polynomial

reopened

Assume q(x)=x³+a₂x²+a₁x+a₀.

Let y be a zero of q(x). Using synthetic division:

y | 1 a₂      a₁                 a₀

1 y        a₂y+y²         a₁y+a₂y²+y³

1 a₂+y  a₁+a₂y+y² | a₀+a₁y+a₂y²+y³=0 remainder

Now replace y with c+d(∛2):

a₀+a₁(c+d(∛2))+a₂(c+d(∛2))²+(c+d(∛2))³=0,

a₀+a₁(c+d(∛2))+a₂(c²+2cd(∛2)+d²(2^⅔))+(c³+3c²d(∛2)+3cd²(2^⅔)+2d³)=0,

a₀+a₁c+a₂c²+c³+2d³+d(∛2)(a₁+2a₂c+3c²)+d²(2^⅔)(a₂+3c)=0.

To remove the irrationals:

a₁+2a₂c+3c²=0, a₂+3c=0, a₂=-3c, a₁-6c²+3c²=0, a₁=3c², if we ignore the trivial case of d=0.

To find a₀, substitute for a₁ and a₂:

a₀+a₁c+a₂c²+c³+2d³=0, a₀+3c³-3c³+c³+2d³=0, a₀=-c³-2d³.

q(x)=x³-3cx²+3c²x-c³-2d³,

which is (x-c)³-2d³=(x-c-d(∛2))((x-c)²+(c+d(∛2))(x-c)+(c+d(∛2))²).

by Top Rated User (1.1m points)
selected
What about p(x) of other degrees?

The question doesn’t mention p(x). It just asks for a(ny) polynomial q(x). I picked the simplest one (a cubic) likely to have a cube root in its zeroes. It’s possible that a degree 6 or higher can be created, but the calculations would be more involved. A multiple of q(x) is also a solution.