Question
Answer
$$(4x-5)^2-2(4x-5)+4=16x^2-48x+39$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4x-5)^2-2(4x-5)+4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16x^2-40x+25-2(4x-5)+4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}16x^2-40x+25-(8x-10)+4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}16x^2-40x+25-8x+10+4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16x^2-48x+39\end{aligned} $$ | |
| ① | Find $ \left(4x-5\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4x } $ and $ B = \color{red}{ 5 }$. $$ \begin{aligned}\left(4x-5\right)^2 = \color{blue}{\left( 4x \right)^2} -2 \cdot 4x \cdot 5 + \color{red}{5^2} = 16x^2-40x+25\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2} $ by $ \left( 4x-5\right) $ $$ \color{blue}{2} \cdot \left( 4x-5\right) = 8x-10 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 8x-10 \right) = -8x+10 $$ |
| ④ | Combine like terms: $$ 16x^2 \color{blue}{-40x} + \color{red}{25} \color{blue}{-8x} + \color{green}{10} + \color{green}{4} = 16x^2 \color{blue}{-48x} + \color{green}{39} $$ |
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