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# Long Division Help!

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What is (2x^5 - 3x^4 + 5x^3 - 10x^2 - 12x + 8) divided by (x^2 + 4)? I’m in pre-calc (in case that helps) and the original problem says

Solve over the complex numbers.

2x^5 - 3x^4 + 5x^3 - 10x^2 - 12x + 8 = 0  if 2i is a root

So what I did was multiply (x - 2i)(x + 2i) since they are factors (if there’s positive 2i, there has to be a negative one to cancel out the i), and I was going to factor out x^2 + 4, but I’m having trouble. Thank you in advance! :)

saseflower  Nov 14, 2017
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#1
+81031
+3

(2x^5 - 3x^4 + 5x^3 - 10x^2 - 12x + 8) divided by (x^2 + 4)?

2x^3   - 3x^2   - 3x    + 2

x^2 + 4  [    2x^5    - 3x^4  + 5x^3  - 10x^2    - 12x   +  8 ]

2x^5                + 8x^3

____________________________________

-3x^4   -  3x^3  - 10x^2

-3x^4                - 12x^2

____________________________________

-3x^3  +  2x^2   - 12x

-3x^3                  -12x

_____________________________________

2x^2               +  8

2x^2               +  8

______________________________________

( No remainder )

2x^5 - 3x^4 + 5x^3 - 10x^2 - 12x + 8 = 0

If   2i is a root...so is  -2i

So  we have this  (x - 2i ) ( x + 2i)  =  x^2 - 4i^2  =   x^2 + 4 is a factor

We already divided this...let's use what we found to find the other three roots

2x^3   - 3x^2   - 3x    + 2  = 0

Note that we can use a factoring trick here, saseflower

Write   -3x^2  as -4x^2 +  x

2x^3 - 4x^2  + x - 3x + 2  =  0  factor

2x^2 ( x - 2)  + ( x - 2) ( x - 1)  = 0

(x -2)  [ 2x^2  + x - 1 ]  = 0

(x - 2) (2x - 1) ( x + 1)  = 0

So...setting these factors to 0 and solving for x gives us the other three roots of

x = 2 , x = 1/2   and x = -1

CPhill  Nov 14, 2017
#3
+91469
+1

You have the setting out of these off pat Chis :)  I am impressed!

As you know I have given up and just do them on paper and take a photo.

Melody  Nov 14, 2017
#2
+1961
+2

Thank you so much! :)

saseflower  Nov 14, 2017

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