Find all complex roots of polynomial

[DryRun]

Homework Statement
Given 2 + 3i is one root of the equation:
x^4 - 6x^3 + 26x^2 -46x +65 = 0
Find the remaining roots.

The attempt at a solution
I am thinking about this as a possible solution although it is too long to be plausible, unless I'm wrong.
let x = a + bi and then replace in the polynomial. Then, expand and finally, divide it by 2+3i

Another possibility (i'm not sure) is this:
x (x^3 - 6x^2 + 26x -46) +65 = 0
So, i try to solve the cubic equation: x^3 - 6x^2 + 26x -46
However, note that one solution to x, is not equal to -65. So, this method is wrong, i believe.

I frankly don't know how else to go about to solve this one.

 

[SteamKing]

What other property about complex roots do you know? (Hint: they always come in pairs)

 

[DryRun]

So, 2-3i is also a root? How do i confirm that? By replacing in the polynomial, it should equal zero, right?

But what about the other roots? Since the maximum power of x is 4, so i assume that there are 4 roots in all.

 

[DryRun]

I still can't figure it out. Can someone please give me a hint?

 

[LCKurtz]

If you have two roots p and q, then (x-p)(x-q) is a factor. Use long division to divide out that factor and see what's left.

 

[SammyS]

If you have two roots p and q, then (x-p)(x-q) is a factor. Use long division to divide out that factor and see what's left.

Notice that when p & q are complex conjugates, (x-p)(x-q) can be written as a sum times a difference which comes out quite nicely as a difference of squares.

(x - (2-3i))(x - (2+3i)) = ((x-2) + 3i)((x-2) - 3i) = ...

 

[DryRun]

OK, here is what i did:
The two known factors are: (2+3i) and (2-3i)
So, the product is: (2+3i)(2-3i) = x^2 - 4x + 13

Using this product as divisor and the polynomial as dividend, i get this:
x^2 - 2x + 5
OK, i got it. Thanks guys. :)

 

[SammyS]

So, what are the two remaining roots?

 

[DryRun]

Initially, i did it like this:
x(x-2) = -5
So x = -5 and x = -3
But then i don't know why but these answers are wrong (from the book).

So, i used quadratic formula
and i got these: 1-2i, 1+2i, which are correct from the book.

But again, i don't understand why i can't use the first set of simpler roots; -5 and -3.

 

[SammyS]

Initially, i did it like this:
x(x-2) = -5
So x = -5 and x = -3
But then i don't know why but these answers are wrong (from the book).

So, i used quadratic formula
and i got these: 1-2i, 1+2i, which are correct from the book.

But again, i don't understand why i can't use the first set of simpler roots; -5 and -3.

You can't use -5 & -3 because they're not solutions of x(x-2) = -5. Plug them into see that.

 

[HallsofIvy]

Basic rule: If ab= 0 then either a= 0 or b= 0. But that is a special property of 0. You cannot say "if ab= -5 then either a= -5 or b= -5" which is what you were trying to do.

 

[DryRun]

I think i have this one wrapped up. Thanks for the help. :)