Polynomial (2 Viewers)

[TNguyen]

Note that the following equations have symmetrical coefficients. Solve these equations (hint: divide by x^2 and reduce to a quadratic).

3x^4 - x^3 + 4x^2 - x + 3 =

Thanks for helping

 

[NT-social]

3x^4 - x^3 + 4x^2 - x + 3 = 0

group 3x^4 + 4x^2+ 3 together and the remainign together

(3x^2 + 3)(x^2 + 1) - x(x^2 + 1) = 0
(x^2 + 1)(3x^2 - x + 3) = 0

x= (+/-) i , x = [1 (+/-) squareroot(-35)] / 2

 

[echelon4]

3x^4 - x^3 + 4x^2 - x + 3 = 0

group 3x^4 + 4x^2+ 3 together and the remainign together

(3x^2 + 3)(x^2 + 1) - x(x^2 + 1) = 0
(x^2 + 1)(3x^2 - x + 3) = 0

x= (+/-) i , x = [1 (+/-) squareroot(-35)] / 2

(3x^2 + 3)(x^2 + 1)=3x^4 + 6x^2 + 3 and not the 3x^4 + 4x^2 + 3 you were looking for

 

[echelon4]

i used the hint and divided the pol by x^2, and then used x=cis@ , and by using X^n+X^(-n)= 2cosn@

x=-1/3 + [(squareroot8)i]/3,

x=-1/3 - [(squareroot8)i]/3

x=1/2 + [(sqaureroot3)/2]i ,

x=1/2 - [(sqaureroot3)/2]i

 

[jyu]

Note that the following equations have symmetrical coefficients. Solve these equations (hint: divide by x^2 and reduce to a quadratic).

3x^4 - x^3 + 4x^2 - x + 3 =

Thanks for helping

:wave:

 

[conman]

Hey jyu I think you made mistake from the third line to the fourth line.

 

[jyu]

Hey jyu I think you made mistake from the third line to the fourth line.

Please correct my mistakes.

 

[toadstooltown]

No, jyu is correct. However, even though we know x=0 isn't a solution, it's still more 'mathematically correct' to take out x² common, rather than divide by it.
ie.
3x^4-x³+4x²-x+3
x²(3x²-x+4-1/x+3/x²)
..
...
x²(3(x+1/x)²-(x+1/x)-2)
x²(3(x+1/x)+2)(x+1/x-1)
(3x²+2x+3)(x²-x+1)
...
...
...
x= jyu's answers

 

[conman]

Sorry my bad, he is correct. But I think I refer to use De Moiver thereom to solve it any way (fast and accurate)