# How do you find the zeroes of F(x) = 1 + 3/(x^2 - 4)?

Jul 19, 2015

Reformulate and factorize to find: $F \left(x\right) = 0$ when $x = \pm 1$

#### Explanation:

$F \left(x\right) = 1 + \frac{3}{{x}^{2} - 4}$

$= \frac{{x}^{2} - 4}{{x}^{2} - 4} + \frac{3}{{x}^{2} - 4}$

$= \frac{{x}^{2} - 4 + 3}{{x}^{2} - 4}$

$= \frac{{x}^{2} - 1}{{x}^{2} - 4}$

$= \frac{{x}^{2} - {1}^{2}}{{x}^{2} - {2}^{2}}$

$= \frac{\left(x - 1\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 2\right)}$

using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So $F \left(x\right) = 0$ when $x = \pm 1$