O-Level Additional Math

[FrekiWang]

suiyuan:

FrekiWang:

5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Hi

Please explain in more details how you get this.

5z^2-30z=5(z^2-6z)=5(z^2-6z+9-9)=5(z^2-6z+9)-45=5(z-3)^2-45

Thank you

I used to believe that 'completing the square' is in Exp Sec 3 syllabus.

Since you asked, here is the explanation.

You should have learnt that a^2-2ab+b^2=(a-b)^2 in Sec 2.

Compare this formula with z^2-6z. we assume z^2 is the a^2 term and 6z is the 2ab term, then b must be 3. Therefore, when we add a 3^2 (which is equal to 9) at the back, the expression z^2-6z+9 will become a complete square (z-3)^2. However, when we introduce an additional +9, we have to subtract a 9 to keep the whole expression unchanged.

[suiyuan]

Hi


Is there a difference between simplification and factorisation of algebraic expressions?

Please help.

Thank you.

[suiyuan]

Sec Two Maths


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

(^2 means to the power of 2)

Please help.

Thank you.

[ADoc]

suiyuan:

Hi


Is there a difference between simplification and factorisation of algebraic expressions?

Please help.

Thank you.

Hi! I understand your concern regarding these two definitions.
Factorisation, I believe, is more intuitive in that it literally just means applying methods such as factorisation by groups to re-write the given expression into multiplication of two or more brackets (i.e. factors).

example, x²+x = (x)(x+1) is factorisation whereas
x²+x = (x+0.5)² - 0.5² isn't factorisation per se


Simplify usually goes with terms such as expand. It means to reduce an ugly looking expression into something simpler by first expanding, group the like terms, and finally add/subtract. All these to be done according to BODMAS or PEMDAS of course.

Some schools implicitly expects the students to factorise as factorised mathematical expressions appear simpler-looking.

To summarise:
(1) factorise is to form a strict multiplication of two or more brackets
(2) simplify is to reduce a long or ugly expression into something simpler or shorter (not always shorter sometimes), and we can leave the final answer in factorised form if we prefer [your teacher should not fault you if he/she is worth his/her two cents]

[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