Question

State how many imaginary and real zeros the function has.
f(x) = x^5 + 7x^4 + 2x^3 + 14x^2 + x + 7

Answer 1

@Euler271

Answer 2

hard question. my bad i didn't notice the mention earlier.

we know it has to have 5 total (whether they be repeated or not)

there are only a few possibilities what could be happening here since 7 is a prime number.

but instead of guessing, i think the way to solve this is by dividing.

you'll want to divide by either (x-1), (x+1), (x-7), ) (x+7)

they won't all work. only one or two will give a new polynomial without a remainder.

you'll want to repeat this process until it's fully factorized.

probably not best method. idk if it'll even help.

Answer 3

Thanks anyways! :D