Question
Answer
$$(x+\dfrac{1}{2})^4-5(x+\dfrac{1}{2})^3-10(x+\dfrac{1}{2})^2+80(x+\dfrac{1}{2})-96=(x+\dfrac{1}{2})^4-5(x+\dfrac{1}{2})^3-10(x+\dfrac{1}{2})^2+\dfrac{160x+80}{2}-96$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+\frac{1}{2})^4-5(x+\frac{1}{2})^3-10(x+\frac{1}{2})^2+80(x+\frac{1}{2})-96& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x+\frac{1}{2})^4-5(x+\frac{1}{2})^3-10(x+\frac{1}{2})^2+80\frac{2x+1}{2}-96 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x+\frac{1}{2})^4-5(x+\frac{1}{2})^3-10(x+\frac{1}{2})^2+\frac{160x+80}{2}-96\end{aligned} $$ | |
| ① | Add $x$ and $ \dfrac{1}{2} $ to get $ \dfrac{ \color{purple}{ 2x+1 } }{ 2 }$. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
| ② | Multiply $80$ by $ \dfrac{2x+1}{2} $ to get $ \dfrac{ 160x+80 }{ 2 } $. Step 1: Write $ 80 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 80 \cdot \frac{2x+1}{2} & \xlongequal{\text{Step 1}} \frac{80}{\color{red}{1}} \cdot \frac{2x+1}{2} \xlongequal{\text{Step 2}} \frac{ 80 \cdot \left( 2x+1 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 160x+80 }{ 2 } \end{aligned} $$ |
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