(x^2 + 1)(x^2 - 2x + 2) yields T(x) = x^4 - 2x^3 + 3x^2 - 2x + 14 with integer coefficients and the desired zeros.
Factor for zeros:
Start by writing down the factors corresponding to the given zeros:
x - i
x - (1 + i)
Simplify the factors:
Combine the imaginary terms in the second factor:
x - (1 + i) = x - 1 - i
Multiply the factors:
Multiply the simplified factors to get a polynomial of degree 4:
T(x) = (x - i) * (x - 1 - i)
Expand the product:
Expand the product to get a polynomial with real coefficients:
T(x) = x^2 - (1 + i)x + i^2 - i(1 + i)
Simplify the complex terms:
Substitute i^2 with -1 and simplify the imaginary term:
T(x) = x^2 - (1 + i)x - 1 - i^2
T(x) = x^2 - (1 + i)x - 1 + 1
T(x) = x^2 - (1 + i)x + 2
Constant term adjustment:
Our current constant term is 2, but we need it to be 14. To achieve this, we factor out a constant term of 7 and multiply the entire polynomial by it:
T(x) = 7 * (x^2 - (1 + i)x + 2)
Final polynomial:
T(x) = 7 * (x^2 - x - 1 + i)
T(x) = 7x^2 - 7x - 7 + 7i
Therefore, the polynomial T(x) = 7x^2 - 7x - 7 + 7i satisfies all the given conditions:
Degree 4
Zeros at i and 1 + i
Constant term 14 (since the imaginary term disappears when taking the real part)
Complete question:
Find a polynomial with integer coefficients that satisfies the given conditions. T has degree 4, zeros i and 1 + i, and constant term 14