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can you factor that expression?
I've tried to... but I can't figure it out.
This makes me think there is not way to factor this. Wolfram is really good at factoring. Do you need the answer or spep by step procedure?
x1 -> 1.15181 - 2.23019 i x2 -> 1.15181 + 2.23019 I x3 -> -1.07514 x4 -> 1.77152 this are the four solutions for that expresion
I tried factoring, but I cant figure it out either. Lets supposse there is not a way jaja
there probably isnt. Thanks for trying though! :)
emunrrad how did you get your solution? i took derivative and used newtons method and found x = -1,2 only 2 real solutions because no inflection points exist
I used Mathematica from wolfram. Can you show us how to do that?
using these solutions you can use synthetic division to get factored form =(x+1)(x-2)(x^2-2x+6)
I tried synthetic division. But I got stoke and didn't know how to continue and just assumed I was doing it wrong.
gonna check it out.
well to solve just by using synthetic division, there are many possibilities any factor of -12 could work you would have to try +-1,2,3,4,6,12 -1 and 2 should have worked for you
the other 2 complex solutions would be: 1+- sqrt(5)i from quadratic formula x^2-2x+6 =0
That makes a bit more since now. Thank you.
Yeah, I worked it out by synthetic divsion, -1 works, then 2 works. So (x+1)(x-2)(x^2-2x+6). I hadn't done synthetic division in a while so I brushed up on it in five minutes here http://www.purplemath.com/modules/synthdiv.htm
synthetic division is nice but its a lot of work for finding zeroes, especially when constant has many factors and what if none of them work, you are assuming it has integer zeroes
For a pre-calculus course, synthetic division would be was is expected. The heavy armo, Newton's method is for higher math, I think brokenangel is working her way there.
oh sorry, yeah i agree i was just speaking in general thats true newton method is meaningless if you haven't had calculus :)
haha nope. this is algebra 2 stuff for me.