Question

What are the zeros of the function f(x) = 3x^2-3x-6/3x-6?

a.–1
b. 1
c. 2
d. –2

Answer 1

Dear, can you please rewrite this with parenthesis to clearly indicate what you meant?

3x^2-3x-6/3x-6 \(\neq\) (3x^2-3x-6)/(3x-6)
\[3x^2-3x-6/3x-6 \neq \frac{3x^2-3x-6}{3x-6}\]

Then do the following:
1. Factor & simplify
2. Set equal to 0, you're looking for when y=0, when your output = 0

Answer 2

\[3x^2-3x-\frac{6}{3}x-6 = 3x^2-3x-2x-6 = 3x^2-5x-6\]

Answer 3

So the answer is..?

Answer 4

Which did you mean?

Answer 5

I think c. 2. but I`m not sure.

Answer 6

No I mean which of the two equations was the original intended problem. They are most certainly NOT the same thing.

Answer 7

It kinda matters, a lot

Answer 8

And I did ask please :-P

Answer 9

If what I think you intended and not what you literally wrote, factor out a 3 first (un-distribute a 3, if you will)

Answer 10

it`s \[3x^2-3x-6 \over 3x-6\]

Answer 11

Yep, see what happens when you get rid of the 3?

\[\frac{x^2−x−2}{x−2}\]

Think you can factor that numerator now? ;-)

Answer 12

\[\frac{(x-2)(x+1)}{(x-2)}\]

( any more and I'll be giving away the answer ;-) )

Make sense?

Answer 13

F.O.I.L. That numerator back if you don't believe me :)

Answer 14

So the answer is -1? (: