Synthetic Division

[vanbeersj]

I have an equation in which I must divide it by the roots provided. I am provided with a double root of j.
From my text it states that the root of

x² + 1= 0 is j or -j which equals

My equation is:
x^6-x^5-2x^3-3x^2-x-2=0

I set up to divide by taking the coeffiecents of x,

1 -1 -2 -3 -1 -2

If I then follow synthetic division with j = sqrt(-1) then I get

1 sqrt(-1)-1 -1+sqrt(-1) -2+sqrt(-1) 2sqrt(-1) 0, So I have a remainder 0. I know I need to divide again by j but I end up with

1 2sqrt(-1)-1 -2+sqrt(-1) -3-sqrt(-1) -sqrt(-1)+1 2, here I'm left with a remainder and I still need to find 4 other roots since I was already provided with two j's as the first two roots.

I don't know how to proceed from here.

 

[Deleted member 4993]

x^6 - x^5 - 2x^3 -3x^2 - x - 2 = (x^2 + 1)(x^4 - x^3 - x^2 - x -2) .....edited

The fourth order function has two irrational roots and two other complex roots.

What methods have you been taught to find such roots? Newton - Raphson method?

 

[vanbeersj]

If I complete this equation (x^2 + 1)(x^4 - x^3 - x^2 -2)
I don't get x^6 - x^5 - 2x^3 -3x^2 - x - 2

I'm suppose to divide x^6 - x^5 - 2x^3 -3x^2 - x - 2 by the double root j.

 

[Denis]

> x^6 - x^5 - 2x^3 -3x^2 - x - 2 = (x^2 + 1)(x^4 - x^3 - x^2 - 2)

Mr Vanbeers, Subhotosh had one of your beers too many and forgot a "x":
x^6 - x^5 - 2x^3 -3x^2 - x - 2 = (x^2 + 1)(x^4 - x^3 - x^2 - x - 2)

You ok now?

"99 bottles of beer on the wall, on the wall..."
Subhotosh, hope you have a hangover and concrete drills are in full action nearby :twisted:

 

[Deleted member 4993]

> x^6 - x^5 - 2x^3 -3x^2 - x - 2 = (x^2 + 1)(x^4 - x^3 - x^2 - 2)

Subhotosh had a beer too many and forgot a "x": ....Indians and firewater don't mix - it was just "paper-to-screen-transfereitis"x^6 - x^5 - 2x^3 -3x^2 - x - 2 = (x^2 + 1)(x^4 - x^3 - x^2 - x - 2)

You ok now?