a) the equation 12x^3-23x^2-3x+2=0 has a solution of x=2. find all other solutions.
b) solve for x, the equation, 12x^3+8x^2-x-1=0 (all solutions are rational and between +/- 1)
okay so im in calculus and i cant remember anything ( i havent taken math in like a year) so i feel retarded and ive been working on these problems for a while and im just frazzled at this point.. help me out please:/ if you could show work itd be really appreciated so i can see how you got your answer. thank you so much to any one who answers my questions.
b) solve for x, the equation, 12x^3+8x^2-x-1=0 (all solutions are rational and between +/- 1)
okay so im in calculus and i cant remember anything ( i havent taken math in like a year) so i feel retarded and ive been working on these problems for a while and im just frazzled at this point.. help me out please:/ if you could show work itd be really appreciated so i can see how you got your answer. thank you so much to any one who answers my questions.
-
factorize the polynomilas and set each factor equal to zero
(12x^3-23x^2-3x+2) = (x -2)(4x -1)(3x +1)
solutions are x = 2, 1/4, -1/3
12x^3+8x^2-x-1 =(2x +1)(3x -1)(2x +1)
solutions are x = -1/2, -1/2, 1/3
(12x^3-23x^2-3x+2) = (x -2)(4x -1)(3x +1)
solutions are x = 2, 1/4, -1/3
12x^3+8x^2-x-1 =(2x +1)(3x -1)(2x +1)
solutions are x = -1/2, -1/2, 1/3