It seems you're asking to simplify or solve the expression:

$\frac{2x^2 + x - 3}{x^2 - 6x + 5}$

Step 1: Factor the numerator and denominator

Numerator: $2x^2 + x - 3$

We need to factor this quadratic expression.

Looking for two numbers that multiply to $2 \times (-3) = -6$ and add to $1$, the middle term:

• The pair $3$ and $-2$ work, because $3 \times -2 = -6$ and $3 + (-2) = 1$.

So we can rewrite the numerator: $2x^2 + x - 3 = 2x^2 - 2x + 3x - 3 = 2x(x - 1) + 3(x - 1)$ Now factor out the common term $(x - 1)$: $2x^2 + x - 3 = (2x + 3)(x - 1)$

Denominator: $x^2 - 6x + 5$

We factor the denominator: $x^2 - 6x + 5 = (x - 5)(x - 1)$

Step 2: Simplify the expression

Now that both the numerator and denominator are factored, we have: $\frac{(2x + 3)(x - 1)}{(x - 5)(x - 1)}$

The $(x - 1)$ terms cancel out (as long as $x \neq 1$), leaving: $\frac{2x + 3}{x - 5}$

Final Simplified Expression:

$\frac{2x + 3}{x - 5}, \quad \text{for} \quad x \neq 1 \quad \text{and} \quad x \neq 5$

Would you like more details on any of these steps?

Here are five related questions to explore:

1. How do you factor quadratics?
2. How do you identify terms that cancel in rational expressions?
3. What happens if $x = 1$ or $x = 5$ in the original expression?
4. How can you apply the quadratic formula if factoring is difficult?
5. What are restrictions in rational expressions?

Tip: Always check for common factors before simplifying a fraction.