What are the least, and most, amount of distinct zeroes of a 7th degree polynomial, given that at least one root is a complex number? 2. what is f(x) = x8 - 1 divided by x - 1? 3. what are the zeroes of f(x) = x3 + 5x2 - 7x + 1?

1. A polynomial has as many zeros as it has degree. A 7th-degree polynomial will have 7 zeros. They may all be distinct, or they may all be the same. Some may be complex. Since the degree is odd, the graph must cross the x-axis, so there must be at least one real zero.

The existence of one complex zero means there will also be another (its conjugate). Thus, the maximum number of distinct real zeros will be 7 - 2 = 5.

2. (x^8 -1)/(x -1) = x^7 +x^6 +x^5 +x^4 +x^3 +x^2 +x +1

(This is the fact that lets us write the sum of a geometric sequence in compact form.) You can get this result from synthetic division or long division as you may desire, or you can write it down based on your knowledge of geometric sequences.

3. A graphing calculator is helpful for finding the zeros of higher degree polynomials. The graph of f(x) shows that x=1 is one zero. Dividing that from f(x) gives the quotient (x+3)²-10, so we know the remaining zeros are -3±√10.

The zeros of f(x) are {-3-√10, -3+√10, 1}.

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Given a quadratic in vertex form y = a(x -h)² +k, we can easily solve for the zeros.

... 0 = a(x -h)² +k

... -k = a(x -h)² . . . . . . . subtract k

... -k/a = (x -h)² . . . . . . divide by a (it is 1 in this problem)

... ±√(-k/a) = x -h . . . . .take the square root

... h ±√(-k/a) = x . . . . . add h

After you have done this a couple of times, you can write the answer directly from the vertex form. We know the vertex form of the quotient of the division because we can read the coordinates of the vertex from the graph.

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