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

Anonymous2016-03-19T10:18:07Z

4 3 3 or 9 2 2

Puzzling2006-11-10T17:12:43Z

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!

an ugly mind2006-11-10T17:39:48Z

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

?2006-11-10T18:29:58Z

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.

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