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mathmate
 one year 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
 one year 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.

evy15
 one year ago
Best ResponseYou've already chosen the best response.0no I dnt know those methods

mathmate
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2So is (x1) a factor to the equation?

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

mathmate
 one year 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.

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

evy15
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, can I use the calculator for that

KingGeorge
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2So that reduces to 12 possibilities only.

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

KingGeorge
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2@evy15 are you able to complete the solution from here?

evy15
 one year ago
Best ResponseYou've already chosen the best response.0can you guide me to what to do next

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

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

evy15
 one year ago
Best ResponseYou've already chosen the best response.0so its 1,+/5i, and 3/2?

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

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

evy15
 one year ago
Best ResponseYou've already chosen the best response.0yes thank you guys so much

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

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

evy15
 one year ago
Best ResponseYou've already chosen the best response.0can i do the same with x^53x^424x^372x^2+12x=12

evy15
 one year ago
Best ResponseYou've already chosen the best response.0sorry wrote the wrong one

evy15
 one year ago
Best ResponseYou've already chosen the best response.0x^53x^424x^372x^225x+75=0

KingGeorge
 one year 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
 one year 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?

evy15
 one year ago
Best ResponseYou've already chosen the best response.0x^4(x3)24x^2(x3)25(x3)

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

evy15
 one year ago
Best ResponseYou've already chosen the best response.0would it be (x^424x^225) and (x3)?

KingGeorge
 one year 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
 one year 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!
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