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anonymous

  • 5 years ago

For each polynomial function, one zero is given. Find all others.. f(x)=4x^3+6x^2-2x-1; 1/2(is given)

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  1. anonymous
    • 5 years ago
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    I need to go to bed, so I'll tell you what you need to do. If you need clarification on how to do anything, I'm sure you can look up what's needed. You're given x=1/2 is a root. This means that (x-(1/2)) is a factor of your equation. Consequently, you can use polynomial division to divide the polynomial by (x-(1/2)) to end up with a quadratic, Q(x), so that\[f(x)=(x-(1/2))Q(x)\]All you need to do then is find the factors of that quadratic, Q(x) and you'll be done. If you do the polynomial division correctly, you should have no remainder.

  2. anonymous
    • 5 years ago
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    Thanks for your help...but I have no idea what your talking about

  3. anonymous
    • 5 years ago
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    Our teacher is teaching syntheic divison and when I did that I came up with 4x^2+8x+2. Im confused because I thought that could be factored

  4. anonymous
    • 5 years ago
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    I can't really explain polynomial division on this site. You can find stuff on YouTube or something...ah, synthetic division...similar thing

  5. anonymous
    • 5 years ago
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    4x^2+8x+2 is what I got too

  6. anonymous
    • 5 years ago
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    I thought the next step would be to factor... to find the answers. I get that (x-1/2) is one answer

  7. anonymous
    • 5 years ago
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    It can be factored. Find the roots of the equation, alpha and beta, say, and then you have \[(x-\alpha)(x-\beta)=4x^2+8x+2\]

  8. anonymous
    • 5 years ago
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    I dont know what that is??

  9. anonymous
    • 5 years ago
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    The discriminant is 32, which is positive, but not a perfect square, so you will get two distinct, real, irrational roots.

  10. anonymous
    • 5 years ago
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    You don't know the quadratic formula?

  11. anonymous
    • 5 years ago
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    Oh, so I use the quadratic instead of factoring??

  12. anonymous
    • 5 years ago
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    Well, you'll still be factoring...I'll show you.

  13. anonymous
    • 5 years ago
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    \[x=\frac{-8{\pm}\sqrt{8^2-4 \times 4 \times 2}}{2 \times 4}=\frac{-2{\pm}\sqrt{2}}{2}\]

  14. anonymous
    • 5 years ago
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    Now, since these x-values give zeros, we must have that\[(x-(\frac{-2+\sqrt{2}}{2}))(x-(\frac{-2-\sqrt{2}}{2}))=0\]

  15. anonymous
    • 5 years ago
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    Those factors on the left-hand side will be your factors. The total thing will then be the left-hand side of the above times (x-1/2).

  16. anonymous
    • 5 years ago
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    I feel like Im wasting your time. I am suppose to have three answers. (x-1/2) is one of the answers and I understand why we did the quadratic. But I still dont know what the other two answers are. Thanks for attempting to help me!!

  17. anonymous
    • 5 years ago
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    The other two answers are what I just gave you. Your polynomial is then \[(x-1/2)(x-(\frac{-2+\sqrt{2}}{2}))(x-(\frac{-2-\sqrt{2}}{2}))\]

  18. anonymous
    • 5 years ago
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    You have your three factors. It's ugly, but that's it.

  19. anonymous
    • 5 years ago
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    Thanks.. I get scared when things are ugly... I tend to think its wrong.

  20. anonymous
    • 5 years ago
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    Wait...I forgot to mention, when I use this method, I expand to check. What you get in this instance is \[\frac{1}{4}(4x^3+6x^2-2x-1)\] which is okay if all we care about are solving for those x's that yield zero, but if you want to get back to the original polynomial, you'll have to multiply the three factors through by 4. Since 4 is 2 x 2, I would give one factor ro the second factor and the other 2 to the last. You'd have\[(x-1/2)(2x-(-2+\sqrt{2}))(2x-(-2-\sqrt{2}))\]

  21. anonymous
    • 5 years ago
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    ^^^ that's what you need. It's so hard to explain this stuff properly online and I'm falling asleep. Good luck, and sorry if I've confused you. I'm better when awake!

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