Sometimes the common factor that we take out when factorising 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$2y−xy, which is easily factorised into $y\left(2-x\right)$y(2−x). So that must mean $2\left(x+1\right)-x\left(x+1\right)$2(x+1)−x(x+1) can be factorised into $\left(x+1\right)\left(2-x\right)$(x+1)(2−x)!
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 factorisation in these examples is all about finding common factors that may be in brackets.
The above technique is very useful in certain situations when factorising four terms. Let's take a look at the example $14-7v+4u-2uv$14−7v+4u−2uv. There is no way we could factorise that by finding an HCF 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 HCF of $7$7, while $4u$4u and $2uv$2uv have an HCF of $2u$2u? That means we can group these four terms into two pairs and factorise them separately. Let's see what happens!
$14-7v+4u-2uv$14−7v+4u−2uv | $=$= | $\left(14-7v\right)+\left(4u-2uv\right)$(14−7v)+(4u−2uv) |
$=$= | $7\left(2-v\right)+2u\left(2-v\right)$7(2−v)+2u(2−v) |
Wow, this is very similar to the example we had before, where we can take out a bracketed term to factorise! Let's try that technique!
$7\left(2-v\right)+2u\left(2-v\right)=\left(2-v\right)\left(7+2u\right)$7(2−v)+2u(2−v)=(2−v)(7+2u)
Isn't that amazing? We managed to factorise four terms with no common factors completely! The key when grouping in pairs is to make sure that each pair can be factorised separately so we end up with $2$2 terms, and then hopefully you can factorise a second time afterwards to get $1$1 term.
Sometimes we need to think in terms of negatives to factorise these kind of expressions. Something like $3\left(y-1\right)-10y\left(1-y\right)$3(y−1)−10y(1−y) may not look like it can be factorised, but that $y-1$y−1 and $1-y$1−y look quite similar don't they? In fact, $1-y$1−y is equal to $-1\times\left(y-1\right)$−1×(y−1)! Let's see if we can use this to rewrite our expression.
$3\left(y-1\right)-10y\left(1-y\right)$3(y−1)−10y(1−y) | $=$= | $3\left(y-1\right)-10y\times\left(-1\right)\times\left(y-1\right)$3(y−1)−10y×(−1)×(y−1) |
$=$= | $3\left(y-1\right)+10y\left(y-1\right)$3(y−1)+10y(y−1) | |
This can then | be | factorised using our usual approach: |
$=$= | $\left(y-1\right)\left(3+10y\right)$(y−1)(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!
Factorise the following expression:
$50+5y+10x+xy$50+5y+10x+xy
Factorise the following expression:
$2x+xz-40y-20yz$2x+xz−40y−20yz
Using the fact that $A\left(B-C\right)=-A\left(C-B\right)$A(B−C)=−A(C−B), or otherwise, factorise $5x-xy+y^2-5y$5x−xy+y2−5y.
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
Apply algebraic procedures in solving problems