Question

degree 3; zeros: 4,-5-i

Answer 1

The policy of openstudy is that we guide the asker not give out answers directly.

with zeros 4, -5 and -i

the roots would be (x-4) , (x+5) (x+i) . to write the function....

f(x)=(x-4)(x+5)(x+i)

Now you would have to expand this.

Answer 2

do you know what a conjugate is?

Answer 3

I definitely did something wrong -:(

Answer 4

ahemmm... well, I was referring to jgermain0427

Answer 5

Well, whatever, you go for it.

Answer 6

jdoe, I don't understand what a conjugate would do here though.

Answer 7

4,-5-i <----

it has a complex root, complex roots do not come all by their lonesome
they come with a "conjugate" companion

thus the 3 roots for the 3rd degree polynomial will be 4, -5-i and -5+i

Answer 8

Oh, I thought -5 and -i are roots and got confused b/c when you expend it, it would be amorphous.

I see it's -4, -x-5 and the other root is conjugate.

Answer 9

yeap, that's when you have a complex root, recall the quadratic formula, if say the discriminant is negative, then you'd end up with some "i" value \(\bf x=\cfrac{-b\pm \sqrt{-something}}{2a}\implies x=\cfrac{-b\pm i\ something}{2a}
\) so you'd get 2 values the +/-