O-Level Additional Math

[ADoc]

suiyuan:

Sec Two Maths


Expand and simplify
(3x+7y)^2 – (-2x-3)^2

(^2 means to the power of 2)

Please help.

Thank you.

Hi!
Just so you understand and subsequently apply the formula on your own, here's a long-winded explanation outlining the mental process you should have. So bear with me... :boogie:

(3x+7y)² – (-2x-3)² = (a + b)² - (c - d)² [you must be familiar with these formula already]

a = 3x
b = 7y
c = -2x
d = 3 (NOT negative 3 so that you won't confuse yourself unnecessarily)

(3x+7y)² – (-2x-3)² = (3x)² + 2(3x)(7y) + (7y)² - [ (-2x)² - 2(-2x)(3) + 3² ] [for inexperienced learners, I recommend inserting the square bracket to avoid committing the mistakes of -/+, else you can just switch the signs as you work] Always be extra careful whenever you see negative signs!!

I'll leave the rest to you.

[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!