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April 16, 2014

April 16, 2014

Posted by **kaitlyn** on Saturday, April 10, 2010 at 3:38pm.

- algerbra 2 -
**Damon**, Saturday, April 10, 2010 at 4:11pmAssume you mean:

3x^4-12x^3+41x^2-8x+26

Well then 2+3i must be one

multiply them

(x -2+3i)(x - 2-3i) = x^2 -4x +13

divide by that and get

3x^2+2

so the other two roots are

x=sqrt(2/3) and x = -2/3

- algerbra 2 -
**Reiny**, Saturday, April 10, 2010 at 4:16pmFirst of all, I will assume you have a typo and the third term should be 41x^2

complex roots always come in conjugate pairs, that is,

if 2-3i is a root, so is 2+3i

sum of those roots = 4

product of those roots = 3-9i^2 = 13

so the quadratic that yields those two roots is

x^2 - 4x + 13

I then did an algebraic long division by x^2 - 4x + 13 and that came out very nicely to an answer of 3x^2 + 2 with no remainder.

so ...

3x^4-12x^3+41x^2-8x+26= 0

(x^2 - 4x + 13)(3x^2 + 2) = 0

x = 2-3i, 2+3i from the first part and

x^2 = -2/3

x = ± √-(2/3)

= ± i√(2/3) or ±(1/3)i √6

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