Rational roots

Question

Given the equation x^3-2x^2+x-3=0, what are the possible rational roots?

 

I determined the possible roots were -1, 1, -3, and 3.  

However, when I tried synthetic substitution to check the roots, none of them worked.  I thought there should be 3 or 1 positive and 0 negative roots, but obviously, I did something incorrectly.  

 

Could you help me discover my mistake?

Answer 1

     The rational root theorem says that if there ARE any rational roots, they will be one of your 4 numbers. It DOES NOT guarantee that a rational root must exist!!!! 

For an easy example ( a good technique when faced with a problem that you're not sure if you're doing right)

x^2 = 5 has 2 answers, neither of which are rational.....namely + or -sqrt5

x^2 - 5 = 0 has the same answers. Using rational root theorem gives you possible rational roots of 5,-5,1,-1 none of which work. I'VE DONE NOTHING WRONG!!!! This equation has no rational roots.

     So, you've done nothing wrong . Out of your 4 numbers, none worked, so no rational roots. 

You are also confusing this theorem with DesCartes rule of signs which deals with REAL NUMBERS. Yes, your equation has one or three REAL answers, you've proven that they must be irrational.

Answer 2

John,

 

It would appear that this equation fails the rational roots test, i.e. there are no rational roots.