Question

f(x)=x^4-3x^3+6x^2-2x-12
find real zeros

Answer 1

can you factor that expression?

Answer 2

I've tried to... but I can't figure it out.

Answer 3

Wolfram presents the following.

Answer 4

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?

Answer 5

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

Answer 6

Ok. Thanks.

Answer 7

I tried factoring, but I cant figure it out either. Lets supposse there is not a way jaja

Answer 8

there probably isnt.
Thanks for trying though! :)

Answer 9

np

Answer 10

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

Answer 11

I used Mathematica from wolfram. Can you show us how to do that?

Answer 12

using these solutions you can use synthetic division to get factored form
=(x+1)(x-2)(x^2-2x+6)

Answer 13

I tried synthetic division. But I got stoke and didn't know how to continue and just assumed I was doing it wrong.

Answer 14

http://en.wikipedia.org/wiki/Newton%27s_method

Answer 15

gonna check it out.

Answer 16

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

Answer 17

the other 2 complex solutions would be: 1+- sqrt(5)i
from quadratic formula
x^2-2x+6 =0

Answer 18

That makes a bit more since now. Thank you.

Answer 19

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

Answer 20

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

Answer 21

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.

Answer 22

oh sorry, yeah i agree i was just speaking in general
thats true newton method is meaningless if you haven't had calculus :)

Answer 23

haha nope. this is algebra 2 stuff for me.