Question

How do you find all the zeros of `f(x)= x^4+x^3+2x^2+4x-8` with its multiplicities?

Answer 1

First, we use the rational root theorem to try and pick out a possible root. I always try 1 first because it is easy to do: just count the sum of the coefficients. Luckily enough, it worked! `1 + 1 + 2 + 4 - 8`

So now we synthetically divide

x1 1 1 2 4 -8
1 2 4 8

result1 2 4 8 0

so now we have `x^3 + 2x^2 + 4x + 8` after taking out `x-1`

Applying guess and check work for the rational root theorem again, we find that `x = -2` works as well!

So we go back to synthetic division

x-2 1 2 4 8
-2 0 -8
Result 1 0 4 0

Now we have `x^2 + 4`

We use the quadratic formula to find that the other two roots are `2i and -2i`

So finally, our zeroes are `2i, -2i, 1, and -2`