your alternative method miss out x=0 because you take 3x / 4x =
3/4. This is BOOBOO!
Mikethm, how did you spot the pattern? perhaps can share
Good job by Mikethm! I can't think of a simpler way than
his.
One way to spot the pattern could be to take note:
3x³ - 2x² - 7x - 2
and
3x^-3 - 4x^-2 - 28x^-1 -16
Note 3*2^0 = 3, 2*2^1 = 4, 7*2^2=28. 2*2^3 = 16
From there, you know the pattern will roughly require the number 2
somewhere in the substitution.
A slight variation/alternative method:
16x^4 + 28x³ + 4x² - 3x = 0
x(16x³ + 28x² + 4x - 3) = 0
Either x = 0 or 16x³ + 28x² + 4x - 3 = 0
for 16x³ + 28x² + 4x - 3, sub x = 1/(2y) ==> it's the same as
Mikethm's of subbing x^-1 = 2y
2y^-3 + 7y^-2 + 2y^-1 - 3 = 0
Multiply by -y³ throughout.
-2 - 7y - 2y² + 3y³ = 0
From here on same as Mikethm's method.