Re: [Math4u] Re: Hi! Questions

thanks

"Brian E. Jensen" <brianejensen@prodigy.net> wrote:

PROBLEM:

Find all integers n for which n^2-x+n divides x^13+x+90

SOLUTION:

We could completely factor x^13+x+90 but if we come up with some complicated factors, we won't know if it is completely factored. I think Jim guessed a solution, but I don't know if it is the only solution.

Let's do the division.

Equation 1

(x^13+x+90) / (n^2-x+n)=x^11+x^10 + (-n+1)x^9 + (-2n+1)x^8 + (n^2-3n+1)x^7 + (3n^2-4n+1)x^6 + (-n^3+6n^2-5n+1)x^5 + (-4n^3+10n^2-6n+1)x^4 + (n^4-10n^3+15n^2-n+1)x^3 + (5n^4-20n^3+21n^2-8n+1)x^2 + (-n^5+15n^4-35n^3+28n^2-9n+1)x + (-6n^5+35n^4-56n^3+36n^2-10n+1) remainder (n^6-21n^5+70n^4-84n^3+45n^2-11n+2)x + (6n^6-35n^5+56n^4-36n^3+10n^2-n+90)

We only care about the remainder which we want to be zero so we get

Equation 2

remainder = 0 = (n^6-21n^5+70n^4-84n^3+45n^2-11n+2)x + (6n^6-35n^5+56n^4-36n^3+10n^2-n+90)

Since x can be zero,

Equation 3

(6n^6-35n^5+56n^4-36n^3+10n^2-n+90) = 0

Since x can be any number such as 1000,

Equation 4

(n^6-21n^5+70n^4-84n^3+45n^2-11n+2) = 0

So what we want to do is find all integral values of n so that equations 3 and 4 are true. There are 3 ways we can do this.

1) Since the last term of equation 4 is 2, perhaps the value of n must be +/1 a factor of 2.

2) Perhaps we can whittle down equations 3 and 4 until we get the answers. Jim told us that 2 is a solution to we can divide equations 3 and 4 by (x-2). We can add together multiples of equations 3 and 4 to get simpler equations.

3) We can graph and use Newton's approximation to find the values of n where the functions cross the x-axis.

Let's solve equation 3

f(n)=0=(6n^6-35n^5+56n^4-36n^3+10n^2-n+90)

f ' (n) = 36n^5 -175n^4 + 224n^3 - 108n^2 +20n - 1

f ' ' (n) = 180n^4 - 700n^3 + 672n^2 - 216n + 20

f ' ' ' (n) = 720n^3 - 2100n^2 + 1342n - 216

f ' ' ' ' (n) = 2169n^2 - 4200n + 1342

f ' ' ' ' ' (n) = 4538n - 4200

f ' ' ' ' ' ' (n) = 4538

So f ' ' ' ' ' (n) is concave up and zero at n = 4200/4538= .9255

So f ' ' ' ' (n) is minimum near (.9255, -687), so it should cross the x-axis twice.

using Newton's approximation:

f ' ' ' ' (n) is zero near (0.403679554672654,0) and (1.53269665556259,0)

So f ' ' ' is maximum near (0.403679554672654,30.8912661289102) and minimum near (1.53269665556259,-499.960363340243)

using Newton's approximation:

f ' ' ' (n) is zero near (0.251176010090297,0), (0.569968474131662,0), (2.0955221824447,0)

f ' ' (n) is horizontal near (0.251176010090297,-2.23407919195235), ( 0.569968474131662, 4.57846603552224), (2.0955221824447, -452.161169250977)

using Newton's approximation:

f ' ' (n) is zero near (0.157980575245044,0), (0.370224983472448,0), (0.719178437514828,0), (2.64150489265657,0)

f ' (n) is horizontal near (0.157981,0.241898440216314), (0.370224983472448, -0.0690722709722777), (0.719178437514828, 0.956830395242577), (2.64150489265657, -463.459270952539)

Notice how these points alternate above and below the x-axis. we are going to have the maximum number of solutions.

using Newton's approximation:

f ' (n) is zero near (.07759,0), (0.298860041260893, 0), (0.436735150770753, 0), (0.860558039906173, 0), (3.18736908964664,0)

f (n) is horizontal near (.07759, 89.96773), (0.298860041260893, 90.00093), (0.436735150770753, 89.99463), (0.860558039906173, 90.2329), (3.18736908964664, 420.19)

so unless I made a mistake, we have a minimum, maximum, minimum, and maximum all above the x-axis and then a minimum below the x-axis. So the function should cross the x-axis twice.

using Newton's approximation:

f (n) is zero near (2, 0) and (3.69086511769383, 0)

Let's see if these work in equation 4:

n=2, ok

n=3.69086511769383 doesn't work

note:

If you stick 0 for n into equation 4, you get 2

If you stick 2 for n into equation 4, you get 0

So 2 is the only real number that works. The only answer is 2, the answer given by Jim.

Regards,

Brian

 

 

 

 

----- Original Message -----

From: w7anf

To: Math4u@yahoogroups.com

Sent: Friday, September 28, 2007 8:01 AM

Subject: [Math4u] Re: Hi! Questions

n=2
x^13+x+90=
(x^2-x+2)*
(x^11+x^10-x^9-3*x^8-x^7+5*x^6+7*x^5-3*x^4-17*x^3-11*x^2+23*x+45)
Jim FitzSimons

--- In Math4u@yahoogroups.com, mustafa ozdemir <namruni@...> wrote:
>
> find all integers n for which x^2-x+n divides x^13+x+90
>
>
>
>
> ---------------------------------
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