O-Level Additional Math

[hot_chocolate]

Hi, just asking, how do I factorise the following sums:


a) a^3 + b^3

b) x^5 + x + 1

Thanks!

[Guan Hui]

a) nothing can be done

b)x(x^4+1) +1??

[OK Lor]

a) (a+b)(a²-ab+b²)

[Guan Hui]

Hi OKlor,

NICE factorization there=)
nv thought of it.=)

[hot_chocolate]

Thanks a lot.

[huaiheanbian]

b) (x^2+x+1)(x^3-x^2+1)

[NormalDistribution]

hot_chocolate:

Hi, just asking, how do I factorise the following sums:


a) a^3 + b^3

b) x^5 + x + 1

Thanks!

your questions seem very out-of-syllabus, are they from IP schools by any chance?

[Eddow]

NormalDistribution:

hot_chocolate:

Hi, just asking, how do I factorise the following sums:


a) a^3 + b^3

b) x^5 + x + 1

Thanks!

your questions seem very out-of-syllabus, are they from IP schools by any chance?

Hi.. the question is not out of syllabus. It's still in the A Maths. I'm a tutor here. hee..

[OK Lor]

Hi Sir,


Need your help for some maths competition questions:

1. When 3^1981 + 2 is divided by 11, the remainder is …
2. The least positive integer which has remainders 1, 1 and 5 when divided by 3, 5 and 7 respectively, is (A)166 (B)151 145 (D)131 (E)none of these.
3. If x and y are integers such that (x-y)² + 2y² = 27, then the only number x can be are (A)3,5 (B)-6,4 0,4,6 (D)0,-4,4,-6,6 (E)0,-2,2,-4, 4,-6,6.

Thanks.

[CoffeeCat]

OK Lor:

Hi Sir,


Need your help for some maths competition questions:

1. When 3^1981 + 2 is divided by 11, the remainder is ...
2. The least positive integer which has remainders 1, 1 and 5 when divided by 3, 5 and 7 respectively, is (A)166 (B)151 (C)145 (D)131 (E)none of these.
3. If x and y are integers such that (x-y)² + 2y² = 27, then the only number x can be are (A)3,5 (B)-6,4 (C)0,4,6 (D)0,-4,4,-6,6 (E)0,-2,2,-4, 4,-6,6.

Thanks.

Hope guanhui won't be angry at me...

2) If the number has the same remainder 1 upon division by 3 and 5, it will leave the same remainder 1 upon division by 15.
the number is of the form 15x + 1 = (14 + 1)x + 1
and 15x will leave a remainder of 4 upon division by 7.
therefore x = 4, or 11, 18, etc
Least integer = 15*4 + 1 = 61

3)notice that the left hand side are all positive so
we are looking at 0, 1, 4, 16, 25 for (x-y)^2
only 25 and 9 fits
and y=1 or -1
x-y= 5 or -5.
or
y= 3 or -3
x-y = 3 or -3
you will get x=6, 4, -6, -4, 0

1) 3^1981 + 2
There are many ways to do this.
for starters use the pattern approach by looking at the remainder of powers of 3 upon division by 11.
3 9 5 4 1 3 9 5 4 1, ...
therefore 3^1981 will leave a remainder of 3.

Since you are doing competition maths, you will eventually come across this method of doing such questions using fermat's little theorem.