Home > Algebra > Factorising into a single bracket

Factorising into a single bracket

When factorising into a single bracket we divide out the highest common factor (H.C.F.) of each term.  This H.C.F. becomes the brackets multiplier and the result of the division becomes the contents of the bracket.


3x+6 \rightarrow 3(x+2)

Here we can see that 3 divides into both terms, but x does not so the most we can divide out is 3.

3x^2-5x \rightarrow x(3x-5)

With this expression there was no number that could be taken out of all terms, but there was an x in both

6x^3-9xy \rightarrow 3x(2x^2-3y)

This time we can divide both terms by 3 and x.  We don’t have to check for y since it is not in the first term

4x^2y+8xy^2-6xy \rightarrow 2xy(2x+4y-3)

Here we can divide out 2, x and y from all three terms.  The number of terms does not effect the problem provided that the factor that is divided out is common to all three.

4(p+q)+(p+q)^2 \rightarrow (p+q)(4+(p+q))=(p+q)(4+p+q)

In the strange example above, both terms had a factor (p+q), therefore we could divide this out.  Make sure you can follow this example, we will be using this idea again in  a later post.

3(x+y)^2+6x(x+y)^3 \rightarrow 3(x+y)^2(1+2x(x+y))=3(x+y)^2(1+2x^2+2xy))

Categories: Algebra
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: