Last night I had a request on my Twitter account to help someone in South Africa with a factorisation problem that he in turn was trying to help his younger brother with. He followed up by sending a photo of the expression that was causing a problem to them. I looked at it and sent a photo back with reasons for each step to solving it. This was what he sent me.
In reality what he was being asked to do was to factorise the sum of two independent expressions either side of the addition symbol. Now this can reasonably be attempted by expanding brackets on both expressions, combining like terms and from there used a tried and tested method to factorise the result. The result of this approach is as follows. It is not necessarily the most efficient however.
The single expression that remains could be quite a challenge for some pupils when asked to factorise it. It would involve finding the factors of the product of the coefficient of ‘x squared’ (7) and the constant (48), so we can then find a pair of factors of 336 which sum to the coefficient of x (-37). Once this is completed and on the condition that we can factorise the expression in the first place, factorisation itself is the relatively easy part.
It is now possible to rewrite the expression using -16x and -21x instead of -37x as we all agree they amount to the same thing but when we do this, factorisations starts to be less of a challenge.
The placing of -21x and -16x is up to the problem solver but the way we have done it here helps with the factorisation process as the pairing of the coefficients 7 with -21 and -16 with 48 clearly show that they share factors. We now add brackets to the first two and last two terms of the expression.
These two sets of brackets can now be factorised.
The above factorisation tells us that x – 3 is being multiplied in turn by both 7x and also -16 and therefore we can put these together in one bracket showing that both x -3 and 7x – 16 are factors of our original expression.
At the beginning of this blog I implied that there are other ways to factorise such expressions. This relies on the ability to realise that an expression of three different terms can be factorised. Look at the original expression again.
You will see that the term x – 3 is now common to both separate expressions between the addition sign that separates them. What we can now do is rename x- 3 some factor n and then substitute into and simplify the expression.
This in turn gives an expression that pupils learning at this level should feel confident enough to factorise.
And finally, we can re-substitute the original value of n and expand where necessary. The result is the same as when using the first method.