The expression you've provided is:

$\frac{x^2 + x - 12}{2x^2 - 7x + 3}$

Step 1: Factor the numerator and denominator

We'll start by factoring both the numerator and the denominator.

Factor the numerator $x^2 + x - 12$:

To factor $x^2 + x - 12$, we need to find two numbers that multiply to $-12$ (the constant term) and add to $1$ (the coefficient of the linear term).

The numbers $4$ and $-3$ work because: $4 \times -3 = -12 \quad \text{and} \quad 4 + (-3) = 1$

Thus, the numerator factors as: $x^2 + x - 12 = (x + 4)(x - 3)$

Factor the denominator $2x^2 - 7x + 3$:

To factor $2x^2 - 7x + 3$, we look for two numbers that multiply to $2 \times 3 = 6$ and add to $-7$.

The numbers $-6$ and $-1$ work because: $-6 \times -1 = 6 \quad \text{and} \quad -6 + (-1) = -7$

We can factor by grouping: $2x^2 - 7x + 3 = 2x^2 - 6x - x + 3$ $= 2x(x - 3) - 1(x - 3)$ $= (2x - 1)(x - 3)$

Step 2: Simplify the expression

Now we substitute the factored forms back into the original expression: $\frac{(x + 4)(x - 3)}{(2x - 1)(x - 3)}$

Notice that $(x - 3)$ appears in both the numerator and denominator, so they cancel out (as long as $x \neq 3$ to avoid division by zero).

The simplified expression is: $\frac{x + 4}{2x - 1}$

Final Answer:

$\frac{x^2 + x - 12}{2x^2 - 7x + 3} = \frac{x + 4}{2x - 1}, \quad \text{for } x \neq 3$

Would you like more details or have any questions?

Here are some related questions:

1. How do you factor quadratic expressions?
2. What is the importance of simplifying algebraic fractions?
3. What happens if $x = 3$ in the original expression?
4. Can all quadratic expressions be factored?
5. How do you handle complex roots when factoring?
6. What is the difference between factoring and expanding?
7. How do you determine the domain of a rational function?
8. What are some common mistakes to avoid when factoring?

Tip: Always check for common factors before starting to factor complex expressions. This simplifies the process.