Final answer:
The polynomial of least degree with integral coefficients and zeros at -3 and 1-i is f(x) = x^3 + x^2 - 8x + 6.
Step-by-step explanation:
To write a polynomial function of least degree with integral coefficients, having zeros at -3 and 1-i, we must first recognize that complex zeros come in conjugate pairs. Hence, if 1-i is a zero, so must 1+i be a zero of the polynomial. A polynomial with these zeros will have factors of (x + 3), (x - (1 - i)), and (x - (1 + i)).
The polynomial function in its factored form is f(x) = (x + 3)(x - (1 - i))(x - (1 + i)). We can expand this to find the polynomial:
- First, expand the complex factors: (x - (1 - i))(x - (1 + i)) = (x - 1 + i)(x - 1 - i) = ((x - 1)^2 - (i)^2) = x^2 - 2x + 2.
- Next, multiply this result by the real factor: f(x) = (x + 3)(x^2 - 2x + 2).
- Finally, distribute and combine like terms to get the polynomial in standard form: f(x) = x^3 + x^2 - 8x + 6.
Therefore, the polynomial of least degree with integral coefficients that has the given zeros is f(x) = x^3 + x^2 - 8x + 6.