The equation you provided is:
x³ - 6x² + 13x - 3 = 7
Adding 7 to both sides gives:
x³ - 6x² + 13x - 10 = 0
Using the Rational Root Theorem:
This theorem states that any rational root (a fraction p/q where p and q are integers) of a polynomial with integer coefficients must be a factor of the constant term (in this case, -10).
Therefore, the possible rational roots are ±1, ±2, ±5, and ±10.
By trying these values and factoring accordingly, you can find that (x - 2) is a factor of the polynomial. Dividing x³ - 6x² + 13x - 10 by (x - 2) gives:
(x - 2)(x² - 4x + 5) = 0
The quadratic factor can be further factored using the quadratic formula or simply by recognizing it as (x - 2)² + 1. Therefore, the roots are:
x1 = 2 (from the first factor)
x2 = 2 + i (from the second factor)
x3 = 2 - i (from the second factor)
Therefore, all the real and complex solutions of the equation x³ - 6x² + 13x - 3 = 7 are:
A. 2, −2 + i, −2 − i
The options B, C, and D do not contain all the correct solutions.