Assume 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
what are the other roots of 3x^4-12x^3+41x-8x+26 that include 2-3i
2 answers
First 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
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
1 answer