How do you find all the complex zeros/roots of a polynomial?

Question

On my homework I have multiple questions asking for all complex zeros/roots. An example that I want answered is if f(x) = 5x^6-30x^5-43x^4+228x^3+22x^2+96x+16 and one of the zeros/roots is 3+sqrt(10) what are all the zeros/roots? Please give a step by step with tons of details because I have a test next week.

Answer 1

If 3+√10 is a root then so is 3-√10. If those are roots, then [x-(3+√10)] and [x-(3-√10)] are factors. These can be written as [(x-3)-√10][(x-3)+√10). If you multiply those factors together and simplify you will get a difference of squares which simplifies to x² -6x -1 . Use long polynomial division into the original function to get a quotient that is quadratic in x². If the long polynomial division has a remainder you made a mistake.

Another option is to try synthetic division with 3+√10 followed by 3-√10, but that will get really messy.

The quotient of the long division should be 5x⁴ - 38x² -16. Visualize letting U = x² and solve 5U² - 38U - 16 = 0. If you replace U with x² and solve the resulting equations you will have all 6 real and complex roots.

Answer 2

You are given one of the zeros as 3 +√10 which means that there is another zero that is 3-√10. Multiply (x-3-√10)(x-3+√10). This gives x² -6x -1. Divide this polynomial into the original polynomial to get another polynomial that contains the other four rots. In doing this you should get 5x⁴ -38x² -16. Use the quadratic equation to get the roots which gives the form x² =(38 ± √1444+320)/10 which = (38 ±42)/10. And so x² = -2/5,8. So the factors are (x² -6x -1)(x² -8)(5x² +2) since x = ±√8 and x=±i√2/5. This polynomial has 4 real roots and 2 complex roots.