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evy15
Find all the zeros of the eq. 2x^4-5x^3+53x^2-125x+75=0
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.
To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.
no I dnt know those methods
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.
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.
So is (x-1) a factor to the equation?
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)=?
Good, now we can rule out the third case (4 complex roots).
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.
When you find another root (i.e. another factor) of the cubic.
oh ok, can I use the calculator for that
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.
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.
So that reduces to 12 possibilities only.
Very true, thanks 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?
@evy15 are you able to complete the solution from here?
can you guide me to what to do next
Good job, (+\- 5i). Now you have all 4 zeroes!
Can you enumerate the four zeroes?
Excellent! There you are! Not -1. Your list is good.
Big thanks to KingGeorge who cut the work in half!
yes thank you guys so much
As an addendum, it's possible to factor the original function by grouping if you're clever. \[\begin{aligned} 2x^4-5x^3+53x^2-125x+75&=2x^4-5x^3+3x^2+50x^2-125x+75 \\ &=x^2(2x^2-5x+3)+25(2x^2-5x+3) \\ &=(x^2+25)(2x^2-5x+3)\\ &=(x^2+25)(x-1)(2x-3) \end{aligned}\]
But this is not an obvious, nor the easiest, way to do it.
Yep, I will look more carefully for these cases in the future.
can i do the same with x^5-3x^4-24x^3-72x^2+12x=12
sorry wrote the wrong one
x^5-3x^4-24x^3-72x^2-25x+75=0
That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.
Your first group will be \[x^4(x-3).\]Knowing this, can you give me the rest of the factorization?
x^4(x-3)-24x^2(x-3)-25(x-3)
Perfect. And can you finish the rest of the factorization?
would it be (x^4-24x^2-25) and (x-3)?
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?
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!