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

1 Answer
Feb 26, 2016

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#