Question

x^3-12x^2+32x=0

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

noted : )
ah oui, the fundamental theory of algebra

Answer 5

still dont get it

Answer 6

how did u get 0 4 8?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

thanks all check it out

Answer 16

good luck