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anonymous
 5 years ago
f(x)=x^43x^3+6x^22x12
find real zeros
anonymous
 5 years ago
f(x)=x^43x^3+6x^22x12 find real zeros

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you factor that expression?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've tried to... but I can't figure it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wolfram presents the following.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x1 > 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried factoring, but I cant figure it out either. Lets supposse there is not a way jaja

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there probably isnt. Thanks for trying though! :)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0emunrrad 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used Mathematica from wolfram. Can you show us how to do that?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0using these solutions you can use synthetic division to get factored form =(x+1)(x2)(x^22x+6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried synthetic division. But I got stoke and didn't know how to continue and just assumed I was doing it wrong.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0well 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

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0the other 2 complex solutions would be: 1+ sqrt(5)i from quadratic formula x^22x+6 =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That makes a bit more since now. Thank you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I worked it out by synthetic divsion, 1 works, then 2 works. So (x+1)(x2)(x^22x+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

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0synthetic 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For a precalculus 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.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0oh sorry, yeah i agree i was just speaking in general thats true newton method is meaningless if you haven't had calculus :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha nope. this is algebra 2 stuff for me.
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