{x^2-5x+6}/{x-1}

{x^2-5x+6}/{x-1}

Hey Sup Forums I'm doing my calculus homework and I'm stuck on this part of the problem. I have to factor this, however, I can't figure it out. I need to try to apply the intermediate value theorem to this, however, I'm stuck on this part.

Also, Pokegirl thread.

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no complete equation (i suppose equal 0?)

formula != equation

that's just the same as (x-2)(x-3)/(x-1)
what exactly is the question?

sorry y only read the first line xD

Sorry, it should read:

{x^2-5x+6}/{x-1}=0

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For this equation, x=2 or 3

/thread

Fucking retard
If your IQ is as low as average nigger IQ, just stop bothering.

Wait, so are you asking us to find a factor of {x^2-5x+6}/{x-1}, a factor of {x^2-5x+6}, whether {x-1} is a factor, or what x=?

because if f(x)=0, x=2 and x=3 since you can remove {x-1} by mulitplying both sides by {x-1}, then you can factorise the quadratic to {x-2}{x-3} = 0. From here you can solve both of these separately by equating each factor to zero and solving for x.

As for the other two, they are slightly more difficult. {x-1} is not a factor and i think it will give {x-4} + 10/{x-1} (it's been a while so idk if that's right). Now I'm not sure if i can find a factor of {x^2-5x+6}/{x-1} as a whole but i don't feel like wasting time to find out.

As far as intermediate value theorum goes, i wouldn''t view that as a thing you can apply necessarily, but rather just a bit of common sense.

x = 7 or -1

this will be the photo

i dont undertand the meaning of the exercise xD

op can you post a pic of the question

Post the question OP

{x^2-5x+6}/{x-1}

A = {x^2-5x+6}

x^2 - 5x + 6 = 0 ~ a^2 + b + c = 0

a = 1
b = -5
c = 6

D = b^2 - 4ac
D = 25 - 4(1)(6)
D = 25-24 = 1

x1 = (-b + sqrt(D)) / 2a
x1 = 6/2 = 3

x2 = (-b - sqrt(D)) / 2a
x2 = 4/2 = 2

x^2 - 5x + 6 = (x-x1)(x-x2)

{x^2-5x+6}/{x-1} = (x-3)(x-2) / (x-1)

solutions : (x-3) = 0 , x = 3
(x-2) = 0 , x = 2
(x-1) is not cuz u cant divide by 0

try this you nigger
x^3-5x^2+6-x/x
just put the x at the end under the stuff before the /

nope I read the question wrong