How do you find all the zeros of #f(x)=x^4-2x-1#?

1 Answer
Aug 9, 2016

Answer:

Reduce to solving two quadratics...

Explanation:

#f(x) = x^4-2x-1#

Since #f(x)# has no term in #x^3# it will factor as the product of a pair of quadratics with opposite middle terms:

#x^4-2x-1#

#= (x^2-ax+b)(x^2+ax+c)#

#= x^4+(b+c-a^2)x^2+(b-c)ax+bc#

Equating coefficients and rearranging a little we get:

#{ (b+c = a^2), (b-c = -2/a), (bc=-1) :}#

Then:

#(a^2)^2 = (b+c)^2 = (b-c)^2+4bc = (-2/a)^2-4 = 4/((a^2))-4#

Multiply both ends by #(a^2)# and rearrange slightly to get:

#(a^2)^3 + 4(a^2) - 4 = 0#

Use Cardano's method to solve this cubic in #(a^2)#...

Let #a^2 = u+v#

#u^3+v^2+(3uv+4)(u+v)-4 = 0#

Add the constraint #v = -4/(3u)# to eliminate the #(u+v)# term:

#u^3-64/(27u^3)-4 = 0#

Multiply through by #27u^3# and rearrange slightly to get:

#27(u^3)^2-108(u^3)-64 = 0#

Use the quadratic formula to find:

#u^3 = (108+-sqrt((-108)^2-4(27)(-64)))/(2*27)#

#=(108+-sqrt(11664+6912))/54#

#=(108+-sqrt(18576))/54#

#=(108+-12sqrt(129))/54#

#=(54+-6sqrt(129))/27#

Hence Real root:

#a^2 = 1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))#

We can use the positive root as the value of #a#:

#a = sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))#

Then:

#b = 1/2(a^2-2/a)#

#= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))#

#c = 1/2(a^2+2/a)#

#= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))#

This leaves us with two quadratics to solve:

#x^2-(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0#

#x^2+(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0#

We can find the roots of these quadratics using the quadratic formula, which are the zeros of the original quartic #f(x)#.