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If I synthetically divide a function by all the real zeros, and am left with 1, is that an x value of 1?

I am supposed to find all the zeros of a polynomial function

Trying to understand...  You did a polynomial division using the synthetic division process and got a remainder of 1...  So whatever factor you divided by (the divisor) therefore is NOT a factor of the original polynomial.  But we need more info to find any of the factors or the original polynomial... have any more to add?

1/2/2013

The question is asking to find all the zeros of a polynomial function with a degree of 4.  I found all the real ones on a graphing calculator, and synthetically divided the original function by each of them continuously.  After that, it left a 1.  I'm not sure if it should be considered a zero or not.

1/2/2013

So I didn't need to factor it at all, and the divisors are the factors.

1/2/2013

Eleanor, the shortest explanation for why 1 can't be another zero is the Fundamental Theorem of Algebra: an nth degree polynomial has at most n real roots. That means, in this case, that your degree 4 polynomial can't have more than 4 real roots --- so if you found 4 real roots on your graphing calculator, you can't possibly get another. The reason you get 1 when you divide through by all the roots is the same reason you get 1 when you divide through by all the prime factors of an integer (like 6): once you "factor out all the prime factors," all you're left with is 1. (The prime factors of 6 are 2 and 3, and when you divide 6 by 2 and then by 3 you get 1: 6 divided by 2 is 3, and 3 divided by 3, in turn, is 1. This is exactly analogous to what's going on with polynomial division, except the "prime factors" aren't numbers anymore; they're polynomials themselves.)

1/3/2013

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

f(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

BruceS