Question

Last one I am stuck on.

The number (1-I) is a root of the equation x^3-8x^2+14x-12=0. Find the other roots of this equation.

Answer 1

hint : complex roots always come in conjugate pairs

Answer 2

So, if \((1-i)\) is one root, its conjugate pair : \((1+i)\), will also be a rot

Answer 3

*root

Answer 4

Should I group and factor (x^3 - 8x^2) and (14x -12)?

Answer 5

are u given options for this q ?

Answer 6

No, any convent method.

Answer 7

Convenient

Answer 8

you already have two roots :
\(1+i\)
\(1-i\)

Answer 9

maybe u can use that info to find the third root ?

Answer 10

familiar with long division ?

Answer 11

or, if u know rational root theorem, u can test and see if any factors of 12 work as roots

Answer 12

Do I divide by -1?

Answer 13

u familiar wid rational root theorem ?

Answer 14

We briefly went over in class, not very sure of my self

Answer 15

okay its simple actually
rational root theorem says this :
```
rationals roots will be of form p/q

where, p = factors of constant term
q = factors of leading coefficient
```

Answer 16

your constant term = 12
leading coefficient = 1

Answer 17

So, u can test and see if u get luck wid any factors of 12 :
1, 2, 3, 4, 6, 12

Answer 18

http://www.wolframalpha.com/input/?i=solve+x%5E3-8x%5E2%2B14x-12%3D0

Answer 19

you will find that "6" is the third root

Answer 20

since this is a cubic, there will be only 3 roots.

Answer 21

I'm going to work the problem and I'll post what I get

Answer 22

okay

Answer 23

Does this look right?
(X-6)(x^2-2x+2)=0
X-6=0
X=6
(X^2-2x+2)
(X^2-2x)=-2
(X^2-2x+1)=-1
(X-1)^2=-1
X='1-i, x=1+i,

Answer 24

yes, but u need to show work on how you got x = 6 as a root

Answer 25

That would just be adding. 6 to get x by its self