Assume you mean:
Well then 2+3i must be one
(x -2+3i)(x - 2-3i) = x^2 -4x +13
divide by that and get
so the other two roots are
x=sqrt(2/3) and x = -2/3
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.
(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
algerbra - If 1 and –6 are two of the roots of 6x^4 + 31 x^3 − 33x^2 ...
Math - How to factor x^3 - 3x^2 + 4 =0 Use D'Alembert's Rational Roots Theorem. ...
Math - Roots Ok, what about roots? Roots of polynomials? Square roots? Cube ...
Algerbra 2 - find the possible roots for: 3x^3-6x^2+11x-6=0
GRADE 11 MATH - Show that the equation 2x^2 - 5x + 2 = 0 has roots that are ...
precalculus - express the roots of unity in standard form a+bi. 1.) cube roots ...
precalc - express the roots of unity in standard form a+bi. 1.) cube roots of ...
Precalculus - "Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the ...
math - You have a difference of squares (x^2-3)(x^2+3) One of these factors to ...
algebra - k is uniformly chosen from the interval (-5,5) . Let p be the ...