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first thing I do in these situations is check if you get lucky and have +/-1 be a solution...
Answer: x= -2 , x= 1 and x = 3
Did you try synthetic division?
No, I haven't tried synthetic division because I was unsure how to start it? Do I just pick number I think will work?
The factors of the constant term, which is 6, are +-1, +-2, +-3, +-6. Would you agree?
The coefficient of the x^3 term is 1. So the possible rational roots are the factors of the constant term divided by the factors of 1.
Yes that makes sense. Okay, so now I try all of those numbers?
\[\pm1, \pm2,\pm3, \pm 6\]
i usually start with the small ones.
Look at that. We stumbled onto one first thing. That means that one factor is (x-1) And the other factor is x^2-x-6 which can then be factored by the usual methods.
So we have: (x-1)(x^2-x-6)= (x-1)(x-3)(x+2)=0 So x=1, 3, or -2
I'm sorry, but could you explain how we got x^2-x-6? I understand how we found the first one but now I am rather lost
I will repost the synthetic drawing and then you will see.
It's the same as when you divide anything. Like 12 divided by 4 = 3 because 3 times 4 = 12
Okay thank you. I understand now. You are a life saver!
S= x^3-2x^2-5x+6 divided by x-1 = x^2-x-6 because (x-1)(x^2-x-6)=x^3-2x^2-5x+6
You are very welcome.