Question

find all zeros:

f(x)=x^4-5x^3+10x^2-3x+3

Answer 1

You'll want to factor this.

Answer 2

Do you know synthetic division and the rational roots theorem?

Answer 3

@Anhvypham

Answer 4

if it factors it will be in this form:
\[(x^2 + ax+3)(x^2 +bx +1)\]
setting up these equations:
\[a +b = -5\]
\[ab + 4 = 10\]
\[a + 3b = -3\]

Answer 5

huh. Would never have thought to approach the problem that way @dumbcow

Answer 6

the answers include imaginary numbers, so that's what is throwing me offf

Answer 7

your equations don't seem to be consistent @dumbcow

Answer 8

@inkyvoyd , its an approach which makes it easier to get exact values

it doesn't work though, function does not factor

use rational root thm and synthetic division, though actually there are no real soltions

Answer 9

@Anhvypham where did you get this problem from? none of the roots are rational

Answer 10

they are also all complex.

Answer 11

its in my math review packet

Answer 12

Well, sorry to say this, but you either made a typo or your teacher did.

Answer 13

are you given all 4 complex answers ? what class are you in?

Answer 14

you sure you don't mean -x^4-5x^3+10x^2-3x+3

Answer 15

i have the answers. they are all complex. i'm in precalc honors

Answer 16

well, what are your answers?

Answer 17

you can reconstruct them that way

Answer 18

you could say:
\[x = a \pm bi, c \pm di\]
plug them in and get 4 equations and try to solve for a,b,c,d
its really messy though

Answer 19

I was thinking for complex roots a, b, c, and d he could just take 0=(x-a)(x-b)(x-c)(x-d) on a CAS system to find out the actual correct equation, then resolve the actual correct equation

Answer 20

if you give me the roots I can give you the equation that it expands to.

Answer 21

though it kinda loses the meaning with the answers since you know what to do

Answer 22

roots: 3i, -3i, \[\frac{ (3i \sqrt{2}) }{ 2 }\], \[-\frac{ (3i \sqrt{2}) }{ 2 }\]

Answer 23

whoa ok you have a typo in your function

Answer 24

well that doesn't seem to check out at all. Your roots are all imaginary?

Answer 25

yea

Answer 26

put that in wolfram alpha and you get x^4+(27 x^2)/2+81/2.... which has nothing to do with your original equation. lol.

Answer 27

this problem makes no sense. it probably was a typo. thanks for your help though guys

Answer 28

yeah I was hoping that through the roots you could reconstruct what the problem was supposed to be, but I guess not lol