# How to factor this polynomial by grouping? -- 2xy - 9cx - 18cy + 81c2^?

Relevance
• 2xy - 9cx - 18cy + 81c^2

group the first two terms and factor out an x:

2xy - 9cx = x(2y - 9c)

group the last two terms and factor out (-9c):

-18cy + 81c^2 = -9c(2y - 9c)

Now you have:

x(2y - 9c) - 9c(2y - 9c)

notice that the binomials inside the ( )'s are the same, so take the whole binomial out as a single common factor:

(2y - 9c) (x - 9c)

hope it helps!

• Login to reply the answers
• 2xy - 9cx - 18cy + 81c^2

= (9 c - x) (9 c - 2 y)

• Login to reply the answers
•  2xy - 9cx - 18cy + 81c^2

= (9 c - x) (9 c - 2 y)

• Login to reply the answers
• I don't get what -- is supposed to mean.

You have 3 variables:  x , y , c.  If we group the first 2 terms and the last 2 terms, we get:

x * (2y - 9c) - c * (18y - 81c) =>

x * (2y - 9c) - 9c * (2y - 9c) =>

(x - 9c) * (2y - 9c)

If we group the 1st and 3rd term and the 2nd/4th term, we get:

2xy - 18cy - 9cx + 81c^2 =>

2y * (x - 9c) - 9c * (x - 9c) =>

(2y - 9c) * (x - 9c)

Which one do you like better?

• Login to reply the answers