1 Answer | Add Yours
Whenever you are given a polynomial of the form:
`f(x) = a_nx^n + a_(n-1)x^(n-1)+...+a_1x + a_0`
it turns out the rational zeros are not that hard to find! If you were to extrapolate the possibilities, you end up finding that the only way rational roots can be extracted is if they are of the following form:
`r = +- p/q`
where `p` is an integer factor of `a_0` and `q` is an integer factor of `a_n`.
So, what we need to realize is that we can for now ignore every term except for the first and last!
Now, let's get our `a_n` and `a_0`.`` Remember, `a_0` is the constant term and `a_n` is the coefficient of the highest power of `x`. This gives us:
`a_n = 6`
`a_0 = 2`
Now, to get our p's and q's. Like we said above, the possible values of `p` are the integer factors of `a_0`, and because `a_0=2` we get the following values for `p`:
`p = 1, 2`
Similarly, because `q` is based on factors of `a_n = 6`, we get the following possibilities for `q`:
`q = 1,2,3,6`
So, to get our possible roots, we find every possible positive and negative combo of p's in the numerator and q's in the denominator:
possible roots = `+-1/1,+-1/2,+-1/3,+-1/6,+-2/1,+-2/2,+-2/3,+-2/6`
Notice now, that there are repeats (i.e. `+-1/1 = +-2/2, +-1/3 = +-2/6`)
So we can simplify to get our final set:
`+-1, +-1/2, +-1/3, +-1/6, +-2, +-2/3`
Based on the rational root theorem, there can be no other possible roots! I hope that helps!
We’ve answered 317,520 questions. We can answer yours, too.Ask a question