# Thread: Grouping Method For Factoring

1. ## Grouping Method For Factoring

Factor By grouping.

(u + v)x - xy + (u + v)^2 - (u + v)y

(u + v)[x - xy + (u + v) - y]

(u + v)[x (1 - y) + (u + v) - y]

Is this correct?

2. ## Re: Grouping Method For Factoring

Factor By grouping.

(u + v)x - xy + (u + v)^2 - (u + v)y

(u + v)[x - xy + (u + v) - y]

(u + v)[x (1 - y) + (u + v) - y]

Is this correct?
No.

The first pair share an x, so

(u + v)x - xy = x[(u + v) - y].

The second pair share (u + v), so

(u + v)^2 - (u + v)y = (u + v)[(u + v) - y].

Putting those together, you have

(u + v)x - xy + (u + v)^2 - (u + v)y =

x[(u + v) - y] + (u + v)[(u + v) - y] =

x(u + v - y) + (u + v)(u + v - y)

What can you factor out next?

3. ## Re: Grouping Method For Factoring

Originally Posted by greg1313
No.

The first pair share an x, so

(u + v)x - xy = x[(u + v) - y].

The second pair share (u + v), so

(u + v)^2 - (u + v)y = (u + v)[(u + v) - y].

Putting those together, you have

(u + v)x - xy + (u + v)^2 - (u + v)y =

x[(u + v) - y] + (u + v)[(u + v) - y] =

x(u + v - y) + (u + v)(u + v - y)

What can you factor out next?
x(u + v - y) + (u + v)(u + v - y)

I can now factor out (u + v - y).

[x + (u + v)][u + v - y]

4. ## Re: Grouping Method For Factoring

Note: I left the site yesterday until now.

x(u + v - y) + (u + v)(u + v - y)

I can now factor out (u + v - y).

[x + (u + v)][u + v - y]
Almost finished. Good work.

Remove the parentheses. Exchange the square brackets for parentheses:

(x + u + v)(u + v - y)

Good work.

5. ## Re: Grouping Method For Factoring

Originally Posted by greg1313
Note: I left the site yesterday until now.

Almost finished. Good work.

Remove the parentheses. Exchange the square brackets for parentheses:

(x + u + v)(u + v - y)

Good work.
Thanks. Glad to know that I almost got it right.