anonymous
  • anonymous
Find all the zeros of the eq. 2x^4-5x^3+53x^2-125x+75=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmate
  • mathmate
This one can be factorized, so you will need to use the following tools: Descartes rule of signs, factorization theorem, fundamental theorem of algebra Are any or some of these famililar to you? Also, this may indicate that you or your teacher have a typo in the other problem. Please re-check for typos on the other problem.
mathmate
  • mathmate
To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.
anonymous
  • anonymous
no I dnt know those methods

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mathmate
  • mathmate
Descartes rule of signs tells us that we have either 4 real positive roots, or two real positive roots and two complex roots, or 4 complex roots.
anonymous
  • anonymous
ok
mathmate
  • mathmate
http://en.wikipedia.org/wiki/Descartes'_rule_of_signs http://en.wikipedia.org/wiki/Factor_theorem http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra Since you have been missing school for at least one day, it may well be advisable to catch up with some of these topics. Hope you got notes from your friends though.
mathmate
  • mathmate
So is (x-1) a factor to the equation?
anonymous
  • anonymous
yes
mathmate
  • mathmate
So you can do a synthetic division, or a long division to reduce the quartic to a cubic, i.e. find (2*x^4-5*x^3+53*x^2+(-125)*x+75)/(x-1)=?
anonymous
  • anonymous
2x^3-3x^2+50x-75
mathmate
  • mathmate
Good, now we can rule out the third case (4 complex roots).
anonymous
  • anonymous
ok
mathmate
  • mathmate
Using the factor theorem, you will need to try factors of the form (ax+b) where a=1 or 2 (from the leading coefficient 2), and b=+/- all possible factors of 75, which are 1,3,5,15,25,75, i.e. 2*2*6=24 possibilities. Once you find that, you can again do the synthetic division or long division to reduce the cubic to a quadratic.
anonymous
  • anonymous
what do I divide it by
mathmate
  • mathmate
When you find another root (i.e. another factor) of the cubic.
anonymous
  • anonymous
oh ok, can I use the calculator for that
KingGeorge
  • KingGeorge
Sorry to bust in here, but \(2x^3-3x^2+50x-75\) is relatively easily factored by grouping, and thus we can skip the whole guess and check portion of finding roots.
mathmate
  • mathmate
In fact, Descartes rule of sign says we don't have to try factors of the form (a+b), only those of the form (a-b) because all real roots are positive.
mathmate
  • mathmate
So that reduces to 12 possibilities only.
mathmate
  • mathmate
Very true, thanks KingGeorge!
KingGeorge
  • KingGeorge
You group like so \[2x^3-3x^2+50x-75=x^2(2x-3)+25(2x-3)\]Can evy15 finish the factoring from here?
mathmate
  • mathmate
@evy15 are you able to complete the solution from here?
anonymous
  • anonymous
can you guide me to what to do next
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
x=i+-5 and 3/2
anonymous
  • anonymous
i mean =-5i
mathmate
  • mathmate
Good job, (+\- 5i). Now you have all 4 zeroes!
mathmate
  • mathmate
Can you enumerate the four zeroes?
anonymous
  • anonymous
so its 1,+/-5i, and 3/2?
anonymous
  • anonymous
or-1
mathmate
  • mathmate
Excellent! There you are! Not -1. Your list is good.
mathmate
  • mathmate
Big thanks to KingGeorge who cut the work in half!
anonymous
  • anonymous
yes thank you guys so much
mathmate
  • mathmate
yw! :)
KingGeorge
  • KingGeorge
As an addendum, it's possible to factor the original function by grouping if you're clever. \[\begin{aligned} 2x^4-5x^3+53x^2-125x+75&=2x^4-5x^3+3x^2+50x^2-125x+75 \\ &=x^2(2x^2-5x+3)+25(2x^2-5x+3) \\ &=(x^2+25)(2x^2-5x+3)\\ &=(x^2+25)(x-1)(2x-3) \end{aligned}\]
KingGeorge
  • KingGeorge
But this is not an obvious, nor the easiest, way to do it.
mathmate
  • mathmate
Yep, I will look more carefully for these cases in the future.
anonymous
  • anonymous
can i do the same with x^5-3x^4-24x^3-72x^2+12x=12
anonymous
  • anonymous
sorry wrote the wrong one
anonymous
  • anonymous
x^5-3x^4-24x^3-72x^2-25x+75=0
KingGeorge
  • KingGeorge
That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.
KingGeorge
  • KingGeorge
Your first group will be \[x^4(x-3).\]Knowing this, can you give me the rest of the factorization?
anonymous
  • anonymous
x^4(x-3)-24x^2(x-3)-25(x-3)
KingGeorge
  • KingGeorge
Perfect. And can you finish the rest of the factorization?
anonymous
  • anonymous
would it be (x^4-24x^2-25) and (x-3)?
KingGeorge
  • KingGeorge
Right again. Now you just need to factor \(x^4-24x^2-25\). You can do this by pretending it's a quadratic equation, and not a quartic. So how does \(x^2-24x-25\) factor?
anonymous
  • anonymous
x-25 and x+1
KingGeorge
  • KingGeorge
Perfect. So that means \(x^4-24x^2-25=(x^2-25)(x^2+1)\). Hence, \[x^5-3x^4-24x^3-72x^2-25x+75=(x^2-25)(x^2+1)(x-3)\]From here, you just have some sums/differences of squares, and should be able to figure out the roots. I've got to go now, but you seem to be catching the hang of this!

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