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anonymous

  • 4 years ago

What are the possible number of negative zeros of f(x) = –2x^7 + 2x^6 + 7x^5 + 7x^4 + 4x^3 + 4x^2 ? Answer: A)4, 2, or 0 B)2 or 0 C)3 or 1 D)7, 5, 3, or 1

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  1. Mertsj
    • 4 years ago
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    Find f(-x) first. Can you do that?

  2. Mertsj
    • 4 years ago
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    It is very difficult to help you if you won't answer me.

  3. anonymous
    • 4 years ago
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    sorry im trying to multi-task

  4. anonymous
    • 4 years ago
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    and idk how to find f(-x):/

  5. Mertsj
    • 4 years ago
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    replace every x with -x

  6. anonymous
    • 4 years ago
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    so you would subtract everything?

  7. anonymous
    • 4 years ago
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    2) f(x) = 2x^7 + 2x^6 + 7x^5 + 7x^4 - 4x^3 + 4x^2 f(-x) = -2x^7 + 2x^5 - 7x^5 + 7x^4 + 4x^3 + 4x^2 There are 3 changes of sign in f(-x), which means there can be possibly either 3 negative zeros or just 1 negative zero. So the answer would be B.

  8. anonymous
    • 4 years ago
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    Sorry, C

  9. Mertsj
    • 4 years ago
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    \[f(-x)=-2(-x)^7+2(-x)^6+7(-x)^5+7(-x)^4+4(-x)^3+4(-x)^2\]

  10. Mertsj
    • 4 years ago
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    Ok kdwndr wants to help you now. Good bye.

  11. anonymous
    • 4 years ago
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    Thank you:)

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