Find a polynomial function of least degree having only real coefficients, a leading of 1, and zeros of 6 and 3 + i.

Since we know that complex solutions ALWAYS come in pairs, the minimal polynomial must include the root of 4-i as an acceptable root..

This leads to a polynomial of ...

p(x) = (x - 5)(x - (4+i))(x - (4-i))

Now we can simplify...

(x - (4+i))(x - (4-i)) --> x2 + [(4+i)+(4-i)]x + (4+i)(4-i) --> x2 + 8x + [ 16 -4i + 4i -i2 ] = x2 + 8x +17

Take this quadratic and multiply by (x-5) to get the final, minimal polynomial...

p(x) = (x-5)(x2 + 8x +17) = x3 + 8x2 + 17x -5x2 + 40x - 85 = x3 + 3x2 + 57x - 85

So p(x) = x3 + 3x2 + 57x - 85

p(x) has only real coefficients, the leading coefficient of 1, and is the smallest to allow for the two solutions requested

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong