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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

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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?

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.

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.)

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2 Answers

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