List all of the possible rational zeros of p(x)=2x^3-6x^2+7x-6.



1 Answer | Add Yours

This answer has been flagged and removed.

1 reply Hide Replies

oldnick's profile pic

Posted on (Reply #1)

Ya'd to add Its the lonely rational solution for: `2x^2-2x+3 ` hasn't solutions. Indeed `Delta=4- 4 xx3xx2<=0`

embizze's profile pic

Posted on (Answer #1)

The rational root theorem states that a polynomial with integer coefficients in standard form with leading coefficient `a_n` and constant term `a_0` can only have rational roots of the form `p/q` where p divides the constant term (or is a factor of the constant term) and q divides the leading coefficient.

Given `p(x)=2x^3-6x^2+7x-6` we note that the coefficients are all integers and the polynomial is in standard form.

Then any possible rational root is of the form `p/q` where p is a factor of -6 and q is a factor of 2.


The possible rational roots are `+-1,+-2,+-3,+-6,+-1/2,+-3/2`

** Note that we left out `+-2/2` since this is really `+-1` and already in the list (also `+-6/2` for a similar reason).


We’ve answered 288,465 questions. We can answer yours, too.

Ask a question