Common factorising in pairs (grouping) – two interesting cases
When we have four terms in an expression which we would like to factorise, we can encounter two interesting cases.
Case 1 You may have to rearrange the order of the terms so that pairs of terms with common factors are beside each other. Example: ab + 2 + a + 2b . Notice the HCF of the fi rst pair is 1: ab + 2 + a + 2b
fi rst pair second pair
We can try pairing the terms without 2s and pairing the two other terms with 2s: ab + a + 2b + 2 Now a is the HCF of the fi rst pair and 2 is the HCF of the second pair: And then, fi nally, take out the common factor, (b + 1 ) :
Case 2 Ensure the common factor (bracket) in each pair is identical by altering the signs. Example: ak − at − 5k+ 5t
fi rst pair second pair
If + 5 is used as the HCF for the second pair, then (a)(k − t) + ( 5 )(− k + t) ✗
Note the second bracket in each pair is not identical. But if − 5 is used as the HCF for the second pair, then we get: (a) (k − t) + (−5 )(k − t) ✓
Then we do get a factor (bracket) identical in each of our two pairs. To complete the factorisation, we take out the common factor: (k − t)(a − 5 )
Worked example 4 Factorise fully the expression 5p − 2q − 2qs + 5ps