+1975 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! :)
+129899 (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
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-3x^4 - 3x^3 - 10x^2
-3x^4 - 12x^2
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-3x^3 + 2x^2 - 12x
-3x^3 -12x
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2x^2 + 8
2x^2 + 8
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( 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
