# How do you find the zeros for the function f(x)=(x^2-x-12)/(x-2)?

Apr 4, 2017

$f \left(x\right)$ has zeros at $- 3$ and $4$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - x - 12}{x - 2}$

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

$\therefore f \left(x\right) = 0 \to \frac{\left(x + 3\right) \left(x - 4\right)}{x - 2} = 0$

Hence the zeros of $f \left(x\right)$ occur when $\left(x + 3\right) \left(x - 4\right) = 0$
Note: $f \left(x\right)$ is undefined at $x = 2$

$\therefore f \left(x\right) = 0$ at $x = - 3$ and $x = 4$

This can be seen from the grapg of $f \left(x\right)$ below:

graph{((x+3)(x-4))/(x-2) [-18.01, 18.04, -9.01, 8.99]}

Apr 4, 2017

X= 4 and x= -3

#### Explanation:

You can find the zeros only when the nominator is equal to 0

So it's not necessary to care about the denominator in here

${x}^{2} - x - 12 = 0$

You can solve by factorisation, quadratic formula, or completing square

$\left(x - 4\right) \left(x + 3\right) = 0$

And you can solve it

You get x= 4 and x= -3