To find the possible rational zeros we will need the factors of the constant term and the leading coefficient. The constant term of P(x) is 8. (Actually it is -8 but since we will be trying both the positive and negative versions of the possible rational zeros it doesn't matter that we ignore the sign. The factors of 8 are 1, 2, 4 and 8. The leading coefficient is 3. The only factors of 3 are 1 and 3.
The possible rational zeros are all the ratios, positive and negative, which can be formed using a factor of the constant term in the numerator and a factor of the leading coefficient in the denominator. With the factors we found above, our list of possible rational zeros are:
+1/1, +2/1, +4/1, +8/1, +1/3, +2/3, +4/3, +8/3
The first four simplify:
+1, +2, +4, +8, +1/3, +2/3, +4/3, +8/3
Now we can go about finding which of these, if any, are actual zeros. We'll start with the integers because they're easier. And we will use synthetic division to test. Testing 1:
1 | 3 -5 -26 -8
---- 3 -2 -28
-------------------
3 -2 -28 -36
The remainder, in the lower right corner, is not zero. So 1 is not a zero of P(x). And since -36 is not very close to zero I am going to skip over 2 and jump to 4. (Note: I may end up coming back to 2 later.)
4 | 3 -5 -26 -8
---- 12 28 8
-------------------
3 7 2 0
Bingo! We have a winner! 4 is a zero (and (x-4) is a factor). Not only that, the rest of the bottom row tells us the other factor. The "3 7 2" translates into . This is a quadratic so we do not need to continue to try the rational zeros. This quadratic factors into (3x+1)(x+2). This gives us additional rational zeros of -1/3 and -2.
Altogether the rational zeros are 4, -1/3 and -2.
P.S. Instead of factoring the trinomial you could (and maybe your teacher wanted you to) continue to go through the list of possible rational zeros. If so, here are some tips on this process:- After finding one zero...
- Test for subsequent zeros using the other factor, not the original polynomial. In this case your next synthetic division would divide into "3 7 2"
- Consider trying the found zero again. Sometimes a number is a zero multiple times.
- Ask your teacher to explain Descartes' Rule of Signs (or look it up on the Internet). It can help save a lot of time. One part of that rule is: The number of positive zeros will be equal to the number of sign changes or less than that by an even number. Your P(x) has just one sign change: As we go from the first term to the second, the sign changes from positive to negative. After that there are no sign changes because all the remaining terms are negative. So according to this rule, there could only be 1 positive zero. (Less than 1 by an even number would be negative so there are no other possibilities.) This means that after we found 4, we would know that there are no other positive roots, including not another 4. So we know to start trying the negative rational zeros.
- Note the logic behind the fact that I chose to skip 2. The remainder for 1 was too far away from 0. The chances of a number near 1 (like 1/3 or 2) being a zero are small (but not impossible). Thinking like this can help you find the zeros more quickly.