I have a slightly harder mathematical puzzle for you...?
Find constants a, b, c, and d such that: 4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)
Rules:
#1: Your factors must be exact. No numerical approximations allowed. #2: Neither the imaginary constant, nor the square root of any negative number may appear in your answer. #3: Your answer must be simple. Nothing more complicated than a trigonometric function may appear in your answer. #4: Yes, this problem is solvable.
2006-11-10T17:22:58Z
Wow... you're all going about it so wrong.
2006-11-10T19:01:19Z
Math_kp has solved it most elegantly. Your check for 10 points is in the mail.
Mein Hoon Na2006-11-10T18:26:56Z
Favorite Answer
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
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.