what are the other roots of 3x^4-12x^3+41x-8x+26 that include 2-3i
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
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
To find the other roots of the given polynomial, we need to use the fact that complex roots occur in conjugate pairs. Since 2-3i is a root of the polynomial, its conjugate, 2+3i, must also be a root.
Now, let's set up the polynomial as an equation and use synthetic division to find the quotient polynomial after dividing by (x - 2+3i)(x - 2-3i). Synthetic division allows us to divide a polynomial by a linear factor and obtain the quotient polynomial.
The factor (x - 2+3i)(x - 2-3i) can be simplified as follows:
(x - 2+3i)(x - 2-3i) = (x - 2)^2 - (3i)^2
= (x - 2)^2 + 9
= x^2 - 4x + 4 + 9
= x^2 - 4x + 13
Using synthetic division, we divide the given polynomial 3x^4 - 12x^3 + 41x - 8x + 26 by x^2 - 4x + 13.
Coefficients: 3 -12 41 -8 26
Root: |____2___3i___
Performing synthetic division:
2+3i | 3 -12 41 -8 26
(multiply)
----------------------------------
3 - 6i -18 + 6i 23 -54i + 26
(subtract)
----------------------------------
3 - 6i -18 + 6i 23 -54i + 26
(multiply)
----------------------------------
3 - 6i -18 + 6i 23 -54i + 26
(subtract)
----------------------------------
0 0 0
The result of the synthetic division is 0, indicating that the factor (x - 2+3i)(x - 2-3i) is a factor of the given polynomial.
Now, we are left with the quotient polynomial, which is 3x^2 - 18x + 23 - 54i.
To find the remaining roots, we can use the quadratic formula since it is a quadratic polynomial. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic polynomial 3x^2 - 18x + 23 - 54i, the coefficients are:
a = 3
b = -18
c = 23 - 54i
Now, we can plug these values into the quadratic formula and solve for x.
x = (-(-18) ± √((-18)^2 - 4(3)(23 - 54i))) / (2 * 3)
= (18 ± √(324 - 4(3)(23 - 54i))) / 6
= (18 ± √(324 - 4(69 - 162i))) / 6
= (18 ± √(324 - 4(69) - 4(-162i))) / 6
= (18 ± √(324 - 276 + 648i)) / 6
= (18 ± √(48 + 648i)) / 6
= (18 ± √(48) √(1 + 648i/48)) / 6
= (18 ± 4√(3) (1 + √(3)i)) / 6
= (3 ± 2√(3) (1 + √(3)i)) / 3
So, the other roots of the polynomial 3x^4 - 12x^3 + 41x - 8x + 26 that include 2-3i are:
x1 = 2-3i
x2 = 2+3i
x3 = (3 + 2√(3) (1 + √(3)i)) / 3
x4 = (3 - 2√(3) (1 + √(3)i)) / 3