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

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 ${x}^{3} + 2 {x}^{2} + 4 x + 8$ 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 $2 i \mathmr{and} - 2 i$

So finally, our zeroes are $2 i , - 2 i , 1 , \mathmr{and} - 2$