Factorising by Grouping in Pairs
Theory
Grouping in pairs factorises an expression with four terms, such as \(xy + 3y + 2x + 6 = (x + 3)(y + 2)\). This Year 10A maths page (Victorian Curriculum 10A) shows how to take out a common bracket, how to group, rearrange and handle a negative pair, how to deal with squared terms, and how to combine grouping with the difference of two squares. Curriculum: VC2M10A04 / AC9M10A01.
A common factor can be a whole bracket. In \(a(x + 1) + 3(x + 1)\) the bracket \((x + 1)\) is in both terms, so it can be taken out: \((x + 1)(a + 3)\).
Grouping in pairs is a way to factorise an expression with four terms that have no common factor all together. Split the terms into two pairs, take the HCF out of each pair, and if the two brackets are the same, take that bracket out.
Sometimes the terms must be rearranged first so that each pair has a common factor. Sometimes the pairs are a difference of two squares and two other terms, as in \(m^2 - n^2 + 5m + 5n\).
The answer is a product of two brackets. Always check whether either bracket can be factorised again.
Grouping in pairs uses the distributive law twice:
With a negative second pair:
The difference of two squares, used when two of the terms are squares:
How to factorise four terms by grouping
- Take out any common factor of all four terms first.
- Group the terms in two pairs. Take the HCF out of each pair. If the third term is negative, take out a negative factor so the brackets can match.
- Compare the brackets. They must be identical. If they are not, rearrange the terms (swap the middle two) and try again.
- Take out the common bracket. What is left of each pair forms the second bracket.
- Factorise again if a bracket is a difference of two squares or has a common factor, then check by expanding.
When two terms are squares
- Pair the two squares together, such as \(a^2 - 9b^2\), and factorise them as a difference of two squares.
- Factorise the other pair so that it contains one of those brackets, then take out the common bracket (Example 4).
- Three terms and one: if three terms form a perfect square, write the expression as a difference of two squares (see Formulas).
(i) \(4a(b + 5) - 3(b + 5)\)
(ii) \(pq + 4p + 3q + 12\)
(i) Both terms contain the bracket \((b + 5)\). Take it out; what is left goes in the second bracket.
\(4a(b + 5) - 3(b + 5)\)
| \(\) | \(=\) | \(\boldsymbol{(b + 5)(4a - 3)}\) |
(ii) Group the first two terms and the last two terms, and take the HCF out of each pair.
\(pq + 4p + 3q + 12\)
| \(\) | \(=\) | \(p(q + 4) + 3(q + 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(q + 4)(p + 3)}\) |
Check: \((q + 4)(p + 3) = pq + 3q + 4p + 12\). ✓
(i) \(2mn - 8m - n + 4\)
(ii) \(3x + 2y + xy + 6\)
(i) The second pair starts with a minus, so take out \(-1\). Then the brackets match.
\(2mn - 8m - n + 4\)
| \(\) | \(=\) | \(2m(n - 4) - 1(n - 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(n - 4)(2m - 1)}\) |
(ii) The pairs \(3x + 2y\) and \(xy + 6\) have no common factor. Rearrange so that each pair shares a pronumeral.
\(3x + 2y + xy + 6\)
| \(\) | \(=\) | \(xy + 3x + 2y + 6\) |
| \(\) | \(=\) | \(x(y + 3) + 2(y + 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(y + 3)(x + 2)}\) |
(i) \(2x^3 - 6x^2 + 5x - 15\)
(ii) \(3x^2 + 6xy + 2x^3 + 4x^2y\)
(i) There is no HCF for all four terms, so group in pairs.
\(2x^3 - 6x^2 + 5x - 15\)
| \(\) | \(=\) | \(2x^2(x - 3) + 5(x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 3)(2x^2 + 5)}\) |
\(2x^2 + 5\) is a sum, not a difference, so it does not factorise further.
(ii) Every term contains \(x\), so take out \(x\) first. Then group the four terms left in the bracket.
\(3x^2 + 6xy + 2x^3 + 4x^2y\)
| \(\) | \(=\) | \(x(3x + 6y + 2x^2 + 4xy)\) |
| \(\) | \(=\) | \(x\big[3(x + 2y) + 2x(x + 2y)\big]\) |
| \(\) | \(=\) | \(\boldsymbol{x(x + 2y)(3 + 2x)}\) |
Group the two squares together and the other two terms together. Each pair has a factor \((a - 3b)\).
\(a^2 - 9b^2 + 2a - 6b\)
| \(\) | \(=\) | \((a - 3b)(a + 3b) + 2(a - 3b)\) |
| \(\) | \(=\) | \((a - 3b)\big[(a + 3b) + 2\big]\) |
| \(\) | \(=\) | \(\boldsymbol{(a - 3b)(a + 3b + 2)}\) |
Grouping \(a^2 + 2a\) and \(-9b^2 - 6b\) gives \(a(a + 2) - 3b(3b + 2)\), which does not share a bracket. Try another pairing when that happens.
Common pitfalls
Frequently asked questions
How do you factorise four terms?
Group them in pairs so that each pair has a common factor, take the HCF out of each pair to get the same bracket twice, then take that bracket out.
What if the two brackets are different?
Rearrange the terms, usually by swapping the middle two, and group again. Also check the sign you took out of the second pair.
Is \(a(x + y) + b(x + y)\) factorised?
No. It is still a sum of two terms. Taking out the common bracket gives the factorised form \((x + y)(a + b)\).
What if two of the terms are squares?
Pair the squares as a difference of two squares, factorise the other pair, then look for a common bracket. If three terms form a perfect square, write the whole expression as a difference of two squares.
How can I check my answer?
Expand the two brackets. You should get back all four terms of the original expression.