HSC 2015 MX2 Marathon (archive) (2 Viewers)
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seanieg89
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Re: HSC 2015 4U Marathon
Well if there is a rational polynomial q(x) of degree < 3 with that root (which we call s), then there must exist such a polynomial m(x) (m for minimal) which divides x^3+3x+1.
Why?
By the division algorithm,
x^3+x+1=q(x)d_1(x)+r_1(x)
q(x)=r_1(x)d_2(x)+r_2(x)
r_1(x)=r_2(x)d_3(x)+r_3(x)
...
with deg(r_j) decreasing and r_j(s)=0 for each j. This process cannot repeat indefinitely, so we must have some r_{n+1}(x)=0, in which case m(x)=r_n(x) satisfies the claim.
So it suffices to show x^3+3x+1 cannot be written as the product of two rational factors of lower degree. (That (x^3+3x+1)/m(x) must also be a rational polynomial follows from that fact that it takes rational values for each integer x. Knowing what a poly does at infinitely many locations allows us to find its coefficients by simultaneous equations which will only involve rational operations.)
One of these factors must be linear, and so x^3+3x+1 must have a rational root. But the rational root theorem tells us this is NOT the case.
Hence x^3+3x+1 is the monic polynomial of least positive degree that annihilates s.
Well if there is a rational polynomial q(x) of degree < 3 with that root (which we call s), then there must exist such a polynomial m(x) (m for minimal) which divides x^3+3x+1.
Why?
By the division algorithm,
x^3+x+1=q(x)d_1(x)+r_1(x)
q(x)=r_1(x)d_2(x)+r_2(x)
r_1(x)=r_2(x)d_3(x)+r_3(x)
...
with deg(r_j) decreasing and r_j(s)=0 for each j. This process cannot repeat indefinitely, so we must have some r_{n+1}(x)=0, in which case m(x)=r_n(x) satisfies the claim.
So it suffices to show x^3+3x+1 cannot be written as the product of two rational factors of lower degree. (That (x^3+3x+1)/m(x) must also be a rational polynomial follows from that fact that it takes rational values for each integer x. Knowing what a poly does at infinitely many locations allows us to find its coefficients by simultaneous equations which will only involve rational operations.)
One of these factors must be linear, and so x^3+3x+1 must have a rational root. But the rational root theorem tells us this is NOT the case.
Hence x^3+3x+1 is the monic polynomial of least positive degree that annihilates s.
dan964
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Re: HSC 2015 4U Marathon
^
here is my attempt at the last part. The easiest way to do it; for the less intelligent people.
To simplify the algebra I substituted a Q in, and resubstituted it back in, to simplify the algebra a little bit.
I find LATEX time consuming to code, so images.
EDIT: The first line is wrong. It is x^2 -bx +c. It has no effect on the proof.
^
here is my attempt at the last part. The easiest way to do it; for the less intelligent people.
To simplify the algebra I substituted a Q in, and resubstituted it back in, to simplify the algebra a little bit.
I find LATEX time consuming to code, so images.
EDIT: The first line is wrong. It is x^2 -bx +c. It has no effect on the proof.
Last edited:
dan964
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Re: HSC 2015 4U Marathon
invalid question, amended question is 2 replies down
to see original q see below
invalid question, amended question is 2 replies down
to see original q see below
Last edited:
dan964
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Re: HSC 2015 4U Marathon
I'll impose a restriction x>y>2
or alternatively you can show that it a unique solution.
okThis isn't true, because.
I'll impose a restriction x>y>2
or alternatively you can show that it a unique solution.
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seanieg89
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PhysicsMaths
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dan964
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Re: HSC 2015 4U Marathon
here is a better question
(A) Solve
)
(B) Hence using part (A) graph}{\pi} )
here is a better question
(A) Solve
(B) Hence using part (A) graph
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Soulful
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Re: HSC 2015 4U Marathon
I'm so confused :S Are we supposed to solve part a) simultaneously? Cause I did and I got x=0 and I have no idea how this can be used in part B)here is a better question
(A) Solve
(B) Hence using part (A) graph
dan964
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Re: HSC 2015 4U Marathon
part (b) is addition of two curves
yes you are supposed to solve it simultaenously...
part (b) is addition of two curves
Drsoccerball
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Re: HSC 2015 4U Marathon
self made question =16 people are to be seated around two square tables each side has 2 seats. In how many ways can two specific people sit on the same side
self made question =16 people are to be seated around two square tables each side has 2 seats. In how many ways can two specific people sit on the same side
dan964
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Re: HSC 2015 4U Marathon
A, B and 14 other people.
A sits anywhere on the table.
B has one option to sit next to him
so the answer is
16 * the permutation of the remaining 14 people in 14 seats
so is 16!/15
A, B and 14 other people.
A sits anywhere on the table.
B has one option to sit next to him
so the answer is
16 * the permutation of the remaining 14 people in 14 seats
so is 16!/15
Drsoccerball
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Re: HSC 2015 4U Marathon
wrong
wrong
Drsoccerball
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Re: HSC 2015 4U Marathon
remember that you can arrange other people aswell
remember that you can arrange other people aswell
Ekman
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Re: HSC 2015 4U Marathon
Group the 2 specific people and put them on one side of the table. Therefore there are 6 seats remaining on Table 1
2! different ways to arrange between the two grouped specific people.
14C6 * 6! different ways to choose and arrange between the 6 other people on the table.
8C8 * 8! different ways to choose and arrange the remaining people on Table 2.
2*14C6*6!*8!*= 1.744 x 10^11 different ways of doing this.
2 Specific people are chosen from 16. Each table has 8 people.
Group the 2 specific people and put them on one side of the table. Therefore there are 6 seats remaining on Table 1
2! different ways to arrange between the two grouped specific people.
14C6 * 6! different ways to choose and arrange between the 6 other people on the table.
8C8 * 8! different ways to choose and arrange the remaining people on Table 2.
2*14C6*6!*8!*= 1.744 x 10^11 different ways of doing this.
Ekman
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Re: HSC 2015 4U Marathon
Or another method is:
2*14! since we lock in 2 people (they are arranged amongst themselves) and there are 14 people remaining to be arranged, thus 2 * 14! = 1.744 x 10^11 (the more easier approach, but I like to show what is going on for permutations)
Or another method is:
2*14! since we lock in 2 people (they are arranged amongst themselves) and there are 14 people remaining to be arranged, thus 2 * 14! = 1.744 x 10^11 (the more easier approach, but I like to show what is going on for permutations)
Drsoccerball
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Re: HSC 2015 4U Marathon
close but you have to select them before arranging. Solution:
Lock people a and b on one side and then arrange them 1x2! but you still have to fill up the other spots... therfore 2!x(14C2)x2!x(12C2)x2!..... take 2 factorial out and then type it into your calculator you get 1362160800 and since there is 8 side you multiply by 8 to get 1.09x10^10 with the assumption that one can differenciate between sides
close but you have to select them before arranging. Solution:
Lock people a and b on one side and then arrange them 1x2! but you still have to fill up the other spots... therfore 2!x(14C2)x2!x(12C2)x2!..... take 2 factorial out and then type it into your calculator you get 1362160800 and since there is 8 side you multiply by 8 to get 1.09x10^10 with the assumption that one can differenciate between sides
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