rare totem
Step B: I factored out a -2
Step C: I think I might be seeing a special factor case of x^3+a^3 but I am not sure since 2 is not a perfect cube? and would it not need to be a perfect cube for that special case to work?
How do I factor this, honestly, I am lost after I factored out the -2.
narrow wadiBOT
rare totem
Am I am going in the right direction with this? I remember doing something like this before in school but I forgot what this method was called. I am not sure if I should continue this line of thought. Please lmk what this method of factoring is called so I can look it up. Would I have to change -6x to some other variable as well?
Also how would the remainder theorem help me factor this? Can it help? I don't get it.
fickle spire
It seems that -4x^3+12x-2 cannot be factored further than 2*(-2x^3+6x-1) unless you will solve -2x^3+6x-1 = 0
winter cave
You cannot solve -2x^3+6x-1=0 by factoring or by testing rational roots.
Zeroes are irrational and, therefore, the function cannot be factorized nicely.
Even though there are three intercepts at f(x) = 0, none of these values are rational. They are all irrational. As such, you cannot factorize. The only way to find a value of x without using the cubic formula (and don't use it), is Newton's Method but it requires the knowledge of derivative in order to apply it.
rare totem
on the graph it appears it has roots but when i plug the x value in Y is not 0
I guess the graph gives an approximate answer.
winter cave
It does give you an approximate answer.
rare totem
ok I have a basic understanding on f` and f``
I see the derivative gives me -1 and 1
-6x^2+6
rare totem
I will have to search it, I am not a student anymore at a school.
Newtons method for calculation roots is the search term?
winter cave
Yep
If you want this is Newton's Method
But if the question is only to factorize your expression, then just leave it to -2(2x^ - 6x + 1)
rare totem
What conditions did you find in -2x^3+6x-1=0 which told you that it cannot be factored nicely is it the lack of GCF?
is -2x^3+6x-1 prime?
winter cave
Factor theorem and Rational Root theorem
rare totem
so the rational root theorem since no division had a remainder of 0
winter cave
Yep
rare totem
word
imma look at newtons method for a bit thanks for putting my eyes in the right direction
winter cave
Np!
If you need help I'm still here
(Btw, just to be sure, was the inital question was just to factor the expression or did you need to solve for x when f(x) = 0?)
Because if it is just to factor, then you already finished it
rare totem
Well all of the above. I wanted to factor in hopes to make solving for x easier when f(x)=0
winter cave
Honestly, I'd tend to say that it is easier to let 2x^3 - 6x + 1 be like it currently is because it doesn't factorise nicely since all the roots are irrational
For the roots, try either Newton's Method, the cubic formula (but I'd like to warn you that it is not the best thing to use because the formula itsef is quite big) or you might use a graphical approch (if it is allowed).
At the end, you'll probably end up with approximations as the exact value is quite long to write.
rare totem
Ok so with newtons method, I pick X that is close to C
what is C? What is C supposed to be close too, so I can guess which number to pick.
I choose 1?
its just a random pick
whats is Xnot plus 1?
would it be 2?
Also for questions like this would it be better to just compute Y by adding .01 to every X and graph it.
aka use a calculator isnt that what the calculator is doing anyways?
winter cave
You cannot use 1 for x subscript n
-6(1)^2 + 6 = 0 (which gives 3 over 0)
You should use another number than 1 or -1, like 0.5, -0.5 or 2
After you got a result, remplace your previous value by the new value
Repete the process infinetely (or at least when you're result is about to get stabilized)
winter cave
Also for questions like this would it be better to just compute Y by adding .01 ...
You can do that. If you're allowed to Desmos or to a graphic calculator, I'm just advising you to use the graphical features that they have. They save a lot of time (anyway, graphing the intial function is actually easy as most important characteristics can be found quite easily).
rare totem
Yea, I am allowed I just trying to understand things better
winter cave
Also, for Newton's Method, the first value that you choose is random. You can pick any number for x except if f'(x) = 0
rare totem
Word, this is a tangent but it would have sucked to live in a time of no desmos. lol
winter cave
Absolutely X))
I mean, that's why many equations like these weren't resolved before the arrival of modern calculus
queen ingot
did you guys get the answer
rare totem
what about the other two just keep plugging away until the answers are the same
winter cave
You plug in another initial number (instead of .5, try maybe 1.5)
The number that you'll plug in at the beginning will approches the root that it is more close to
So, if you plugged .2 instead of .5, you'll still obtain 0.16825... because .2 is closer to that value that it is closer to the other roots.
rare totem
any number between -1 and 3 and that is not 1
I saw a video on youtube showing to try 0 first and then 1
and it said the roots will be between those numbers
and another x = -1.810091078, I get it, it appears this may be easier than trying each x value individually. ok im done this is solved.
rare totem
just the last zero for good measure. x=1.642361111 
broken patrol
.close
