# Where is the hole in this rational function f(x) = (x^2 + 2x - 8) / (x^2 - x - 2)?

Jun 5, 2015

Hole is a 'common' term for removable discontinuities for a rational function $f \left(x\right)$ which can be expressed as a quotient of two polynomial functions in the form of $f \left(x\right) = \frac{p \left(x\right)}{q \left(x\right)}$ . The following tutorial discusses the concept in detail.

Step I : We need to factorize the polynomials in the numerator and denominator.

Given $f \left(x\right) = \frac{{x}^{2} + 2 x - 8}{{x}^{2} - x - 2}$
$\implies f \left(x\right) = \frac{{x}^{2} + 4 x - 2 x - 8}{{x}^{2} + x - 2 x - 2}$
$\implies f \left(x\right) = \frac{x \left(x + 4\right) - 2 \left(x + 4\right)}{x \left(x + 1\right) - 2 \left(x + 1\right)}$
$\implies f \left(x\right) = \frac{\left(x - 2\right) \left(x + 4\right)}{\left(x - 2\right) \left(x + 1\right)}$

Step 2 : We need to identify the common factor with same multiplicity in numerator and denominator, elimination of which from both the numerator and denominator makes the function defined for that particular value of $x$.

In our present case, both numerator and denominator contains the factor $\left(x - 2\right)$ with a multiplicity of 1, elimination of which makes the function defined for $x - 2 = 0$ .

$\therefore x - 2 = 0$ is a removable discontinuity.

So, the hole of our function is $x = 2$.