please help!!! how to find all zeros of a polynomial function!!!?

Question

i need detailed steps into how to fin all zeros. i dont understand how u find imaginary numbers. for example

f(x)=x^3-x^2+49x-49


the answer is 1+/- 7i but how did u get that? please help!!! and if u have your own equation please use it to show me!

Answer 1

use remainder theorem and you will know (x-1) is one of the factors
(x-1)(x^2+49)=0
roots are 1,+-7

Answer 2

First of all, there is no absolutely general method for finding all the zeros of a polynomial function of degree 5 or more. Second, the general method for finding the solution of an equation of degree 4 is so complicated almost nobody knows it. And the general solution for finding all the zeros of an equation of degree 3 is so complicated people usually look it up rather than memorize it. (It's called Cardano's Method.)

That said, you already know the general method for solving equations of degree 1 (trivial) and degree 2 (quadratic formula). You also know that if R is a root, then (x-R) divides f(x), and the quotient, g(x)/(x-R), is of lower degree -- and the roots of f are exactly the roots of g, plus R is added to the list.

So if you see a cubic or higher-degree polynomial, you're usually supposed to do one of a few things:

1. Find a root somehow, thereby reducing the problem to one with a polynomial of lower degree.

2. Notice that it is really a polynomial in x^2, x^3, or whatever, solve for that, and then take the appropriate root.

3. Notice that it fits a pattern like (ax +/- b)^n or x^n - y^n.

2&3 don't apply in this case, so let's try #1.

The first thing you should always do is check whether 0, 1, or -1 is a root. In this case, 1 is. Hooray. So divide f(x) by x-1, and you're left with g(x) = x^2 +49. Well, the roots of that are obvious -- 7i and -7i. So the three roots of f(x) are 1, 7i, and -7i.