anonymous
  • anonymous
How do I solve this binomial (factoring) : x^8-256x^4. Thank you :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mertsj
  • Mertsj
First factor out the common factor which is x^4
anonymous
  • anonymous
ok :)
anonymous
  • anonymous
x^4 *(x^2 - 16) *(x^2 + 16) = x^4 *(x-4)(x+4)(x^2 + 16)

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More answers

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

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