Question
Answer
$$\dfrac{y^2+6y+5}{-4y^2+24y+28}=\dfrac{y+5}{-4y+28}$$
Explanation
| $$ \begin{aligned}\frac{y^2+6y+5}{-4y^2+24y+28}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{y+5}{-4y+28}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{y^2+6y+5}{-4y^2+24y+28} $ to $ \dfrac{y+5}{-4y+28} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y+1}$. $$ \begin{aligned} \frac{y^2+6y+5}{-4y^2+24y+28} & =\frac{ \left( y+5 \right) \cdot \color{blue}{ \left( y+1 \right) }}{ \left( -4y+28 \right) \cdot \color{blue}{ \left( y+1 \right) }} = \\[1ex] &= \frac{y+5}{-4y+28} \end{aligned} $$ |
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