Final answer:
To find the polynomial f(x) of degree 3 with given zeros -1, 1+2i, and its complex conjugate 1-2i, one can write down the factors (x + 1), (x - (1 + 2i)), and (x - (1 - 2i)). Expanding and multiplying these gives f(x) = x^3 - x^2 + 3x + 5.
Step-by-step explanation:
To find a polynomial f(x) of degree 3 with real coefficients that has the given zeros -1 and 1+2i, we must also include the complex conjugate of 1+2i, which is 1-2i, to ensure the coefficients are real. A polynomial with real coefficients will have complex roots (zeros) that come in conjugate pairs.
The factors associated with these zeros are (x + 1), (x - (1 + 2i)), and (x - (1 - 2i)). We can write the polynomial as the product of these factors:
f(x) = (x + 1)(x - (1 + 2i))(x - (1 - 2i))
Expanding the quadratic term first:
(x - (1 + 2i))(x - (1 - 2i)) = ((x - 1) - 2i)((x - 1) + 2i) = (x - 1)2 - (2i)2 = x2 - 2x + 1 + 4 = x2 - 2x + 5
Now we can find the full polynomial:
f(x) = (x + 1)(x2 - 2x + 5) = x3 - 2x2 + 5x + x2 - 2x + 5 = x3 - x2 + 3x + 5
Therefore, the polynomial f(x) of degree 3 with the given zeros -1, 1+2i is f(x) = x3 - x2 + 3x + 5.