What are the factors for #6y^2 - 5y^3 - 4#?

1 Answer
Jun 16, 2015

Answer:

#6y^2-5y^3-4 = -5(y-y_1)(y-y_2)(y-y_3)#

#y_1 = 1/(u_1+v_1)#

#y_2 = 1/(omega u_1+omega^2 v_1)#

#y_3 = 1/(omega^2 u_1 + omega v_1)#

as explained below...

Explanation:

Attempt to solve #f(y) = -5y^3+6y^2-4 = 0#

First divide through by #-y^3# to get:

#5-6/y+4/y^3 = 0#

Let #x = 1/y#

Then #4x^3-6x+5 = 0#

Now let #x = u + v#

#0 = 4(u+v)^3 - 6(u+v) + 5#

#=4u^3+4v^3+(12uv-6)(u+v)+5#

#=4u^3+4v^3+6(2uv-1)(u+v)+5#

Let #v = 1/(2u)#

#=4u^3+1/(2u^3)+5#

Multiply through by #2u^3# to get:

#8(u^3)^2+10(u^3)+1 = 0#

#u^3 = (-10+-sqrt(100-32))/16#

#=(-10+-sqrt(68))/16#

#=(-5+-sqrt(17))/8#

Write:

#u_1 = root(3)((-5+sqrt(17))/8)#

#v_1 = root(3)((-5-sqrt(17))/8)#

Then real root of #4x^3-6x+5 = 0# is

#x = u_1 + v_1#

The other two (complex) roots are:

#x = omega u_1 + omega^2 v_1#

#x = omega^2 u_1 + omega v_1#

where #omega = -1/2+sqrt(3)/2i#

#y = 1/x#

So the real root of #f(y) = 0# is #y_1 = 1/(u_1+v_1)#

and the complex roots are:

#y_2 = 1/(omega u_1 + omega^2 v_1)#

#y_3 = 1/(omega^2 u_1 + omega v_1)#

#f(y) = -5(y - y_1)(y - y_2)(y - y_3)#