find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 2, and zeros 4-i and 3i

Answer 1

(Edited 4/24 to correct forgotten leading coefficient)

Carlos, construct the polynomial by multiplying together the leading coefficient by simple monomials made from the roots you’ve got:

2* (x -3i) * (x-(4-I))

and make two more because of the complex roots, (4 - I) and (3i).

Whenever your root is (Integer) - (integer) i , you need to make a term with the opposite sign

( x - (4 + i))

you have to have both: if you have the one negative i, you need to make the one with positive i , and vice versa. The two roots are called “complex conjugates”.

Same as true with the 3i, its complex conjugate is -3i, so you need

(x - (-3i)) which is ( x + 3i)

so the final product is:

2 * (x -3i) * (x+ 3i) * (x-(4-I)) * ( x - (4 + i))

now multiply these out ( it’s easier to multiply a pair of complex conjugates together first, as the I’s go away).

Then you’ll have your polynomial

Its pretty late but here’s the answer built up

(x-3i) * (x+3i) = x²-9i² = x² + 9

(x-(4-i))*(x-(4+i)) = x²-x(4+i) -x(4-i)-(4²-i²)

= x²-8x+0xi+17

= x²-8x+17

2(x²+9)(x²-8x+17) = 0

You can finish multiplying it out now that the instances of i are all gone.