Question

Find all zeros of f(x)=x^3-8 x^2+18 x-12.

I know to take the first and last coefficients (in this case 1 and 12) and take their factors (1 having only a factor of 1 and 12 having 1, 2, 3, 4, 6, 12).

I then use synthetic division to find a zero. I have found 2 to be a zero. Then I take the new quadratic which is: x^2-8x+18 and input it into the quadratic formula.

So I got these as zeros: 2, 4+i√2, 4-i√2

My math site for school kicked me back as wrong. Where is my error?

Answer 1

eh okay so you did the Descartes rule of sign
perhaps you made an error

Answer 2

redo the process and be more careful

Answer 3

According to wolfram there are 3 real zero
no complex one

Answer 4

Yeah I've done it twice, I'm about to try it again.

Answer 5

Does wolfram explain the algebra of the problem?

Answer 6

No you have to pay to get the details, but you don't need that
you need to work it out! it is just about errors in your steps

Answer 7

http://www.wolframalpha.com/input/?i=f(x)%3dx%5e3-8+x%5e2%2b18+x-12
this is the graph

Answer 8

i think you have got one zero correct
2 is right
but find the other two

Answer 9

Thank you. I'm up to the synthetic division and so far I'm good. Still only getting 2.

Answer 10

that's good! stop there!
find the two remaining roots

Answer 11

well I found one error. used the dang initial polynomial numbers instead of the x^-6x+6 that I would get after the SD. But that can't be factored so I have to use the quadratic formula.

Answer 12

Yeah! you need quadratic to find the other zeros

Answer 13

welcome!
did you mean 3 plus or minus root(3)

Answer 14

Yes I did.

Answer 15

okay good!

Answer 16

I think there is an alternative way
when it comes to degree 3 poly

Answer 17

\(\large \rm f(x)=x^3-8x^2+18x-12=x^3-2x^2-6x^2+18x-12\\=x^2(x-2)-6(x^2-3x+2)=x^2(x-2)-6(x-2)(x-1)\)

Answer 18

So what is happening there?

Answer 19

factor by grouping?

Answer 20

you factor x-2 and you will get the other part that you get by SD
yes! this by grouping (but adding one step)
i play with one term to make it factorable

Answer 21

it takes some practice to use this technique, but it doesn't work every time lol
but i guarantee that it will work for degree 3 or even degree 4
but 5 up things get crazy

Answer 22

I see now. oh I like it this way much better, but it won't always work. Which is why we are using the quadratic formula.

Answer 23

one other way i go with this is test some number like f(1)
f(-1)
f(2)
simple cases to know if they are potential zeros

Answer 24

So plug and solve using the factors of the first and last coefficients right?

Answer 25

no it will work all the time for degree 3

Answer 26

i plug in a potential zero and try to use that to factor

Answer 27

but i would say that SD or LD are what you need to think of more than this

Answer 28

ehh long division, I love it. /sarcasm

Answer 29

yeah whenever something looks hard to factor
go for SD or LD
but make sure you are doing it carefully! or otherwise you will be thrown off