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mathmate Group TitleBest ResponseYou've already chosen the best response.2
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 recheck for typos on the other problem.
 one year ago

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

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

mathmate Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

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

mathmate Group TitleBest ResponseYou've already chosen the best response.2
So 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)=?
 one year ago

evy15 Group TitleBest ResponseYou've already chosen the best response.0
2x^33x^2+50x75
 one year ago

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

mathmate Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

evy15 Group TitleBest ResponseYou've already chosen the best response.0
what do I divide it by
 one year ago

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Sorry 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.
 one year ago

mathmate Group TitleBest ResponseYou've already chosen the best response.2
In 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.
 one year ago

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

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

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

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

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

evy15 Group TitleBest ResponseYou've already chosen the best response.0
x=i+5 and 3/2
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Right 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?
 one year ago

evy15 Group TitleBest ResponseYou've already chosen the best response.0
x25 and x+1
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Perfect. 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!
 one year ago
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