Question

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

Answer 1

You should definitely check my work, but here is how I went about it. First I plugged the negative of the constant term in just to see if I would get lucky. Since it is 7, I tried letting x = -7 and it turned out it was a zero (whew!)

Know that we know -7 is a zero, I divided it from the polynomial using synthetic division. The result I came up with was x^4 + 2x^2 + 1, which looks quadratic(ish). To solve it as a quadratic, I let u = x^2, so now I had to deal with u^2 + 2u + 1. From here, you can use whatever factoring method you feel comfortable with if your intuition didn't kick in to tell you it factored as (u + 1)^2. Since we started in x, lets now go back to it by back subbing x^2 for u, the result looking something like (x^2 + 1)^2. We can write it as (x^2 +1)(x^2 +1) if we want to, but since they are both the same you just need to solve for the zeros of one of them and then use the results twice. x^2 + 1 = 0 becomes x^2 = -1. Applying the square root property should give you x = +/- i.

After all that, it looks like the answer to your question is there is one real zero and four imaginary zeros. Seriously though, check my work.

Answer 2

Refer to the attached normal plot and a log plot in that order.