Question
Answer
$$\dfrac{-x^2+6x-5}{x^2+2x-3}=\dfrac{-x+5}{x+3}$$
Explanation
| $$ \begin{aligned}\frac{-x^2+6x-5}{x^2+2x-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-x+5}{x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{-x^2+6x-5}{x^2+2x-3} $ to $ \dfrac{-x+5}{x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{-x^2+6x-5}{x^2+2x-3} & =\frac{ \left( -x+5 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{-x+5}{x+3} \end{aligned} $$ |
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