Answer:
To find an equation with a solution of x = 3 of multiplicity 1, and a solution of x = 1 of multiplicity 2, we can use the concept of polynomial roots and their multiplicities.
First, let's consider the solution x = 3 of multiplicity 1. This means that (x - 3) is a linear factor in the equation.
Next, for the solution x = 1 of multiplicity 2, we need a quadratic factor (x - 1)^2.
Multiplying these factors together, we obtain an equation that satisfies the given conditions:
(x - 3)(x - 1)^2 = 0
Expanding this equation gives:
(x - 3)(x - 1)(x - 1) = 0
Simplifying further, we have:
(x - 3)(x^2 - 2x + 1) = 0
Distributing the terms:
x^3 - 2x^2 + x - 3x^2 + 6x - 3 = 0
Combining like terms:
x^3 - 5x^2 + 7x - 3 = 0
Thus, the equation x^3 - 5x^2 + 7x - 3 = 0 has a solution of x = 3 with multiplicity 1 and a solution of x = 1 with multiplicity 2.