when trying to determine rational roots for a polynomial using synthetic division, use the The Rational Roots Test
Sorry, I wasn't entirely sure how to title this thread.
I am working from a book that is showing how to use synthetic division to find factors of a polynomial in the form x-r
Lets look at two examples...
The book I am working from used synthetic division for r=-4 to 4.
In this problem, r=-4, -2, or 4 yield a remainder of zero and therefore, x+4, x+2, and x-4 are all factors of the polynomial
Since the original polynomial has a degree of 3, we know that there are exactly 3 complex roots. In this case, they are all real roots. (this is correct, right?)
My question is how does one know where to begin testing values of r given a particular polynomial expression. In other words, why not start testing r=-5 and increasing the value of r by increments of 1 until they find a number of roots equal to the degree of the polynomial.
The book also used synthetic division to find factors of the polynomial
This time, they also started at r=-4 and work their way up to 2. I understand why they stopped at 2 and did not go up to r=4 since the polynomial has a degree of 3 and since they found their third root at r=2, there was no need to continue up to r=4.
I suspect it has something to do with the factors of the constant. but for , -4 is not a factor of 6. So, if i'm on the right track here, why didn't they just start at -3 to save some time? As it turns out, for , factors are x+3, x-1, and x-2. I noticed that r= -4 and/or 4 were not factors of the polynomial. I also noticed that -4 and 4 are not factors of the constant 6.
So if a polynomial has a constant of 1 or -1, can i just test r= -1, 0, and 1 and save myself the trouble of testing for a wider range of r values?