I am supposed to find all the zeros of a polynomial function
If I synthetically divide a function by all the real zeros, and am left with 1, is that an x value of 1?
Nope! If I understand your question correctly, being "left with 1" means a factor of 1, not a remainder of 1 and not the expression "x−1". For example:
Maybe you needed to find all the zeros of x4−15x2−10x+24. You determined that there was a real zero at x=4, so you divided the original polynomial by x−4 using synthetic division to obtain the cubic x3+4x2+x−6. Then you found another zero at x=−2, so you divided that cubic by x+2 to obtain x2+2x−3. Next you divided this quadratic by x−1 because that quadratic has a zero at x=1. This division gave you the linear expression x+3 which has a zero at x=−3, and the final synthetic division of x+3 by x+3 gives you 1.
It's true that (1)(x+3)(x-1)(x+2)(x-4) = x4−15x2−10x+24.
But it's also true that (1)(1)(1)(1)(x+3)(x-1)(x+2)(x-4) = x4−15x2−10x+24. And it's a bit silly.
That's because 1 is the mutiplicative identity. When the only factor left is one, your fun is done...assuming that you think solving a quartic is fun...which would make you a very unusual person. I prefer to solve quartics using Wolfram Alpha. Try http://www.wolframalpha.com/input/?i=Solve+x^4%E2%88%9215x^2%E2%88%9210x%2B24%3D0 .
Anyway, if that's what happened then you're in for a treat because, coincidentally, the great mathematics educator Vi Hart came out with a video on this exact topic! See http://www.youtube.com/watch?v=GFLkou8NvJo
The Remainder Theorem provides the answer:
The Division Algorithm states:
f(x) = (x-c) * q(x) + r
q is the quotient of the synthetic division
r is the remainder of the synthetic division
Set x=c to get a value for r:
f(c) = (c-c)*q(c) + r
f(c) = ( 0 )*q(c) + r
This tells you that the remainder, r, is the value of f(x) at x=c
If r = 1, then c is not a root of the function f.
If r = 0, then c is a root of the function f, i.e. f(c)=0