What are all the possible rational zeros for #f(x)=x^6-64#?

2 Answers
Dec 20, 2016

Answer:

"Possible" rational zeros:

#+-1, +-2, +-4, +-8, +-16, +-32, +-64#

Actual zeros:

#+-2#, #-1+-sqrt(3)i#, #1+-sqrt(3)i#

Explanation:

By the rational roots theorem, any rational zeros of #f(x) = x^6-64# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-64# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only "possible" rational zeros are:

#+-1, +-2, +-4, +-8, +-16, +-32, +-64#

In practice we can factor this sextic as a difference of squares, difference of cubes and sum of cubes, allowing us to find the Real zeros and quadratics for the Complex zeros.

We will use the following identities:

#a^2-b^2=(a-b)(a+b)#

#a^3-b^3=(a-b)(a^2+ab+b^2)#

#a^3+b^3=(a+b)(a^2-ab+b^2)#

So:

#x^6-64 = ((x^3)^2-8^2)#

#color(white)(x^6-64) = (x^3-8)(x^3+8)#

#color(white)(x^6-64) = (x^3-2^3)(x^3+2^3)#

#color(white)(x^6-64) = (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)#

Hence rational zeros:

#x = +-2#

The remaining quadratics have negative discriminants, so Complex zeros, but we can factor them using the difference of squares identity as follows:

#x^2+2x+4 = x^2+2x+1+3#

#color(white)(x^2+2x+4) = (x+1)^2-(sqrt(3)i)^2#

#color(white)(x^2+2x+4) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(x^2+2x+4) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Similarly:

#x^2-2x+4 = (x-1-sqrt(3)i)(x-1+sqrt(3)i)#

Hence Complex zeros:

#x = -1+-sqrt(3)i" "# and #" "x = 1+-sqrt(3)i#

These zeros form the vertices of a regular hexagon in the Complex plane:

graph{((x-2)^2+y^2-0.01)((x+2)^2+y^2-0.01)((x-1)^2+(y-sqrt(3))^2-0.01)((x-1)^2+(y+sqrt(3))^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01)(x^2+y^2-4) = 0 [-5, 5, -2.5, 2.5]}

Dec 21, 2016

Answer:

#+-2#. See graph

Explanation:

graph{x^6-64 [-3, 3, -500, 500]}

We can sort out rational zeros, from real zeros.

Use #(a^3-b^3)=(a-b)(a^2+ab+b^2)#

Any #x^(2N) - a^(2N)=0# has exactly two real roots #x = +-a#,

Here, #(x^6-64)=((x^2)^3-4^3)=(x^2-4)(x^4+4x^2+16)#

So, the zeros are given by

#x^2-4=0 to x = +-2# and

#(x^2)^2+4x^2+16=0 to x^2=-(2+-i2sqrt3)=4e^(+-i2/3pi)#

#x=(4e^(+-i2/3pi))^(1/2)=+-1+-isqrt3#,

upon simplification, using De Moivre;e theorem

So, the real ( and , of course, rational ) zeros are #+-2#.

The complex zeros ( occurring in conjugate pairs ) are

#=+-1+-isqrt3#

Direcly, all-inclusive zeros can be listed as

#x= 2e^(ik/3pi), k = 0, 1, 2, 3, 4, 5.#.

The real roots #+-2# appear for k = 0 and k = 3..