1. 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

Question

1. 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?

Answer 1

Hi Ekene,

1. Two pieces of information. First, every polynomial of degree n has n roots total, but some of them may be complex. So the number of real + complex roots always equals n. (in this case n is 7)

Second, complex roots always come in pairs. If a polynomial has root (3 + i), it will also have root (3 - i).

Your question says you have a 7th-degree polynomial with a complex root. So it will have at least two complex roots. It could either have:

2 complex roots and 5 real roots,
4 complex roots and 3 real roots, or
6 complex roots and 1 real root.

I'm pretty sure your teacher is asking about real roots. If I'm right the answer is: at least one real root and at most five distinct real roots. (I could mention multiplicity of real roots but it's unnecessary.)

2. For this you need to do polynomial long division or synthetic division. Answer: x⁷+x⁶+x⁵+x⁴+x³+x²+x+1

3. Again you need polynomial long division. You need to guess the first root, which is 1.
Divide: you factor to (x - 1)(x2 + 6x - 1).
For the quadratic factor you need the formula. It will give you x = -3 ± √(10).
So the zeroes are: x = 1, x = (-3 + √(10)), x = (-3 - √(10)).