How do i solve this? 6z³-27z²+12z=0 ?

Question

I know to factor out the 3z but i get confused after that

Answer 1

Simple:

Take out 3z : 3z(2z²-9z+4) = 0

Factorise those within brackets : 3z(2z-1)(z-4)=0

Solve: z=0, z= 0.5, z= 4

Answer 2

3z(2z^2 - 9z + 4)=0

Quadratic equation.

3z(2z-1)(z-4)=0

Therefore either 2z - 1=0 or z-4 =0

Hence,

z=1/2 or 4

Answer 3

takin z common

6z(sq) - 27 z + 12 =0

taki 3 common

2z(sq) - 9z + 4 = 0

2z(sq) -8z - z + 4 =0

2z(z-4) -1 (z - 4) =0

2z-1 =0 and z-4=0 and z=0

z=1/2, z=4 and z =0

Answer 4

OK, after you take 3z out then it will become like this

3z(2z^2 - 9z + 4) = 0

Now after this you have to factorize the remaing bracket also, by splitting the middel term of -9z (way you just do it in quadratic equations), like this

3z(2z^2 - 8z - z + 4) = 0

Now take out 2z common from the first 2 terms in bracket and -1 from last 2 terms in the bracket, like this

3z.[ 2z(z- 4) - 1(z-4) ]

Now take z-4 as common factor from the bracket, you got it!

3z(z - 4)(2z -1) = 0

Now easily, z = 0, z = 4 and z = 1 / 2.

Answer 5

first u have to take 3z common

=> 3z{ 2z^2 - 9z+4} =0

=> { 2z^2 -8z -1z + 4} =0            [ 3z get cancelled  when taking to the right side]

=> { 2z(z - 4) - 1(z - 4) } = 0

=> { (2z -1)(z - 4) } = 0

=> z = 1/2 ; z = 4 ; z= 0

Answer 6

6z³-27z²+12z = 3z(2z^2 - 9z + 4) = 3z(2z-1)(z-4) = 0

z = 0, 1/2 and 4