I have a slightly harder mathematical puzzle for you...?

let p(x) be given polynomial

we reallize that sqrt(3)/2 = cos pi/6

so 4x^3-3x + cos pi/ 6= 0

or - cos pi/6 =-3x + 4x^3 = cos(3t) if x= cos t

so cos 3t = - cos pi/6 = cos 5pi/6

t = 5pi/18 or cos (2pi/3-5pi/18) = cos(7pi/18)

or cos 17pi/18

so a = 4

b= cos 5pi/18

c= cos 7pi/18

d = cos 17pi/18

as these are zeroes of polynomial

Please check the calculation steps

4 3 3 or 9 2 2

4x^3 - 3x + sqrt(3/2) = a * (x-b) * (x-c) * (x-d)

Multiply through on the right hand side:

ax^3 – adx^2 – acx^2 + axcd – abx^2 + abxd + abxc + abcd

Now group your x terms:

ax^3 - (ad + ac + ab)x^2 + (acd + abd + abc)x + abcd

Now match up coefficients and solve...

a = 4

a(b + c + d) = 0

a(bd + bc + cd) = -3

abcd = sqrt(3/2)

a = 4

b + c + d = 0

bc + bd + cd = -3/4

bcd = sqrt(3/2)/4

b(c + d) + cd = -3/4

b(-b) + cd = -3/4

b(-b) + [ sqrt(3/2)/4 ] / b = -3/4

-b^3 + 3/4b + sqrt(3/2)/4 = 0

b is approximately 1.0738817282479558

Yuck -- I've definitely done this the hard way and the wrong way!

doing the multiply operation on the right side gives

4x^3-3x+sqrt(3)/2=ax^3-a(b+c+d)x^2+a(bc+bd+cd)x-abcd

now

a=4 *

0=-a(b+c+d) **

-3=a(bc+bd+cd) ***

-abcd=sqrt(3)/2 ****

then

frm **

-b=c+d

frm ***

b(c+d)+cd=-3/4

-b^2+cd=-3/4

cd=-3/4+b^2

frm ****

bcd=-sqrt(3)/8

b(-3/4+b^2)=-sqrt(3)/8

b^3-3/4b+sqrt(3)/8=0

solve for b

b=-0.9848 or 0.6428 or 0.3428

choos any to find c and d

however the solution for b is done numerically. I cant remebber a formula to solve this equation

a = 4

{b,c,d} = {cos(5π/18), cos(17π/18), cos(29π/18)}

= {.642787609, -0.984807753, .342020143}

OR ANY PERMUTATION THEREOF.

{i was planning on typing the answer, submitting, then editing with the derivation, but math_kp beat me to it. His analysis is correct, knowing of course that cos(29π/18) = cos(7π/18)}

Additional proof using casus irreducibilis of the cubic equation:

4x³ - 3x + √3/2 = 0

x³ - (3/4)x + √3/8 = 0

p = -3/4, q = √3/8

r = √(-p^3/27) = 1/8

cosφ = -q/2r = -√3/2

φ = 5π/6

y1 = 2r^(1/3)cos(φ/3) = cos(φ/3) = cos(5π/18)

y2 = 2r^(1/3)cos(φ/3 + 2π/3) = cos(17π/18)

y2 = 2r^(1/3)cos(φ/3 + 4π/3) = cos(29π/18)

are the roots, thus b,c,d, and we know a=4 by inspection.

Let's figure out the factors for each power of x.

x^3 ---> a=4.

x^2 ---> a(-b-c-d) = 0 ---> b+c+d = 0.

x ---> a(bc+bd+cd) = -3 ---> bc+bd+cd = -3/4.

constant ---> -abcd = sqrt(3)/2 ---> bcd = -sqrt(3)/8.

Solve for b, c, d.

you said in your answer that you would start by showing that span(S) is closed under addition and scalar multiplication. And you told me how easy that was suppost to be well it is not very easy if I (sorry) don't even know how to do that. Basically I am asking you how would you show that a span(S) is closed under addition and scalar multiplication.

4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)

4x³ - 3x + √3/2=ax-ab (x^2 – xd – xc + cd)

4x³ - 3x + √3/2=ax^3 – adx^2 – acx^2 + axcd – abx^2 + abxd + abxc + abcd

I get:

a = 4

b = cos(50 deg)

c = -(cos(50 deg) + sqrt(3) * sin(50 deg))/2

d = -(cos(50 deg) - sqrt(3) * sin(50 deg))/2

There are methods for solving cubics. I'd check wikipedia.

Here it is:

http://en.wikipedia.org/wiki/Cubic_function