Answer:
x = 2, 1 + 3i, 1 − 3i
Step-by-step explanation:
Find the Roots (Zeros)
x^4 − 6x^3 + 22x^2 − 48x + 40
Set x^4 − 6x^3 + 22x^2 − 48x + 40 equal to 0. x^4 − 6x^3 + 22x^2 − 48x + 40 = 0
Solve for x.
Factor the left side of the equation.
Factor x^4 − 6x^3 + 22x^2 − 48x + 40 using the rational roots test.
(x − 2) (x^3 − 4x^2 + 14x − 20) = 0
Factor x^3 − 4x^2 + 14x − 20 using the rational roots test.
(x − 2) (x − 2) (x2 − 2x + 10) = 0
Combine like factors.
(x − 2)2 (x^2 − 2x + 10) = 0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
(x − 2)^2 = 0
x^2 − 2x + 10 = 0
Set (x − 2)^2 equal to 0 and solve for x.
Set (x − 2)^2 equal to 0.
(x − 2)^2 = 0
Solve (x − 2)^2 = 0 for x.
x = 2
Set x^2 − 2x + 10 equal to 0 and solve for x.
Set x^2 − 2x + 10 equal to 0. x^2 − 2x + 10 = 0
Solve x^2 − 2x + 10 = 0 for x.
Use the quadratic formula to find the solutions.
−b ± (√b^2 − 4 (ac) )/2a
Substitute the values a = 1, b = −2, and c = 10 into the quadratic formula and solve for x.
2 ± (√(−2)^2 − 4 ⋅ (1 ⋅ 10))/2 ⋅ 1
Simplify.
Simplify the numerator.
x = 2 ± 6i/ 2.1
Multiply 2 by 1
x = 2 ± 6i/2⋅1
Simplify
2 ± 6i/2
x = 1 ± 3i
The final answer is the combination of both solutions.
x = 1 + 3i, 1 − 3i
The final solution is all the values that make (x − 2)2 (x2 − 2x + 10) = 0 true.
x = 2, 1 + 3i, 1 − 3i
![\bf cos({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[sin({{ \alpha}}+{{ \beta}})\quad -\quad sin({{ \alpha}}-{{ \beta}})]](https://tex.z-dn.net/?f=%5Cbf%20cos%28%7B%7B%20%5Calpha%7D%7D%29sin%28%7B%7B%20%5Cbeta%7D%7D%29%3D%5Ccfrac%7B1%7D%7B2%7D%5Bsin%28%7B%7B%20%5Calpha%7D%7D%2B%7B%7B%20%5Cbeta%7D%7D%29%5Cquad%20-%5Cquad%20sin%28%7B%7B%20%5Calpha%7D%7D-%7B%7B%20%5Cbeta%7D%7D%29%5D)