solve for x: x^3-x^2 -4x +4 =0?

x^3 - x^2 - 4x + 4 = 0
x^2(x-1) - 4(x-1) = 0
(x^2 - 4)(x-1) = 0
(x-2)(x+2)(x-1) = 0
x-2 = 0, x+2 = 0, x-1 = 0
x = 2, x = -2, x = 1

first yu check some numbers like 1 or 2 so yu will got 2 is answer after that yu must divide yur problem to X-2 that is:
x^3-x^2 -4x +4 =(x-2)*(x^2+x-2) so the answers are -2 ,2,1

f (1) = 1 - 1 - 4 + 4 = 0
Thus (x - 1) is a factor .
To find other factors , use synthetic division:-
1 |1____-1___ - 4_____- 4
_ |______1____0______4
_ |1_____0___- 4______0

(x - 1)(x² - 4) = 0
(x - 1)(x - 2)(x + 2) = 0
x = 1 , x = 2 , x = - 2

x^3 - x^2 - 4x + 4 = 0
(x^3 - x^2) - (4x - 4) = 0
x^2(x - 1) - 4(x - 1) = 0
(x - 1)(x^2 - 4) = 0

x - 1 = 0
x = 1

x^2 - 4 = 0
x^2 = 4
x = ±√4
x = ±2

∴ x = 1 , ±2

A. x^3 goes to infinity as x-->infinity and to negative infinity as x-->negative infinity. B. x^3 +x^2 -4x -4 = (x^2 -4)(x+1) = 0 if x = -1, -2, 2. Crosses the x axis at each of these. C. (0,-4) D. Neither E. Turns between -1 and -2. Turns between 0 and 1.

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RE:
solve for x: x^3-x^2 -4x +4 =0?

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

x^3-x^2 -4x +4 =0
(x - 1)(x² - 4) = 0
(x - 1)(x - 2)(x + 2) = 0
x = 1 , x = 2 , x = - 2

break apart in the middle and such, then you'll get -2, 1, and 2.

i think x = 4/3
but i only passed algebra with a 71 for the year
and that was the 2nd time i took it so idk