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.

Examples

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

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.

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