Question

How do I solve this binomial (factoring) : x^8-256x^4. Thank you :)

Answer 1

First factor out the common factor which is x^4

Answer 2

ok :)

Answer 3

x^4 *(x^2 - 16) *(x^2 + 16) = x^4 *(x-4)(x+4)(x^2 + 16)

Answer 4

Is this the answer: x^4(x^2+16)?

Answer 5

\[x ^{4}(x ^{4}-16) = x ^{4}(x ^{2}-4)(x ^{2}+4)=x ^{4}(x-2)(x+2)(x ^{2}+4)\]

Answer 6

So its the last one?

Answer 7

x4(x−2)(x+2)(x2+4)

Answer 8

Is a formula used to solve this?

Answer 9

The difference of two squares.

Answer 10

how did you get (x^2+4)

Answer 11

Use it twice. Once on the x^4-16 and again on x^2-4

Answer 12

\[x ^{4}-16=(x ^{2}-4)(x ^{2}+4)\]

Answer 13

@Mertsj .. it should be (x-4)(x+4) instead of (x-2)(x+2) .. you have to factor 256 which is 16^2

Answer 14

x^4 - 256 = (x - 4)(x+4)(x^2 + 16)

Answer 15

dont you take x^2 out of x^4 Mertsj

Answer 16

Add a factor of x^2+16)

Answer 17

\[x ^{4}-256=(x ^{2}-16)(x ^{2}+16)=\]

Answer 18

so wouldnt the final answer not have x^4, what is the forumula exactly?

Answer 19

\[(x-4)(x+4)(x ^{2}+16)\]

Answer 20

Then put the x^4 that we factored out in the front of it.

Answer 21

\[x ^{4}(x-4)(x+4)(x ^{2}+16)\]

Answer 22

That's the final answer.

Answer 23

cant you break down the 16 tho?

Answer 24

No. The sum of two squares is not factorable in the set of real numbers.

Answer 25

The difference of two squares is factorable but not the sum.

Answer 26

Thanks want to help me with another? :)

Answer 27

its set up different

Answer 28

Sure

Answer 29

9x^2+64

Answer 30

That is the sum of two squares and it cannot be factored.

Answer 31

So its prime?

Answer 32

yep

Answer 33

ok, be right back :3

Answer 34

im going to start a new question, this time i want to check an answer i got :)