HELP! please! so confused:(
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HELP! please! so confused:(

[From: ] [author: ] [Date: 11-09-07] [Hit: ]
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.solutions are x = 2, 1/4,solutions are x = -1/2, -1/2,......
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.

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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
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