A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Find all the zeros of the eq. 2x^45x^3+53x^2125x+75=0
anonymous
 3 years ago
Find all the zeros of the eq. 2x^45x^3+53x^2125x+75=0

This Question is Closed

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2This 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 recheck for typos on the other problem.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2To make things simpler, when the sum of coefficients = 0, it means that (x1) is a factor.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no I dnt know those methods

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Descartes 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.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2http://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
 3 years ago
Best ResponseYou've already chosen the best response.2So is (x1) a factor to the equation?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2So you can do a synthetic division, or a long division to reduce the quartic to a cubic, i.e. find (2*x^45*x^3+53*x^2+(125)*x+75)/(x1)=?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Good, now we can rule out the third case (4 complex roots).

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Using 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
 3 years ago
Best ResponseYou've already chosen the best response.0what do I divide it by

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2When you find another root (i.e. another factor) of the cubic.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok, can I use the calculator for that

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry to bust in here, but \(2x^33x^2+50x75\) is relatively easily factored by grouping, and thus we can skip the whole guess and check portion of finding roots.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2In fact, Descartes rule of sign says we don't have to try factors of the form (a+b), only those of the form (ab) because all real roots are positive.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2So that reduces to 12 possibilities only.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Very true, thanks KingGeorge!

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1You group like so \[2x^33x^2+50x75=x^2(2x3)+25(2x3)\]Can evy15 finish the factoring from here?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2@evy15 are you able to complete the solution from here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you guide me to what to do next

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Good job, (+\ 5i). Now you have all 4 zeroes!

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Can you enumerate the four zeroes?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so its 1,+/5i, and 3/2?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Excellent! There you are! Not 1. Your list is good.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Big thanks to KingGeorge who cut the work in half!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes thank you guys so much

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1As an addendum, it's possible to factor the original function by grouping if you're clever. \[\begin{aligned} 2x^45x^3+53x^2125x+75&=2x^45x^3+3x^2+50x^2125x+75 \\ &=x^2(2x^25x+3)+25(2x^25x+3) \\ &=(x^2+25)(2x^25x+3)\\ &=(x^2+25)(x1)(2x3) \end{aligned}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1But this is not an obvious, nor the easiest, way to do it.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.2Yep, I will look more carefully for these cases in the future.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can i do the same with x^53x^424x^372x^2+12x=12

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry wrote the wrong one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^53x^424x^372x^225x+75=0

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Your first group will be \[x^4(x3).\]Knowing this, can you give me the rest of the factorization?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^4(x3)24x^2(x3)25(x3)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Perfect. And can you finish the rest of the factorization?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would it be (x^424x^225) and (x3)?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Right again. Now you just need to factor \(x^424x^225\). You can do this by pretending it's a quadratic equation, and not a quartic. So how does \(x^224x25\) factor?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Perfect. So that means \(x^424x^225=(x^225)(x^2+1)\). Hence, \[x^53x^424x^372x^225x+75=(x^225)(x^2+1)(x3)\]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!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.