Question

Can you help me find my mistake? I'm factoring.

Answer 1

\[-2x^6=16\]

Answer 2

That's the equation, I'll show my work.

Answer 3

I subtracted 16 to the other side to get
-2x^6-16=0
then i factored the GCF
-2(x^6+8)=0
-2=0 makes 0 so that's one answer I got.

Then i commenced factoring by cubes and got
(x^2+2) (x^4-2x^2+4)

which ends up being the same answers and i ended up with the answers

x-0, \[\pm i \sqrt{2}\]

Answer 4

x=0 isn't one of the answers though so i did something wrong and don't know what.

Answer 5

Alright so you can use the Foil method or distribution method...Which do you prefer?

Answer 6

i have to factor these out

Answer 7

These are factoring polynomials

Answer 8

I know

Answer 9

What do you mean by foil or distribution?

Answer 10

Foil=First Outside Inside Last... meaning that you multiply the two things inside the parenthesis. (x^2 +2) (x^4-2x^2+4)

I am going to use distribution

x^2(x^4-2x^2+4)
Can you do this?

Answer 11

Yes, I know how to. But i was told to take the equations and make them equal to zero to find the x-ints

Answer 12

but i factored out the 2 which made the 16 equal to 8

Answer 13

I subtracted 16 to the other side to get
-2x^6-16=0
then i factored the GCF
-2(x^6+8)=0
-2=0 makes 0 so that's one answer I got.

Then i commenced factoring by cubes and got
(x^2+2) (x^4-2x^2+4)

which ends up being the same answers and i ended up with the answers

x-0,
±i2√

Answer 14

\[\pm i \sqrt{2}\]

Answer 15

when you get to
-2(x^6+8)=0
you divide both sides by -2 to get
x^6 + 8 = 0

now you can factor it.

(note: -2=0 is not true, and does not mean x=0 is a root)

Answer 16

oh because there is no x so -2=0 cannot happen

Answer 17

it just cancels out

Answer 18

in a way

Answer 19

Thank you, I know understand

Answer 20

x^4-2x^2+4=0

let y= x^2, and write this as
y^2 -2y + 4 = 0

the roots are y= \(1 \pm \sqrt{3} \)
and x = \( \pm \sqrt{1 \pm \sqrt{3} }\)

Answer 21

Thank you