In terms of x, what is the polynomial P of least degree, with integral coefficients for which P(rad3 +rad2)=rad3-rad2?

Question



Answer 1

constant function will not work...

linear:

C(rad3+rad2) + k = rad3 - rad2

C*rad3 + C*rad2 + k = rad3 - rad2

(C-1)*rad3 + (1+C)*rad2 + k = 0

C-1 = 0 = 1+C

-1 = 1

no solution

quadratic:

A(rad3+rad2)^2 + B(rad3+rad2) + C = rad3 - rad2

A(3 + 2 + 2*rad3*rad2) + B( rad3 + rad2) + C - rad3 + rad2 = 0

A(5 + 2*rad3*rad2) + B(rad3+ rad2) + C - rad3 + rad2 = 0

5A + 2A*rad3*rad2 + B*rad3 + B*rad2 + C - rad3 + rad2 = 0

5A + 2A*rad3*rad2 + (B-1)*rad3 + (B+1)rad2 + C = 0

A(5 + 2*rad3*rad2) + (B-1)*rad3 + (B+1)rad2 + C = 0

A=C=0, which is not quadratic

B-1 = B+1 = 0 ---> -1 = 1 --> no solution

cubic:

Ax^3 + Bx^3 + Cx^2 + D = 0

A(rad3+rad2)^3 + B(rad3+rad2)^2 + C(rad3+rad2) + D = rad3-rad2;

But...

(rad3+rad2)^2 = (3 + 2 + 2 * rad3rad2) = 5 + 2 *rad3rad2

(rad3 + rad2)^3 = (5 + 2*rad2rad3)(rad3 + rad2)

= 5 rad3 + 5 rad2 + 6 rad2 + 4 * rad3

= 9 rad3 + 11 rad2

plugs into the polynomial...coefficient table is:

coefficients of rad3 coefficients of rad2 coefficients of cross term rad3*rad2 constants

9A 11A 2b 5B

C C D

9A + C = 1

11A+C = -1

2b = 0

D+5b = 0

The last two equations imply that b=d=0 since b=0 forces d=0

Subtracting first equation FROM the second...

2A = -2

A=-1

which forces C=10

the cubic polynomial is -x^3 + 10x

check:

-(rad3 + rad2)^3 + 10(rad3 + rad2) =

-(9 rad3 + 11 rad2) + 10rad3 + 10rad2 =

-9rad3 - 11rad2 + 10rad3 + 10rad2 =

rad3 - rad2

YES

the cubic polynomial is -x^3 + 10x