^3} \color{orange}{-36x^2} \color{blue}{-36x} -12 = \\ = 4x^6 \color{blue}{-8x^5} \color{orange}{-20x^4} + \color{green}{16x^3} + \color{orange}{28x^2} \color{blue}{-8x} -12 $$ |