korosh23
  • korosh23
Pre-cal 12 Question! Polynomial equation. Determine an equation of a polynomial of degree 4 that has -1/2 as a root of multiplicity 3, and 2x^2-x-1as a factor.
Mathematics
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SOLVED
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katieb
  • katieb
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korosh23
  • korosh23
p(x)=a(x+1/2)^3 (2x^2-x-1) p(x)= a (x+1/2)^3 (x-1)(2x+1) How is it possible as a degree of 4? This would be degree of 5.
zepdrix
  • zepdrix
Try to examine the (2x+1) factor. What root does that lead to? :)
korosh23
  • korosh23
x= -1/2

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zepdrix
  • zepdrix
Good good good. Which means you only need `two more` to get multiplicity 3, ya?
korosh23
  • korosh23
why multiplicity of 3?
zepdrix
  • zepdrix
Because the question is insisting that it have exactly multiplicity 3 at that root, yes? :o
korosh23
  • korosh23
right :) sorry
korosh23
  • korosh23
so (x+1/2)^3
zepdrix
  • zepdrix
You had originally thought of something like this:\[\large\rm p(x)= a\left(x+\frac{1}{2}\right)^3 (x-1)\left(2x+1\right)\]Let's try something interesting. Let's factor a 2 out of each term in the last bracket.\[\large\rm p(x)= 2a\left(x+\frac{1}{2}\right)^3 (x-1)\left(x+\frac{1}{2}\right)\] You are correct that this is a quintic equation, which is too much! So it turns out we have too many (x+1/2) powers, yes? Because the second degree polynomial that they gave us contained `one of those roots`.
zepdrix
  • zepdrix
Ignore the 2 I pulled out, just absorb it into the `a` out front.\[\large\rm p(x)= a\left(x+\frac{1}{2}\right)^{\color{red}{3}} (x-1)\left(x+\frac{1}{2}\right)\]Do you see how you can fix your polynomial?
korosh23
  • korosh23
I believe it is p(x)=a(x+1/2)^2 (x−1) (x+1/2)
zepdrix
  • zepdrix
Ah that looks better! \c:/
korosh23
  • korosh23
but how to take out that extra degree? @zepdrix
lochana
  • lochana
yes. try x = -1/2 for 2x^2 - x -1. you will get zero again. that means (2x -1) also a factor of 2x^2 - x -1
korosh23
  • korosh23
@lochana yes you are right
zepdrix
  • zepdrix
I think you're misunderstanding the question :) It's not saying `root -1/2 multiplicity 3` AND SEPARATELY `2x^2-x-1` No no no ^ it's not saying that. There is overlap in this information. It's saying `root -1/2 multiplicity 3` some of which might come from this `2x^2-x-1`
korosh23
  • korosh23
oh I see what you mean. I have to adjust the degree of the factors to achieve the polynomial degree of 4. I get it know. :) Thank you!
zepdrix
  • zepdrix
ya you didn't have to do anything fancy to fix it, just use your eraser ^^ hehe
korosh23
  • korosh23
lol

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