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Grade 10

Grouping in pairs


Sometimes the common factor that we take out when factoring is a lot more complicated than just a single algebraic term. Have a look at this example: $2\left(x+1\right)-x\left(x+1\right)$2(x+1)x(x+1). In algebra we kind of think of 'terms' as chunks that are separated by addition and subtraction symbols, so in our example here we actually have two terms: $2\left(x+1\right)$2(x+1) and $-x\left(x+1\right)$x(x+1). That $x+1$x+1 is quite pesky and complicated isn't it? Try to think of it as another variable, completely different from $x$x to simplify things. In fact, let's replace it with something like $y$y! Now we have $2y-xy$2yxy, which is easily factored into $y\left(2-x\right)$y(2x). So that must mean $2\left(x+1\right)-x\left(x+1\right)$2(x+1)x(x+1) can be factored into $\left(x+1\right)\left(2-x\right)$(x+1)(2x)!

Can you see that $x+1$x+1 here is a common factor of both terms even though it's a bracketed term? This is important as factoring in these examples is all about finding common factors that may be in brackets.


Grouping in Pairs

The above technique is very useful in certain situations when factoring four terms. Let's take a look at the example $14-7v+4u-2uv$147v+4u2uv. There is no way we could factor that by finding an GCF of all four terms! But let's have a look at what factors there are in each term anyway:

Can you see that $14$14 and -$7v$7v have an GCF of $7$7, while $4u$4u and $2uv$2uv have an GCF of $2u$2u? That means we can group these four terms into two pairs and factor them separately. Let's see what happens!

$14-7v+4u-2uv$147v+4u2uv $=$= $\left(14-7v\right)+\left(4u-2uv\right)$(147v)+(4u2uv)
  $=$= $7\left(2-v\right)+2u\left(2-v\right)$7(2v)+2u(2v)

Wow, this is very similar to the example we had before, where we can take out a bracketed term to factor! Let's try that technique!


Isn't that amazing? We managed to factor four terms with no common factors completely! The key when grouping in pairs is to make sure that each pair can be factored separately so we end up with $2$2 terms, and then hopefully you can factor a second time afterwards to get $1$1 term.


Tricky negatives

Sometimes we need to think in terms of negatives to factor these kind of expressions. Something like $3\left(y-1\right)-10y\left(1-y\right)$3(y1)10y(1y) may not look like it can be factored, but that $y-1$y1 and $1-y$1y look quite similar don't they? In fact, $1-y$1y is equal to $-1\times\left(y-1\right)$1×(y1)! Let's see if we can use this to rewrite our expression.

$3\left(y-1\right)-10y\left(1-y\right)$3(y1)10y(1y) $=$= $3\left(y-1\right)-10y\times\left(-1\right)\times\left(y-1\right)$3(y1)10y×(1)×(y1)
  $=$= $3\left(y-1\right)+10y\left(y-1\right)$3(y1)+10y(y1)
This can then   be  factored using our usual approach:
  $=$= $\left(y-1\right)\left(3+10y\right)$(y1)(3+10y)

So when you see two binomials that are exactly the same but with inverted signs, you can use our trusty friend $-1$1 to help you out!

Worked Examples

Question 1

Factor the following expression:


Question 2

Factor the following expression:


Question 3

Using the fact that $A\left(B-C\right)=-A\left(C-B\right)$A(BC)=A(CB), or otherwise, factor $5x-xy+y^2-5y$5xxy+y25y.




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