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

2 Answers
Apr 4, 2017

f(x) has zeros at -3 and 4

Explanation:

f(x) =(x^2-x-12)/(x-2)

= ((x+3)(x-4))/(x-2)

:. f(x) = 0 -> ((x+3)(x-4))/(x-2) =0

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

:.f(x) = 0 at x=-3 and x=4

This can be seen from the grapg of f(x) 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

(x-4)(x+3) = 0

And you can solve it

You get x= 4 and x= -3