How do you find the holes for (2x^3-5x^2-9x+18) / (x^2 - x - 6)?

Jul 28, 2016

There are holes at $x = - 2$ and $x = 3$

Explanation:

Factoring the denominator, we find:

${x}^{2} - x - 6 = \left(x + 2\right) \left(x - 3\right)$

So the denominator is zero when $x = - 2$ or $x = 3$

At any hole, both the numerator and denominator are zero. Note this condition is required but not sufficient.

Let $f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} - 9 x + 18$

Then $f \left(- 2\right) = - 16 - 20 + 18 + 18 = 0$.

So $x = - 2$ is a zero of the numerator and $\left(x + 2\right)$ a factor:

$2 {x}^{3} - 5 {x}^{2} - 9 x + 18$

$= \left(x + 2\right) \left(2 {x}^{2} - 9 x + 9\right)$

$= \left(x + 2\right) \left(x - 3\right) \left(2 x - 3\right)$

So we find:

$\frac{2 {x}^{3} - 5 {x}^{2} - 9 x + 18}{{x}^{2} - x - 6} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 3\right)}}} \left(2 x - 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 2\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 3\right)}}}} = 2 x - 3$

excluding $x = - 2$ and $x = 3$

The singularities at $x = - 2$ and $x = 3$ are holes, since the function is defined for any other Real value of $x$ and is continuous at these points.