find a polynomial with rational coefficients of minimal degr

find a polynomial with rational coefficients of minimal degree passing through (-2,7) and having zeros 1, 1/2, 1/2. 1- squar 2 i, 2+squar three i

zeros 1, 1/2, 1/2. 1- squar 2 i, 2+squar three i

Conjugate complex zeros: for 1 - sqrt (2) i another zero would be 1+sqrt(2)*i

Conjugate complex zeros: for 2+sqrt(3) *i another zero would be 2- sqrt(3) *i

Now P(x) = a(x -1)( x-1/2)^2(x -1+ isqrt2)(x -1 -i*sqrt2)(x -2+isqrt3)(x-2 - isqrt3)

(x-1)(x- 1/2)^2 = (x^2+1/4 -x)(x-1) = x^3-x^2 +x/4 -1/4 -x^2 +x = x^3 -2x^2 +5x/4 -1/4

(x -1+ isqrt2)(x -1 -i*sqrt2) =x^2 + -x -ixsqrt2 -x +1 +isqrt2 + ixsqrt2 -isqrt2 +2 = x^2 -2x +3

(x -2+isqrt3)(x-2 - isqrt3) = x^2 -2x -ixsqrt3 - 2x +4 +2isqrt3 + xisqrt3 -2isqrt3 + 3 = x^2 - 4x +3

So, P(x) = a(x -1)( x-1/2)^2(x^2 -2x +3)(x^2 -4x +3)

where a is a constant

Given P(x) passes through ( -2. 7) we can find constant a by plugging (x=-2 and P(x) =7 and solve for a

7 = a( -2-1)(-2-1/2)^2( 4 +4 +3)(4+8+3)

7 = a(-3)(25/4)(11)(15)

a = - 28/ 12375

So, P(x) = -28/12375{(x -1)( x-1/2)^2(x^2 -2x +3)(x^2 -4x +3)}