anonymous 5 years ago x^3-12x^2+32x=0

1. anonymous

x(x^2 - 12x + 32) = 0 x(x - 4)(x - 8) = 0 x = 0, 4, 8

2. Regan

divide by x, then solve the remaining quadratic. x^2-12x+32=0=(x-8)(x-4) x=8,2

3. anonymous

regan, you have to account for the fact that you divided by x, meaning that x could be zero as well you see this is true if you plug zero back into the equation, you'll get 0 = 0, which is true, hence, x = 0 is also a solution to this equation

4. Regan

noted : ) ah oui, the fundamental theory of algebra

5. anonymous

still dont get it

6. anonymous

how did u get 0 4 8?

7. anonymous

x(x^2 - 12x + 32) = 0 x(x - 4)(x - 8) = 0 do you get this part?

8. anonymous

after that part you just set each part equal to 0 so x = 0 x - 4 = 0 x - 8 = 0 solve for each x x = 0, 4, 8

9. anonymous

No i dont get how u got the x(x-4) and (x-8)?

10. anonymous

x(x^2 - 12x + 32) = 0 this part? i factored out the x to get to this part x(x - 4)(x - 8) = 0 you have to factor do you know how to factor?

11. anonymous

so how would u do 4x^3-x=0

12. anonymous

Ya i know how 2 factor i jus dont get how x factored 12 and got 4?

13. anonymous

factor out the x x(4x^2 - 1) = 0 x = 0 4x^2 - 1 = 0 4x^2 = 1 x^2 = 1/4 $x = \pm 1/2$

14. anonymous

i factored x out of x^3-12x^2+32x=0 to get x(x^2 - 12x + 32) = 0 than i factored INSIDE the parenthesis to get x(x - 4)(x - 8) = 0 go to this site http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx it teaches you how to factor

15. anonymous

thanks all check it out

16. anonymous

good luck

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