Question

x^2-5x+6/x-2 identify the holes

Answer 1

Now, the first thing you need to do is factor the numerator. Just to see stuff more clearly...

Answer 2

I mean, clearly, the denominator can no longer be factored...

Answer 3

could you like step by step show me how to solve this please

Answer 4

I'm not really good at teaching how to factor... suffice it to say, the numerator factored is
(x - 2)(x - 3)

Answer 5

\[\Large =\frac{(x-2)(x-3)}{x-2}\]
right?

Answer 6

yeah

Answer 7

Now, a 'hole' is a point where BOTH the numerator and denominator are zero.
So... for what value(s) of x is the *denominator* equal to zero?

Answer 8

im not sure is it?

Answer 9

What is the denominator?

Answer 10

x-2

Answer 11

And when is this equal to zero?
when x = ...?

Answer 12

-2

Answer 13

When x = -2
then...
x - 2
becomes
-2 - 2
= -4
which isn't zero :P

...try again

Answer 14

-3 as one of them

Answer 15

No...
lol
For
x - 2 = 0
Just solve for x...

Answer 16

idk

Answer 17

2

Answer 18

Yes, 2.
So the denominator becomes zero when x = 2
Now what about the numerator?
When is the numerator (x-2)(x-3) = 0?

Answer 19

2,3

Answer 20

That's right.
So, numerator is zero when x is 2 or 3
denominator is zero when x is 2...

so.... when are they BOTH zero?

Answer 21

when x is 2 or 3?

Answer 22

No... because when x = 3, the denominator is not zero :P

Answer 23

=/

Answer 24

2,3 is wrong

Answer 25

Yup. I told you, holes would be the values of x that would make BOTH the numerator and denominator zero.

Answer 26

ok so what do i have to do

Answer 27

Think of it this way...
the set of values for x that would make the numerator zero is {2,3} while the set of x-values that make the denominator zero is {2}

The elements they have in common, that's the answer.