Using the Integer Root Theorem to find the roots of the cubic equation f(x) = x³ - x² - x + 1 shows that the roots of the equation are 1, -1, (1 + √5)/2, and (1 - √5)/2 with two integer roots and two irrational roots.
The Integer Root Theorem states that if a polynomial has an integer root, then that root must divide the constant term of the polynomial.
The constant term here is 1. We know that the only integers that divide 1 are 1 and -1. So, if the equation has integer roots, they must be 1 or -1.
Substituting these possible roots into the equation:
If we substitute x = 1, we get f(1) = 1³ - 1² - 1 + 1 = 0. So, x = 1 is a root of the equation.
If we substitute x = -1, we get f(-1) = (-1)³ - (-1)² - (-1) + 1 = -1 - 1 + 1 + 1 = 0. So, x = -1 is also a root of the equation.
So, the equation f(x) = x³ - x² - x + 1 has two integer roots: 1 and -1.
Find the remaining root, we divide the polynomial by (x - 1)(x + 1) = x² - 1, which gives us a quadratic equation x² - x - 1 = 0.
Using the quadratic formula gives us the remaining roots, which are (1 ± √5)/2. These are irrational numbers.
Thus, the roots of the equation are 1, -1, (1 + √5)/2, and (1 - √5)/2.