Use the factor theorem to find a linear factor of x3 - 5x2 + 7x - 3. Hence solve the equation x3 − 5x2 + 7x − 3 = 0.

Let $P(x) = x^3 - 5x^2 + 7x - 3$. According to the Factor Theorem, if $P(a) = 0$, then $(x-a)$ is a factor of $P(x)$. We test integer factors of the constant term -3, which are $\pm 1, \pm 3$.

Test $x=1$:

Since $P(1) = 0$, $(x-1)$ is a linear factor of $P(x)$.

Divide $x^3 - 5x^2 + 7x - 3$ by $(x-1)$. We can use synthetic division or long division.
Using synthetic division with root 1:

1 | 1  -5   7  -3
  |    1  -4   3
  ----------------
    1  -4   3   0

The quotient is $x^2 - 4x + 3$.
So, $x^3 - 5x^2 + 7x - 3 = (x-1)(x^2 - 4x + 3)$.

We need to solve $x^2 - 4x + 3 = 0$.
This quadratic equation can be factored:

The roots from this quadratic factor are $x-1=0 \implies x=1$ and $x-3=0 \implies x=3$.

The solutions for $x^3 - 5x^2 + 7x - 3 = 0$ are the roots found from both the linear factor and the quadratic factor.
The roots are $x=1$ (from the initial factor), $x=1$ (from the quadratic factor), and $x=3$.

The linear factor is $\boxed{x-1}$ and the solutions to the equation are $\boxed{x=1, x=3}$.