Question
Answer
$$(2x+3)^2(x-1)=4x^3+8x^2-3x-9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x+3)^2(x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(4x^2+12x+9)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^3-4x^2+12x^2-12x+9x-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^3+8x^2-3x-9\end{aligned} $$ | |
| ① | Find $ \left(2x+3\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(2x+3\right)^2 = \color{blue}{\left( 2x \right)^2} +2 \cdot 2x \cdot 3 + \color{red}{3^2} = 4x^2+12x+9\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{4x^2+12x+9}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{4x^2+12x+9}\right) \cdot \left( x-1\right) = 4x^3-4x^2+12x^2-12x+9x-9 $$ |
| ③ | Combine like terms: $$ 4x^3 \color{blue}{-4x^2} + \color{blue}{12x^2} \color{red}{-12x} + \color{red}{9x} -9 = 4x^3+ \color{blue}{8x^2} \color{red}{-3x} -9 $$ |
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