## anonymous 4 years ago Solve: x^3-2x^2-5x+6=0

1. anonymous

2. TuringTest

first thing I do in these situations is check if you get lucky and have +/-1 be a solution...

3. anonymous

Answer: x= -2 , x= 1 and x = 3

4. Mertsj

Did you try synthetic division?

5. anonymous

No, I haven't tried synthetic division because I was unsure how to start it? Do I just pick number I think will work?

6. Mertsj

The factors of the constant term, which is 6, are +-1, +-2, +-3, +-6. Would you agree?

7. Mertsj

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.

8. anonymous

Yes that makes sense. Okay, so now I try all of those numbers?

9. Mertsj

$\pm1, \pm2,\pm3, \pm 6$

10. Mertsj

11. Mertsj

|dw:1327802691317:dw|

12. Mertsj

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.

13. Mertsj

So we have: (x-1)(x^2-x-6)= (x-1)(x-3)(x+2)=0 So x=1, 3, or -2

14. anonymous

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

15. Mertsj

I will repost the synthetic drawing and then you will see.

16. Mertsj

|dw:1327802958637:dw|

17. Mertsj

It's the same as when you divide anything. Like 12 divided by 4 = 3 because 3 times 4 = 12

18. anonymous

Okay thank you. I understand now. You are a life saver!

19. Mertsj

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

20. Mertsj

You are very welcome.