Question
Answer
$$2(3x-1)^2(5x^2-x+1)=90x^4-78x^3+40x^2-14x+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2(3x-1)^2(5x^2-x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(9x^2-6x+1)(5x^2-x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(18x^2-12x+2)(5x^2-x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}90x^4-78x^3+40x^2-14x+2\end{aligned} $$ | |
| ① | Find $ \left(3x-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}{ 3x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(3x-1\right)^2 = \color{blue}{\left( 3x \right)^2} -2 \cdot 3x \cdot 1 + \color{red}{1^2} = 9x^2-6x+1\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2} $ by $ \left( 9x^2-6x+1\right) $ $$ \color{blue}{2} \cdot \left( 9x^2-6x+1\right) = 18x^2-12x+2 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{18x^2-12x+2}\right) $ by each term in $ \left( 5x^2-x+1\right) $. $$ \left( \color{blue}{18x^2-12x+2}\right) \cdot \left( 5x^2-x+1\right) = 90x^4-18x^3+18x^2-60x^3+12x^2-12x+10x^2-2x+2 $$ |
| ④ | Combine like terms: $$ 90x^4 \color{blue}{-18x^3} + \color{red}{18x^2} \color{blue}{-60x^3} + \color{green}{12x^2} \color{orange}{-12x} + \color{green}{10x^2} \color{orange}{-2x} +2 = \\ = 90x^4 \color{blue}{-78x^3} + \color{green}{40x^2} \color{orange}{-14x} +2 $$ |
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