Synthetic division with complex number

[jkristia]

I have been doing some homework on synthetic division today.
Most of it is pretty straight forward, until I got to a problem with complex number

I was able to solve the first problem
(x^3 – 3x^2 + x – 3) / (x-i)
By using +i for the ‘synthetic divisor constant’ (sorry don’t know the proper name)
Q = x^2 + (-3+i)x + 3i, R = 0

But I cannot figure out the next problem. It says

“Let P(x) = x^2 + 2ix – 10, Use synthetic division to find P(2-i)”

I think I have to use (-2+i) for the divisor (??), but the answer in the book is just -5, so it seems like they simply evaluated the polynomial for x = (2-i), which of course is what P(2-i) mean, but is there any way to get to this result using synthetic division? The book does not give a single example on how to perform the division using complex number, and it is a pure online class, so no teacher to ask.

Thanks
Jesper

 

[eumyang]

I think I have to use (-2+i) for the divisor (??),

No, not for synthetic division. Finding P(2-i) is the same as finding the remainder when dividing
$$\frac{x^2 + 2ix - 10}{x - (2 - i)}$$

but the answer in the book is just -5, so it seems like they simply evaluated the polynomial for x = (2-i), which of course is what P(2-i) mean, but is there any way to get to this result using synthetic division? The book does not give a single example on how to perform the division using complex number, and it is a pure online class, so no teacher to ask.

Synthetic division is no different, really, with complex numbers. Just remember that when you add two complex numbers, you add the corresponding real and imaginary coefficients, and when you multiply complex numbers, you can "FOIL." Here is an example of using synthetic division to verify that 1-i is a root of the cubic x³ - 3x² + 4x - 2:

1-i|  1   -3     4    -2
---+       1-i  -3+i   2
    ---------------------
      1   -2-i   1+i   0

 

[jkristia]

Ahh, I see, all they are asking is to find the remainder for (2-i). Thank you very much

2-i |  1   0+2i   -10
    |      2-1i   (2-i)(2+i) = 5
-----------------------------------
       1   2+1i   -5 <- remainder