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# How do you simplify (x^2 + 5x + 4 )/ (x^2 - 16)?

Mar 9, 2018

$\frac{x + 1}{x - 4} , \text{ ""if } x + 4 \ne 0$

#### Explanation:

Step 1: Factor both the numerator and denominator.

$\frac{{x}^{2} + 5 x + 4}{{x}^{2} - 16} = \frac{\left(x + 1\right) \left(x + 4\right)}{\left(x - 4\right) \left(x + 4\right)}$

Step 2: Cancel common factors.

$\frac{\left(x + 1\right) \cancel{\left(x + 4\right)}}{\left(x - 4\right) \cancel{\left(x + 4\right)}} = \frac{x + 1}{x - 4} , \text{ ""if } x + 4 \ne 0$

Why write this "if $x + 4 \ne 0$" part? Because in the original quotient, if $x + 4 = 0$ (that is, if x=–4), then the denominator would equal zero, and division by zero is undefined.

On its own, $\frac{x + 1}{x - 4}$ does not give the same problem when x=–4. Since we want our simplified expression to reflect the original one in its entirety, we make sure it has all the same restrictions, including $x + 4 \ne 0.$