To find the polynomial function of lowest degree with a lead coefficient of 1 and specific roots, we can start by identifying the given roots.
The roots provided are 1 and 1 i. Since we are working with real polynomials, we should remember that complex roots come in conjugate pairs. Therefore, if 1 i is a root, its conjugate -1 i must also be a root. So, the complete set of roots for our polynomial is 1, 1 i, and -1 i.
Using these roots, we can construct the polynomial. The polynomial can be expressed as:
(x - r_1)(x - r_2)(x - r_3)
Where r_1 = 1, r_2 = 1 i, and r_3 = -1 i.
Thus, the polynomial becomes:
(x - 1)(x - 1 i)(x + 1 i)
Next, we can simplify this expression. Starting with the complex roots:
(x - 1 i)(x + 1 i) = x^2 + 1
Therefore, our polynomial expression becomes:
(x - 1)(x^2 + 1)
Now, we’ll distribute:
x(x^2 + 1) - 1(x^2 + 1) = x^3 + x - x^2 - 1 = x^3 - x^2 + x - 1
Finally, we arrive at the polynomial function:
f(x) = x^3 – x^2 + x – 1
This polynomial has the desired properties: it has a lead coefficient of 1, and it has the roots 1, 1 i, and -1 i.