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

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

Mathematics
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First factor out the common factor which is x^4
ok :)
x^4 *(x^2 - 16) *(x^2 + 16) = x^4 *(x-4)(x+4)(x^2 + 16)

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Other answers:

Is this the answer: x^4(x^2+16)?
\[x ^{4}(x ^{4}-16) = x ^{4}(x ^{2}-4)(x ^{2}+4)=x ^{4}(x-2)(x+2)(x ^{2}+4)\]
So its the last one?
x4(x−2)(x+2)(x2+4)
Is a formula used to solve this?
The difference of two squares.
how did you get (x^2+4)
Use it twice. Once on the x^4-16 and again on x^2-4
\[x ^{4}-16=(x ^{2}-4)(x ^{2}+4)\]
@Mertsj .. it should be (x-4)(x+4) instead of (x-2)(x+2) .. you have to factor 256 which is 16^2
x^4 - 256 = (x - 4)(x+4)(x^2 + 16)
dont you take x^2 out of x^4 Mertsj
Add a factor of x^2+16)
\[x ^{4}-256=(x ^{2}-16)(x ^{2}+16)=\]
so wouldnt the final answer not have x^4, what is the forumula exactly?
\[(x-4)(x+4)(x ^{2}+16)\]
Then put the x^4 that we factored out in the front of it.
\[x ^{4}(x-4)(x+4)(x ^{2}+16)\]
That's the final answer.
cant you break down the 16 tho?
No. The sum of two squares is not factorable in the set of real numbers.
The difference of two squares is factorable but not the sum.
Thanks want to help me with another? :)
its set up different
Sure
9x^2+64
That is the sum of two squares and it cannot be factored.
So its prime?
yep
ok, be right back :3
im going to start a new question, this time i want to check an answer i got :)

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