Question

(x^2+8) factored

Answer 1

i got \[0\pm \sqrt{-32}\]------------
2

Answer 2

then i have\[\pm 2\iota \sqrt{2}\]

Answer 3

but the answer in the back just has +/- 2i

Answer 4

(x^2+8) factored is not the same question as (x^2+8)=0, find x.
I assume you are factoring so here we go.

Answer 5

(x^2+8) factored over the set of Complex numbers is the following:

( x + i *sqrt(8) ) * ( x - i * sqrt(8) =

( x + 2i sqrt(2) * ( x - 2i sqrt(2) )

Answer 6

you lost me on the 2nd part

Answer 7

This is what I think of as an extension of the difference of two squares pattern for Real numbers.

Consider the trinomial: x^2 + 1.

Over the set of real numbers, it does not factor.

Answer 8

Over the set of complex numbers, it does.

Answer 9

Ok

Answer 10

x^2 + 1 = (x + sqrt(-1) ) * ( x - sqrt(-1)) = (x + i) * (x - i).

Recall that i^2 = -1.

Answer 11

that sqrt is confusing

Answer 12

When you felt lost, was it at this step: i * sqrt(8) = 2i sqrt(2)
That comes from sqrt(8) = sqrt(2*2*2) = 2*sqrt(2).

Answer 13

Yup!

Answer 14

ok, hold on my brain just blew up.

Answer 15

Here's the original question. FInd all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.
f(x)=2x^4-x^3+7x^2-4x-4

Answer 16

Then I got
f(x)=(x-1)(x+1/2) (x^2+8)

Answer 17

But, you wrote this --> (x^2+8) factored as the question so that is what I worked on.

Answer 18

Yeah because that was the part I couldn't figure out how to factor

Answer 19

I put the +/-2i sqrt 2 as a zero, but the answer in the back is only +/-2i

Answer 20

2x^4 -x^3 +7x^2 -4x -4 =

(x - 1) (2x + 1) ( x^2 + 4) and then ...

Answer 21

(x - 1) (2x + 1) ( x^2 + 4) =

(x - 1) (2x + 1) (x + 2i) (x - 2i)

Answer 22

To get the roots, set that equal to zero.

(x - 1) (2x + 1) (x + 2i) (x - 2i) = 0

x = 1 or x = -1/2 or x = -2i or x = 2i

Answer 23

Oh no i see where I went wrong!!! AHHHHH I left out the [-->(2)<----x^2+8]

Answer 24

Sorry!!!

Answer 25

That is the way it goes with factoring. There's nothing to be sorry about. The good thing is that you recognized your error.

Answer 26

Thank you. I would've totally just gave up!

Answer 27

Never give up.
"Problems worthy of attack prove their worth by hitting back."
(Piet Hein)
And, this one hit you. Just smile and move on to the next "number battle."