anonymous
  • anonymous
A little help here please? :( The function of the polynomial is x^3-2x^2-4x+8 and its zero is 2, -2, and 2. And the degree of P(x) is 3. I don't understand how they got the zeros (2, -2 and 2) please explain me this. This is for my report this coming friday. :(
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[(x-2)(x-2)(x+2)\]
cwrw238
  • cwrw238
given th zeros we can find the factored form of the function as satellite as done. solving (x - 2)(x - 2)(x + 2) = 0 gives the zeros. the above would have been found by factoring x^3-2x^2-4x+8
cwrw238
  • cwrw238
using the rationalr oot theorem its a good better that 1 or more zeros are 2 because factors of the last term 8 are 2 and 4 find f(2) to confirm this (by checking if f(2) = 0)

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cwrw238
  • cwrw238
so one of the facors is (x-2) now divide f(x) by x-2 using long or synthetic division finally factorize the quotient
cwrw238
  • cwrw238
* a good bet
anonymous
  • anonymous
I think your are explaining me the other when the zeros are guven the remaining zeros can be found. I'm asking, how 2, -2, and 2 was got by the polynomial function P(x)=x^3-2x^2-4x+8 @cwrw238 :) @satellite73 : is that the factor of x^3-2x^2-4x+8 ? :)
anonymous
  • anonymous
\[\Large x^3-2x^2-4x+8=x^2(x-2)-4(x-2)\\ \Large (x-2)(x^2-4)=0\\ \Large => x-2=0, or,x^2-4=0\\ \Large => x=2,or x=\mp2\]
anonymous
  • anonymous
Hmm. Thank yooou so much @cinar , now i understand :)
anonymous
  • anonymous
by the way, why is it ∓2? @cinar
anonymous
  • anonymous
@campbell_st : Please, let me understand why is it ∓2? in the solution if @cinar
anonymous
  • anonymous
it is +/- because you have x^2 -4 = 0 by null factor. and therefore x = sqrt 4 sqrt 4 is posotibve or negative 2 (-2)^2 = 4 just as 2^2 = 4
campbell_st
  • campbell_st
ok... so here is another slant on the above answers... the zeros or roots of a polynomial are where the polynomial cuts the x-axis so if your polynomial is \[P(x) = x^3 -2x^2 - 4x + 8\] you are told the zeros... to show it substitute x = 2 into your polynomial \[P(2) = (2)^2 -2(2)^2 -4(2) + 8\] calculate it out and you get zero... so when x = 2 P(2) = 0 hence the name... hope this makes sense you can substitute x = -2 and find P(-2) = 0 ... as well
campbell_st
  • campbell_st
and as for the solutions... there are only 2, x = 2 and x = -2... the 2nd solution is repeated.... (x-2)(x-2) = (x-2)^2 this is normally described as multiplicity of 2.... the zero at x = -2 occurs twice
DebbieG
  • DebbieG
@campbell_st I don't think she was given the zeros, I think she has a problem and answer and wants to understand how to get there. Like @cinar showed, this polynomial can conveniently be factored by grouping. That might be as far as you've gotten in this lesson. But in general, you can use the rational root theorem as was discussed above, OR a graphing calculator if allowed, to find a first rational root. Then pull out the factor associated with that root by synthetic division, and what you have left is a quadratic. Solve using methods for solving quadratic equations and that will give you the other 2 roots.
anonymous
  • anonymous
@DebbieG : debbieg's right. I need to understand how to get there of the answer. :3 cause i don't get it. Why in the solution of @cinar is ∓2? why not positive 2? :3
DebbieG
  • DebbieG
Because once you reach this point: \(\Large (x-2)(x^2-4)=0\\ \) Now you have to factor the \(\Large (x^2-4)\\ \) That's a difference of squares. What do you get when you factor it?
DebbieG
  • DebbieG
You get 2 roots there - so he is just combining them using \(\pm\) notation. You could write them each separately too. {2, -2} Of course, the 2 is already a root of the polynomial, so that one is a repeated root.
DebbieG
  • DebbieG
When all is said and done, what you end up with is: \(\Large y= (x-2)^2(x+2)\\ \)
anonymous
  • anonymous
Ahhhh. :) nooow, i understand. Thank yooooou so much @DebbieG your really helped me a lot. ^_^v and thanks also to @campbell_st for sharing your idea and @cinar for sharing the solution. Thank yoooooou so much! :)
DebbieG
  • DebbieG
You're welcome. :)

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