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anonymous

  • 4 years ago

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

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  1. Mertsj
    • 4 years ago
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    First factor out the common factor which is x^4

  2. anonymous
    • 4 years ago
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    ok :)

  3. anonymous
    • 4 years ago
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    x^4 *(x^2 - 16) *(x^2 + 16) = x^4 *(x-4)(x+4)(x^2 + 16)

  4. anonymous
    • 4 years ago
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    Is this the answer: x^4(x^2+16)?

  5. Mertsj
    • 4 years ago
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    \[x ^{4}(x ^{4}-16) = x ^{4}(x ^{2}-4)(x ^{2}+4)=x ^{4}(x-2)(x+2)(x ^{2}+4)\]

  6. anonymous
    • 4 years ago
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    So its the last one?

  7. Mertsj
    • 4 years ago
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    x4(x−2)(x+2)(x2+4)

  8. anonymous
    • 4 years ago
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    Is a formula used to solve this?

  9. Mertsj
    • 4 years ago
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    The difference of two squares.

  10. anonymous
    • 4 years ago
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    how did you get (x^2+4)

  11. Mertsj
    • 4 years ago
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    Use it twice. Once on the x^4-16 and again on x^2-4

  12. Mertsj
    • 4 years ago
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    \[x ^{4}-16=(x ^{2}-4)(x ^{2}+4)\]

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

  14. anonymous
    • 4 years ago
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    x^4 - 256 = (x - 4)(x+4)(x^2 + 16)

  15. anonymous
    • 4 years ago
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    dont you take x^2 out of x^4 Mertsj

  16. Mertsj
    • 4 years ago
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    Add a factor of x^2+16)

  17. Mertsj
    • 4 years ago
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    \[x ^{4}-256=(x ^{2}-16)(x ^{2}+16)=\]

  18. anonymous
    • 4 years ago
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    so wouldnt the final answer not have x^4, what is the forumula exactly?

  19. Mertsj
    • 4 years ago
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    \[(x-4)(x+4)(x ^{2}+16)\]

  20. Mertsj
    • 4 years ago
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    Then put the x^4 that we factored out in the front of it.

  21. Mertsj
    • 4 years ago
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    \[x ^{4}(x-4)(x+4)(x ^{2}+16)\]

  22. Mertsj
    • 4 years ago
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    That's the final answer.

  23. anonymous
    • 4 years ago
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    cant you break down the 16 tho?

  24. Mertsj
    • 4 years ago
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    No. The sum of two squares is not factorable in the set of real numbers.

  25. Mertsj
    • 4 years ago
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    The difference of two squares is factorable but not the sum.

  26. anonymous
    • 4 years ago
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    Thanks want to help me with another? :)

  27. anonymous
    • 4 years ago
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    its set up different

  28. Mertsj
    • 4 years ago
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    Sure

  29. anonymous
    • 4 years ago
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    9x^2+64

  30. Mertsj
    • 4 years ago
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    That is the sum of two squares and it cannot be factored.

  31. anonymous
    • 4 years ago
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    So its prime?

  32. Mertsj
    • 4 years ago
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    yep

  33. anonymous
    • 4 years ago
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    ok, be right back :3

  34. anonymous
    • 4 years ago
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    im going to start a new question, this time i want to check an answer i got :)

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