anonymous
  • anonymous
For each polynomial function, one zero is given. Find all others.. f(x)=4x^3+6x^2-2x-1; 1/2(is given)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Thanks for your help...but I have no idea what your talking about
anonymous
  • anonymous
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

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anonymous
  • anonymous
I can't really explain polynomial division on this site. You can find stuff on YouTube or something...ah, synthetic division...similar thing
anonymous
  • anonymous
4x^2+8x+2 is what I got too
anonymous
  • anonymous
I thought the next step would be to factor... to find the answers. I get that (x-1/2) is one answer
anonymous
  • anonymous
It can be factored. Find the roots of the equation, alpha and beta, say, and then you have \[(x-\alpha)(x-\beta)=4x^2+8x+2\]
anonymous
  • anonymous
I dont know what that is??
anonymous
  • anonymous
The discriminant is 32, which is positive, but not a perfect square, so you will get two distinct, real, irrational roots.
anonymous
  • anonymous
You don't know the quadratic formula?
anonymous
  • anonymous
Oh, so I use the quadratic instead of factoring??
anonymous
  • anonymous
Well, you'll still be factoring...I'll show you.
anonymous
  • anonymous
\[x=\frac{-8{\pm}\sqrt{8^2-4 \times 4 \times 2}}{2 \times 4}=\frac{-2{\pm}\sqrt{2}}{2}\]
anonymous
  • anonymous
Now, since these x-values give zeros, we must have that\[(x-(\frac{-2+\sqrt{2}}{2}))(x-(\frac{-2-\sqrt{2}}{2}))=0\]
anonymous
  • anonymous
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).
anonymous
  • anonymous
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!!
anonymous
  • anonymous
The other two answers are what I just gave you. Your polynomial is then \[(x-1/2)(x-(\frac{-2+\sqrt{2}}{2}))(x-(\frac{-2-\sqrt{2}}{2}))\]
anonymous
  • anonymous
You have your three factors. It's ugly, but that's it.
anonymous
  • anonymous
Thanks.. I get scared when things are ugly... I tend to think its wrong.
anonymous
  • anonymous
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\[(x-1/2)(2x-(-2+\sqrt{2}))(2x-(-2-\sqrt{2}))\]
anonymous
  • anonymous
^^^ 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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