First, apply the Rational Roots Test to get a list of possible zeroes.
Then test these possible zeroes using synthetic division.
Once you've found one root (and thus one factor), you can apply the Quadratic Formula to what remains.
First, apply the Rational Roots Test to get a list of possible zeroes.
Then test these possible zeroes using synthetic division.
Once you've found one root (and thus one factor), you can apply the Quadratic Formula to what remains.
You can also use the following:
Vičte's formulas - Wikipedia, the free encyclopedia
There are all the other possible zeroes, as provided by the denominator (the factors of the 2). (When you study the lesson in the link provided earlier, you'll learn how correctly to apply the Rational Roots Tests, you'll see how to find all of the possible zeroes.)
But it turns out that your conclusion in this case is right: there are no rational zeroes for this particular polynomial. I hadn't checked before, since of course they "always" give you factorable stuff when they ask you to do factorizations.
At a guess, there may be a typo here. You might want to ask your instructor.
And from this formula it easily follows that - Wolfram|Alpha