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secondary 4 | A Maths
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Kathy
Kathy

secondary 4 chevron_right A Maths chevron_right Singapore

Thank you

Date Posted: 3 months ago
Views: 34

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Eric Nicholas K
Eric Nicholas K's answer
2297 answers (Tutor Details)
1st
5a to 5g

Will do the rest at a later time when I can spare the time.
Kathy
Kathy
3 months ago
Can I know if I don’t u test out to find the factor got any way to do it
Kathy
Kathy
3 months ago
Thank you so much pls help ah
Eric Nicholas K
Eric Nicholas K
3 months ago
At the secondary level, this test out of a factor is the ONLY way to factorise the expression.

Unlike quadratic expressions, cubic expressions are much harder to factorise (or even to spot the pattern at all). Only a small minority of such cubic equations can be factorised by observation and grouping of terms (in a very similar way to factorising ab + ac + bd + cd by grouping expressions into twos).

Once you proceed to polynomials of degree 4 and more, factorising gets even harder.

Fortunately for your exams, the expressions are factorisable with at least one number which is guessable. And the numbers to guess are predictable. Typically, the numbers to guess are factors of the constant term (and their negatives). For example, if the constant term is 30 or -30 (eg x3 - 29x + 30 or x3 - 29x - 30), then the numbers to guess are 1, 2, 3, 5, 6, 10, 15, 30 and their negatives. Guessing of fractions such as 1/2 is more difficult, but should not be seen in exams at all.

There is another method to do this, but it's way beyond the secondary level and I have no idea how to execute the other method at all (it's way too advanced for me).

I am currently working on the remaining questions which will take me another 45 mins to complete.
Eric Nicholas K
Eric Nicholas K
3 months ago
In the exams, there is at least ONE guessable integer factor. The other factors may be unguessable or do not even exist (a cubic equation can have one, two or three solutions).
Kathy
Kathy
3 months ago
For e and f I can’t understand
Eric Nicholas K
Eric Nicholas K
3 months ago
For e and f, there is a special factorisation formula for it:

a3 + b3 = (a + b) (a2 - ab + b2)
a3 - b3 = (a - b) (a2 + ab + b2)

where I recognise that the two terms there can be expressed as cubes.

Take 8x6 - 343 for example.

If you identify 8x6 as (2x2)^3 and 343 as 7^3, then you would be able to factorise them using the above formula, noting that a = 2x2 and b = 7. Then you can replace the a in the formula by 2x2 and the b by 7.

It's the same idea to factorise terms like a2 - b2 = (a + b) (a - b) where you recognise the difference of two 'perfect squares'.

You can also factorise the expression the normal way by guessing numbers as well, but possibly much harder (and I am very sure that the number to guess here is not an integer).
Kathy
Kathy
3 months ago
Ok. Understand,
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Eric Nicholas K
Eric Nicholas K's answer
2297 answers (Tutor Details)
6d, 6e, 6f, 6g.

6f and 6g are not exactly in your syllabus as fractional guesses are required.

I will remain online until around 1 am.
Kathy
Kathy
3 months ago
Ok thank you so much
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Eric Nicholas K
Eric Nicholas K's answer
2297 answers (Tutor Details)
6h, which again requires fractional guess.

Also, an important note is that the quadratic expression which you factor out may not have any solutions at all. Do not be surprised if no solutions appear out of the quadratic expression.
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Eric Nicholas K
Eric Nicholas K's answer
2297 answers (Tutor Details)
5h, 6a, 6b, 6c

(Lack of space on paper)
Kathy
Kathy
3 months ago
Only one way to do this?
Eric Nicholas K
Eric Nicholas K
3 months ago
For question 6, the first thing to do is to guess (the only way at your level).

Subsequently, to obtain the Ax2 + Bx + C, you can perform different things, any one of this

- Long division
- Comparing coefficients (which I have used)
- Sub in random x like you would for solving identities (after all, both sides are identities since we are just factorising from the LHS to the RHS). This method is the hardest of the three.
Kathy
Kathy
3 months ago
If Guess wrongly how?
Eric Nicholas K
Eric Nicholas K
3 months ago
So in 6a, my first try was x = 1, and it leads to LHS not equal to RHS, so x = 1 does not work.

I moved on to x = 2 and found that x = 2 works. Had it not worked, I would have needed to try other numbers (may be even up to x = -30).

Typically the numbers are factors of the constant term.

For your syllabus, at least one of these integer number guesses must work. The exams will not ask questions which does not meet this criteria.
Kathy
Kathy
3 months ago
Ok thanks.
Can help more tonight, having a test this Friday