O-Level Additional Math

[Herbie]

Hi! Are there rules in solving qn on factorisation?? Tq

[Herbie]

(3t+4s)(6q-7r)-(t-3s)(7r-6q)

Can some show the steps to such qn? Tq

[ADoc]

Herbie:

Hi! Are there rules in solving qn on factorisation?? Tq

Hi there! There aren't any rules really, just a few common techniques and of cos experience, if that's what your question is asking. For lower Secondary, [1] factorisation by groups, [2] factorisation by extracting common factors. And finally proceeding onto quadratic/cubic factorisation, and higher orders of factorisation. All of which are simply based on observing and extracting common factors.

A key use of factorisation is to solve quadratic and/or cubic equations for Sec syllabus. While factor and remainder theorems are required topics, most students have opted for a much easier and surest technique, that's using the calculator to solve and penning down the workings a priori.

[ADoc]

Herbie:

(3t+4s)(6q-7r)-(t-3s)(7r-6q)

Can some show the steps to such qn? Tq

Hi there are a few variations. Here's one. Do note the steps aren't necessarily this long. Just wanna be clear every step of the way for clarity.

(3t+4s)(6q-7r)-(t-3s)(7r-6q)
= (3t+4s)(6q-7r)-(t-3s)(-1)(-7r+6q) [extract the common factor of negative 1; this is an important technique so as to switch the sign from plus to minus and vice versa]

= (3t+4s)(6q-7r)+(t-3s)(-7r+6q) [negative-negative --> positive]
= (3t+4s)(6q-7r)+(t-3s)(6q-7r) [rewrite as 6q-7r]
= (6q-7r)(3t+4s+t-3s) [extract common factor of 6q-7r]
= (6q-7r)(4t+s)

[Herbie]

hi adot, thanks for ur explamation. can give example on the factorisation by group and factors? Can? Yq

[ADoc]

Herbie:

hi adot, thanks for ur explamation. can give example on the factorisation by group and factors? Can? Yq

Here are two trivial examples. You will be able to find plenty of worked examples from the lower sec textbooks and the school notes (if your child is from one of the IP schools).

By Group:

a + ac + b + bc = a(1+c) + b(1+c) [by factoring \"a\" & \"b\", we have created new \"grouped\" factor of (1+c)]

= (1+c)(a+b)
Factorisation by group usually involves at 3 unknowns.


\"Normal\" Factorisation
x + xy = x(1+y)

[Herbie]

Hi Adot,


Thanks for yr reply.
Can help to show the step in solving the qn below? tq

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)

I have a qn on mode.

The no.of bags owned by a group of 9 students are 5,5,47,9,2,6 5 and 2.

State the no. of bags owned by a new member of the group such that there are now 2 modes for the group.

can explain what the qn meant by 2 modes??

[CoffeeCat]

Herbie:

I have a qn on mode.

The no.of bags owned by a group of 9 students are 5,5,47,9,2,6 5 and 2.

State the no. of bags owned by a new member of the group such that there are now 2 modes for the group.

can explain what the qn meant by 2 modes??

Currently there is only 1 mode, \"5\" which occured 3 times. The next candidate is \"2\" which occured 2 times. 2 modes just means there are 2 objects that occured the highest number of times (means they are sorta tied together for the winner). So the new member of the the group must owned \"2\" bags, so that now the frequency of \"5\" and \"2\" are 3.

[ADoc]

Herbie:

Hi Adot,


Thanks for yr reply.
Can help to show the step in solving the qn below? tq

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)

Hi. This question is similar to the previous that you have posted. The technique is to \"switch\" the sign of (5d-3c) to become (3c-5d).

I will run through the first couple of steps again. I'll leave the rest for you to solve.

(7a+5b)(3c-5d)-(5d-3c)(2a+3b)
= (7a+5b)(3c-5d)-(-1)(-5d+3c)(2a+3b) [take note of the plus & minus signs]
= (7a+5b)(3c-5d)+(3c-5d)(2a+3b) [we can now proceed to extract the factor (3c-5d)]
= ... ...

cheers!

[ADoc]

Hi, I am no guru. Just wanna to share a little something with our aspiring Sec Ones. It is assumed that you are already comfortable with product of plus & minus, i.e. [plus] x [minus] = [minus], etc.


(a). Invisible Plus
Every number or unknown (for algebra) is always \"paired\" with either a positive or negative sign, such as (+)3, -10, etc. Notice that I wrote the + in parenthesis, because we don't, and it's not necessary, write positive as such. It's common knowledge that 3 means positive 3.

(b). What is the negative or minus sign?
Consider 10 - 5. It can mean two things: one, positive ten minus positive five, (+)10 - (+5); two, positive ten plus, negative five, (+)10 + (-5).

The \" - \" in case one is the mathematical operator for subtraction, whereas in case two, it is to denote that the number 5 is a negative instead of a positive number.

You must be wondering why I am going through these seemingly trivial concepts. The understanding of (a) & (b) will help you in your factorisation, as well as rationalising when and how you can manipulate the plus & minus sign when expanding and simplifying those hideous brackets. Foundation at this early stage is critical so you won't find yourself having difficulty in your subsequent years of algebra. Our teachers nowadays may not always appreciate the importance of making this distinction to our Sec1. There's only a short little para in the textbooks.

(c) The Invisible Negative 1
Having understood (b), it shouldn't be difficult to appreciate that, say -5 is made up of two very important factors, which are (-1) & (+5), i.e. (-1)(5). More importantly, any positive number can be re-written as a \"negative\".

5 = (-1)(-5)

Example 1
Trying reordering x - y such that y comes first on the left.
Again, it may sound trivial but please bear with me.

From above, we know every number or unknown is always paired with a sign. Hence whenever we manipulate an equation, we must always carry the signs together with the unknowns.

x - y = (+)x + (-)y = -y + x

Example 2
x - y = (-1)(-x) + (-1)(y) [factorising the invisible negative 1]
= (-1)(-x + y)
=-(-x + y) or -(y - x)

This technique is particular useful for Factorisation by Group

Example 3
Factorise 2a(b - c) + d(c - b)

2a(b - c) + d(c - b) = 2a(b - c) + d[(-1)(-c) + (-1)(b)]
= 2a(b - c) + d(-1)(-c + b) = 2a(b - c) - d(b - c) [+ve & -ve = -ve]
= (b - c)(2a - d)

Note that these steps aren't required in your workings. They are just for explanation.

Do leave a note if you think it's useful. And if it isn't, do post a reply saying so as well. No hard feelings at all... :xedfingers: Else I will continue to post a few more of such explanations on other operations. And that was Part [1]...boys & girls...

My students found these trivial explanations useful in guiding their algebra. Hope you'll find them useful too. Cheers! Basics are super important in order to breeze through the rest of your mathematics career for the next few years.