# Extraction and Grouping Common Factors help!

• January 11th 2009, 12:38 PM
Mathtard
Extraction and Grouping Common Factors help!
Hey guys, I am really confused with this work, I missed some time from college as I was ill and missed the work. Can anyone help me?

Question says "Factorise the following expression by extraction and grouping of common factors". Could someone please explain how to to this for me? Thank you!

$2a (x+5) +b(x+4)+2ax+3bx$
• January 11th 2009, 05:14 PM
stapel
Quote:

Originally Posted by Mathtard
Question says "Factorise the following expression by extraction and grouping of common factors". Could someone please explain how to to this for me? Thank you!

$2a (x+5) +b(x+4)+2ax+3bx$

My guess that they want you to factor "in pairs":

. . . . .4x^2 + 3xy + 8x + 6y

. . . . .4x^2 + 8x + 3xy + 6y

. . . . .4x(x + 2) + 3y(x + 2)

. . . . .(x + 2)(4x + 3y)

In your case, though, you'll be able to simplify inside at least one of the sets of parentheses. :wink:

Have fun! :D
• January 12th 2009, 04:57 AM
Factorising
[/quote]
Quote:

Originally Posted by stapel
. . . . .4x^2 + 3xy + 8x + 6y

. . . . .4x^2 + 8x + 3xy + 6y

. . . . .4x(x + 2) + 3y(x + 2)

. . . . .(x + 2)(4x + 3y)

In your case, though, you'll be able to simplify inside at least one of the sets of parentheses. :wink:

Have fun! :D

3xy on the second line (as I have amended above).

In case you're still not sure what to do:

$2a(x+5)$ and $2ax$ have a common factor $2a$, so you can 'take it out' and get:

$2a(x+5) + 2ax = 2a((x+5) + x)$

$= 2a(x+5+x)$

$= 2a(2x+5)$

Now do something similar with $b(x+4)$ and $3bx$, simplifying the terms inside the brackets (parentheses) as far as you can.

But there isn't then a further factorising that you can do, as stapel has been able to do in the example he gave you. You will have to leave the answer as two separate pairs of brackets with an ' $a$' term outside the first and a ' $b$' term outside the second.