Question

x^3-4x^2+4x=0 Solve for x

Answer 1

Have you considered factoring?

Answer 2

\[x(x^2-4x+4) = x(x-2)^2 so x = 0, 2\]

2 is a solution with a mutliplicity of 2 meaning that 2 is a solution twice

Answer 3

so then x =2

Answer 4

I don't know where to begin I need step by step instructions

Answer 5

Will someone please help me

Answer 6

Let me type it out for you

Answer 7

ok

Answer 8

so first you factor out an x like I did above so you get
\[x(x^2−4x+4)\]
then you factor by grouping in this case you get
\[x(x-2)^2\]
set each of those equal to 0
\[x=0\] & \[(x-2)^2=0\]
solve
\[x=0,2\]
and 2 with a multiplicity of 2 meaning that 2 is a solution twice.

Make sense?

Answer 9

yes thank you

Answer 10

Glad to help! :)