1 and -1 are possible roots of any polynomial. And, since raising 1 or -1 to powers is pretty easy, it is pretty easy to check if 1 or -1 are roots with some mental math. In this case, neither 1 nor -1 are roots of the function.
For other roots we must look to factoring the polynomial. For this polynomial the Greatest Common Factor is 1 (which we rarely bother factoring out). The polynomial does not fit any of the factoring patterns and there are too many terms for trinomial factoring. And since I don't see how factoring by grouping will work, we seem to be left with just factoring by trial and error of the possible rational roots.
The possible rational roots of a polynomial are all the ratios that can be formed, positive and negative, using a factor of the constant term (at the end) over a factor of the leading coefficient. The constant term of this polynomial is 18 and the factors of 18 are 1, 2, 3, 6, 9 and 18, The leading coefficient (in front of the ) is 1. So the possible rational roots of this polynomial are all the ratios/fractions, positive and negative, that can be formed using a factor of 18 as the numerator and a factor of 1 in the denominator:
1/1, 2/1, 3/1, 6/1, 9/1 and 18/1 and the negative of each of these
All these possible rational roots simplify to integers since the denominators are all 1's:
1, 2, 3, 6, 9 and 18 and the negative of each of these
So there are 12 possible rational roots. We have already checked 1 and -1 so we are down to 10 possible rational roots.
Now we start the trial and error part of the process. The "trial" part is to see if (x-r), where "r" is one of the possible rational roots, divides evenly into the polynomial. (If it divides evenly then it must be a factor!) So we try divisions until we find a factor. If you know synthetic division, then this is the easiest/fastest way to divide polynomials. Let's try the the possible rational root 2. So we want to divide the polynomial by x-2 to see if it divides evenly. With synthetic division it would look like this:
2 | 1 -2 -7 18 -18
2 0 -14 8
1 0 -7 4 -10
The -10 in the lower right corner is the remainder. Since it is not zero then x-2 did not divide evenly. Next we'll try 3:
3 | 1 -2 -7 18 -18
3 3 -12 18
1 1 -4 6 0
This time we got a zero remainder! So x-3 is a factor. And not only that, the rest of the bottom row tells us what the other factor is. The
1 1 -4 6
tells us that
1x^3 + 1x^2 + (-4)x + 6
is the other factor. So now we have:
We keep trying to factor. Only now we are looking to factor just
Its possible rational roots are 1, 2, 3 and 6 and the negatives of these. We have already found that 1, -1 and 2 are not roots of P(x). So they cannot be a root of a factor of P(x) either. So we are left with 3, 6 -2, -3 and -6. (Note: We already found that 3 was a root of P(x). But numbers can be roots multiple times. So 3 could be a root of , too.)
Skipping to a root that works:
-3 | 1 1 -4 6
-3 6 -6
1 -2 2 0
We've found another factor, (x-(-3)) or (x+3), and another root, -3. And the other factor is given to us by the 1 -2 2 in the rest of the bottom row. So now
The last factor is quadratic. So we don't have to keep trying the possible rational roots. We can use patterns or trinomial factoring and, if those don't work, the Quadratic Formula. The last factor will not factor further so we'll use the Quadratic Formula:
which simplifies as follows:
With the negative number inside the square root we now know that the roots will not be real. They will be complex. And since the problem asks only for real roots we will not bother to find out what these complex roots will be.
So we found only 2 real roots: 3 and -3
P.S. If you don't know synthetic division, then you'll have to use long division.
P.P.S. Rational functions are functions with variables in denominators. P(x) is not a rational function. It is just a polynomial with some rational roots.