Question

How do I solve x^3-4x?

Answer 1

you want to factor?

Answer 2

Yes, its equal to 0

Answer 3

x^3-4x = 0

x(x^2 - 4) = 0 ... factor out x

x(x-2)(x+2) = 0 ... use the difference of squares rule

Answer 4

You can then break up x(x-2)(x+2) = 0 into

x = 0 or x-2 = 0 or x+2 = 0

basically you set each factor equal to zero

Answer 5

now solve these 3 equations: x = 0 or x-2 = 0 or x+2 = 0

well one is already solved for you

Answer 6

Thank you so much, I didn't see that.

Answer 7

can you help me with one more please?

Answer 8

sure

Answer 9

okay, x^4-1

Answer 10

x^4 - 1 = x^4 - 1^2

use the difference of squares rule to get ???

Answer 11

huh?

Answer 12

That one is equal to zero also.

Answer 13

ok try thinking of it like this (x^2)^2 - 1^2

if you factored that, what would you get?

Answer 14

why is it 1^2?

Answer 15

because 1 = 1^2

Answer 16

I'm formatting the given expression as a difference of squares

Answer 17

so you can factor using that rule

Answer 18

so how do I solve it? I'm sorry, I'm just really confused.

Answer 19

say you had x^2 - 1

how would you factor that?

Answer 20

(x-1)(x+1) ? right?

Answer 21

so (x^2)^2 - 1 factors to what?

Answer 22

I don't know.

Answer 23

let z = x^2

Answer 24

z^2 - 1 factors to what?

Answer 25

x^4 - 1

Answer 26

so what does that haveto do with solving it?

Answer 27

I'm helping you factor x^4 - 1

Answer 28

z^2 - 1 = (z+1)(z-1)

replace every z with (x^2)

z^2 - 1 = (z+1)(z-1)

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

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

now factor x^2-1 to get (x-1)(x+1)

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