Find x, if, x^3 -3x-2=0!?

Question

I mean x^3 - 3x -2 = 0
First correct, full solution, get 10 points!

Answer 1

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

x+1 is a factor

By division:
x^3 - 3x - 2
= (x+1)(x^2 - x - 2)
= (x+1)(x^2 - 2x + x - 2)
= (x+1)(x(x-2)+1(x-2))
= (x+1)(x+1)(x-2)
= (x-2)(x+1)^2

x = {-1, -1, 2} Double root at x=-1

Answer 2

X 3-3x 2

Answer 3

OK, I'm guessing you don't want Tartaglia's formula for a cubic. This one can be solved pretty easily by inspection, -1 is a solution since
(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0.

After that, you divide x^3-3x-2 by x+1 to see what's left, you get x^2 - x - 2, and that factors as (x+1)(x-2), so you get x = -1 (again) and x = 2.

By the way, the rational root theorem says the only rational numbers that could work are 1, 2, -1 or -2. Then you can check those until you found one, and go from there. (that's how I found x=-1 in the first place!)

Answer 4

x^3 -3x-2=0

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

x = 2 or x = -1

Answer 5

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RE:
Find x, if, x^3 -3x-2=0!?

Answer 6

Since x=2 is one obvious answer (or, if not obvious, can be found with factor theorem), you know that (x-2) is a factor, so you can divide your original function by (x-2) to see if you get any other answers.

You then get that

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

So the other answer is x=-1, and there are no others.

x= 2 and x = -1

Answer 7

f(-1) = - 1 + 3 - 2 = 0
Thus x - (-)1 is a factor
x + 1 a factor
Find other factors by synthetic division:-
-1|1__0___-3___-2
_|___-1____1___2
_|1__-1___-2___0

(x + 1) (x² - x - 2) = 0
(x + 1) (x - 2) (x + 1) = 0
(x + 1)² (x - 2) = 0
x = - 1 , x = 2

Answer 8

x^3 -3x-2=0
=>
x^3 -x-2x-2=0
=>
x*(x^2-1)-2*(x+1)=0
x*(x+1)*(x-1)-2*(x+1)=0
[x*(x+1)-2]*(x+1)=0
(x^2+x-2)*(x+1)=0
=>
(x^2+x-2)*=0 or (x+1)=0=>x=-1

d=1-4(-2)=9

x=(-1+-3)/2=-2 or 1


so you have x=1 or x=-2 or x=-1

Answer 9

huh!

x^3-3x-2=0
x(x^2-3)-2=0
(x-2)= 0
x = 2
________

(x^2-3) = 0
x^2 = 3
x = sq root (3)

So x =2 or x= sq root (3)

Answer 10

The answer is x=2... just substitue values until you find one which works....There are other more general method,s but they arent necessary here.