Maths Topics and Concepts

Firstly, factorisation can be solve through grouping, cross-multiplication method, algebraic rules, as all of you should have known and figure it out by now, and lastly, the most easiest method, through highest common factor.

We will just recap how to factorise through grouping. When given this question: 3x + 12x - 6, we may factorise like this: 3(x + 4x - 2). This is same for those with 4 values and more, and it is when there is a no need to put the like terms together.

When in such circumstances when we need to put the like terms together, then the question will be: 4x + 3 + 9z + 8y. In this case, we have to regroup so that we can factorise. There should be more information about grouping when we learn in lesson, this is a summary, like from this question below:

2xy - 5y - 15 + 6x = y(2x - 5) - 3(5 - 2x)

                            = y(2x - 5) + 3(2x - 5)

                            = (y - 3)(2x - 5) 

Now, for the highest common factor, it is definitely more easy to factorise. We do not really have to group the like terms together or rather the terms which have a common factor together in order to factorise. No matter how many terms there are, which is usually 3, we still will factorise using the same method. When they give this question: 4x + 16y+ 12z, then it will become: 4(x + 4y + 3z).

Now, for the cross multiplication method, it will become more easier if you master it and keep doing practice. Firstly, we have to find the 2 factors that can multiply to form the answer as the one with the x², y², z² and so on and so forth. We have to find the 2 factors which can multiply together to form the answer of the constant. Then, what we will do is draw the table that can help us try to find the value of the number with normally a coefficient and a algebraic term together. Like, for example, if given this question: x² + 5x + 6, how do we solve it?

Factors of x² is only: x multiply to x.         

Factors of 6 is: 1 x 6, 2 x 3, -1 x -6, -2 x - 3 

I cannot draw the table over here but I will try my best to do so:

x² = x*x         6 = 2 x 3  

When we multiply x to 2 and x to 3, we will get 2x and 3x.

When we add 2x and 3x together, we will get 5x,correct. 

So, the answer will be (x + 2)(x + 3), table.

Lastly, is the method of algebraic rules. There are 3 algebraic rules which we had learnt, and they are: a² + 2ab + b², a² - 2ab + b², a² - b². For these rules:

a² + 2ab + b² = (a + b)²

a² - 2ab + b² = (a - b)²

a²- b² = (a + b)(a - b)

So, when we factorise, it is the other way of expand, totally opposite. We must look at the algebraic rules and see whether which is the one that looks like our questions, or the expression we did work out. Then, we factorise according to the algebraic rules.

That is it for this topic on solving quadratic expressions and simplifying through factorisation. I think what we should consider is that all these may somewhat be applied on the addition and subtraction of fractions, algebraic fractions. And we must be able to do the fractions and master our fractions well before we do this and go into that particular topic.

So, I can see that these topics may be the combination of other smaller topics and these topics are likely to be tested in the tests. I will try to put more revision cards for myself so that I will not go confused and be unable to understand the main concepts, or sometimes if I do understand, I will clarify, but I must also not forget my basic rules, especially the algebraic rules, as mentioned above.