These problems can take a significant amount of time so I am not going to do all three. I'm just going to do one and hope that from that example you will be able to do the others on your own. Note: A "root" and a "zero" are pretty mcuh the same thing. So consider them to be synonyms.
Finding zeros is a test of your factoring skills, especially for the polynomials you've posted. None of your polynomials have a greatest common factor (other than 1) and they have too many terms for any of the factoring patterns or for trinomial factoring. So we are left with just factoring by grouping and factoring by trial and error of the possible rational roots. And only the first one will factor by grouping.
So the only way to factor the last two polynomials is with the trial and error method which, as the name implies, can be time-consuming. I will do one of these so you can see this method of factoring. I will find the zeros of:
First we list the possible rational roots. The possible rational roots of a polynomial are all the possible ratios, positive and negative, that can be formed using a factor of the constant term over a factor of the leading coefficient. This polynomial's constant term is -1. Its factors are 1 and 1. (Note: we always include all positive and negative ratios so the sign of these numbers does not matter at this point.) Our leading coefficient is 3. Its factors are 1 and 3. So the possible rational roots of our polynomial are:
1/1 (or 1), -1/1 (or -1), 1/3 and -1/3.
We a lucky that there are only 4 possible rational roots. Sometimes there are quite a few.
Now we try these to see which, if any, are actually roots. While 1 and -1 can be checked mentally, I'm going to use synthetic division to test them so that you can see how that works. Checking it 1 is a root:
1 | 3 -7 5 -1 <== Note: these numbers came from
--- 3 -4 1
----------------
3 -4 1 0
The zero in the lower right-hand corner is the remainder. If it had not been a zero, then 1 would not be a root and we would have to try one of theother possible roots. But since the remainder is zero, it means that 1 is a root and that (x-1) is a factor of the polynomial. Not only that, the rest of the bottom line tells us what the other factor is. The "3 -4 1" translates into .
So at this point we have factored
into
We continue to factor. Since (x-1) is fully factored, we are only looking to factor . Since it is a quadratic, we can use patterns or trinomial factoring. (If neither of these methods work, use the Quadratic Formula to find the other roots.) This will factor using trinomial factoring:
With the polynomial fully factored we are ready to find the zeros/roots. The zeros are the value(s) of x that make a factor zero. So we solve each of the following:
x-1 = 0 or 3x-1 = 0 or x-1 = 0
Solving each of these we get:
x = 1 or x = 1/3 or x = 1.
These are the three zeros of your polynomial: 1, 1 and 1/3. (I know 1 is listed twice. But it counts twice as a root because (x-1) is a factor twice. It is called a "double root" or a "root of multiplicity 2".)
For the other problems:- Make a list of the possible rational roots.
- Use synthetic division to test the various possible roots. I guarantee that in these two problems at least one of the possible roots will be a root.
- Once you find a root, write the polynomial in factored form using the form:
(x - root)(bottom line from synthetic division without the zero remainder) - The second factor will be a quadratic in these two problems. Use other factoring techniques or the Quadratic Formula to find the remaining roots.