**Determine a list of all possible rational zeroes**

Let's see how to come up along a list of possible rational zeroes for a polynomial.

**Example** Find a list of all possible rational zeroes for following polynomials.

P ( x ) = x^{4} - 7 x^{3} + 17 x^{2} -17 x = 6

**Solution**

if x =(b/c) is to be a zero of P ( x ) then b have to be a factor of 6 and c have to be a factor of 1. Also, as we illustrated in the previous example we can't forget negative factors.

Thus, the first thing to do is really to list all possible factors of 1 & 6. Following they are.

6 : ±1, ± 2, ± 3, ± 6

1: ±1

Now, to obtain a list of possible rational zeroes of the polynomial all we have to do is write down all possible fractions which we can compose from these numbers where the numerators have to be factors of 6 & the denominators have to be factors of 1. Actually this is easier than it might at first seem to be.

There is extremely simple shorthanded way of doing this. Let's go through the first one thoroughly then we'll do the rest earlier. Firstly, take the first factor from the numerator list, by including the ± , and divide this through the first factor (only factor in this case) from the denominator list, again involving the ± . It gives,

±1 /±1

It looks like a mess, however it isn't too bad. Here are four fractions. They are,

+1 / +1 =1 +1 / -1 = -1 -1/ + 1 = -1 -1 /- 1= -1

However Notice that the four fractions all reduce down to two possible numbers. It will always happen with these kinds of fractions. What we'll do from now is make the fraction, do any simplification of the numbers, avoiding the ± , and then drop one of the ± .

Thus, the list possible rational zeroes for this polynomial is,

±1 / ±1 = ±1 ±2 / ±2 = ±1 ±3 / ± 3 = ±1 ±6 / ±6 = ±1

Thus, it looks there are only eight possible rational zeroes & in this case they are all integers. Notice as well that any rational zeroes of this polynomial will be somewhere in this list, even though we haven't found them still.