how do I factor 2xy-12xz+3y-18z compleyely?

Answer 1

2xy-12xz+3y-18z= -6z(2x+3)+y(2x+3)=(2x+3)(y-6z)

Answer 2

Group and factor and inspect.

Note that in the first 2 terms, you can factor out x and a factor with a y and a z.

And the 3rd and 4th term have a y and z term. This gives you a clue that they can have a common factor by inspection

as, 2x(y-6z) + 3(y-6z).. Now since y-6z is a common factor,

you can rewrite it as (y-6z) (2x +3).

Now, it is always a good idea to check your work by FOILing. You should get the original expression.

Answer 3

By GRouping ( put two sets to a group - get two groups)

(2xy-12xz)+(3y-18z)
2x(y-6z)+3(y-6z)

Pull common factor (y-6z)

(2x+3)(y-6z)

Answer 4

Start by factoring a 2x out of the first two terms:
2x(y - 6z) + 3y - 18z

Now factor a 3 out of the last two terms:
2x(y - 6z) + 3(y - 6z)

Now notice the common (y - 6z) and factor that out:
(2x + 3)(y - 6z)

Answer 5

group into 2 groups

(2xy + 3y) - (12xz + 18z)

fator out the commons,

y(2x + 3) - 6z(2x + 3)

(2x + 3) is now the common, factor it out

therefore,

(2 - 6z) ( 2x + 3). Ans.

Answer 6

Here, you have a potential pattern in the four terms some multiple of y - some multiple of z, some multiple of y - some multiple of z. So let's find the multiples in that pattern. 2x divides both of the first terms, and 3 divides both of the latter terms. putting this together, your expression reads:

2x(y-6z) + 3(y-6z).

Now, we see that (y-6z) divides the entire expression, and the complete factorization is:

(2x+3)(y-6z).