Question
Answer
$$\dfrac{4x^4+x^3+28x^2+3x+47}{(x^2+4)^2}\cdot\dfrac{1}{x+1}=\dfrac{4x^4+x^3+28x^2+3x+47}{x^5+x^4+8x^3+8x^2+16x+16}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4x^4+x^3+28x^2+3x+47}{(x^2+4)^2}\cdot\frac{1}{x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4x^4+x^3+28x^2+3x+47}{x^4+8x^2+16}\cdot\frac{1}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4x^4+x^3+28x^2+3x+47}{x^5+x^4+8x^3+8x^2+16x+16}\end{aligned} $$ | |
| ① | Find $ \left(x^2+4\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^2 } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(x^2+4\right)^2 = \color{blue}{\left( x^2 \right)^2} +2 \cdot x^2 \cdot 4 + \color{red}{4^2} = x^4+8x^2+16\end{aligned} $$ |
| ② | Multiply $ \dfrac{4x^4+x^3+28x^2+3x+47}{x^4+8x^2+16} $ by $ \dfrac{1}{x+1} $ to get $ \dfrac{ 4x^4+x^3+28x^2+3x+47 }{ x^5+x^4+8x^3+8x^2+16x+16 } $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{4x^4+x^3+28x^2+3x+47}{x^4+8x^2+16} \cdot \frac{1}{x+1} & \xlongequal{\text{Step 1}} \frac{ \left( 4x^4+x^3+28x^2+3x+47 \right) \cdot 1 }{ \left( x^4+8x^2+16 \right) \cdot \left( x+1 \right) } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 4x^4+x^3+28x^2+3x+47 }{ x^5+x^4+8x^3+8x^2+16x+16 } \end{aligned} $$ |
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