# How do you simplify (x^2-8x+12)/(x^2+7x-18) and then find the excluded values?

Apr 16, 2017

$= \frac{x - 6}{x + 9}$$\text{ }$ For all $x \in \left(- \infty , - 9\right) \cup \left(- 9 , 2\right) \cup \left(2 , + \infty\right)$

#### Explanation:

To simplify a fraction we think about factorizing the numerator and denominator .
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Factorizing the numerator:
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We think about the trial and error method that says:
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Given" "color(red)(x^2+Sx+P
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If we find two real numbers $\textcolor{b l u e}{a}$ and $\textcolor{b l u e}{b}$
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Such that $S = a + b \text{ "and " } P = a \times b$
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Then
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$\textcolor{red}{{x}^{2} - S x + P = \left(x + \textcolor{b l u e}{a}\right) \left(x + \textcolor{b l u e}{b}\right)}$
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${x}^{2} - 8 x + 12$
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$= \left(x - 2\right) \left(x - 6\right)$
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Factorizing the denominator:
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${x}^{2} + 7 x - 18$
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$= \left(x - 2\right) \left(x + 9\right)$
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From the above factorization of numerator and denominator we
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recognize the common factor $\left(x - 2\right)$
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$\frac{{x}^{2} - 8 x + 12}{{x}^{2} + 7 x - 18}$
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$= \frac{\left(x - 2\right) \left(x - 6\right)}{\left(x - 2\right) \left(x + 9\right)}$
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$x - 2 \ne 0 \Rightarrow x \ne 2$
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$x + 9 \ne 0 \Rightarrow x \ne - 9$
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Therefore,
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$\frac{{x}^{2} - 8 x + 12}{{x}^{2} + 7 x - 18}$
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$= \frac{x - 6}{x + 9}$$\text{ }$ For all $x \in \left(- \infty , - 9\right) \cup \left(- 9 , 2\right) \cup \left(2 , + \infty\right)$