The images you provided contain three quadratic trinomials. Let's factor each one by finding two binomials whose product is equal to the given trinomial.

1. $a^2 - 5a + 6$

To factor this trinomial, we need to find two numbers that multiply to $6$ (the constant term) and add to $-5$ (the coefficient of $a$).

The numbers that work are $-2$ and $-3$:

$a^2 - 5a + 6 = (a - 2)(a - 3)$

2. $y^2 - y - 12$

Here, we need to find two numbers that multiply to $-12$ (the constant term) and add to $-1$ (the coefficient of $y$).

The numbers that work are $-4$ and $3$:

$y^2 - y - 12 = (y - 4)(y + 3)$

3. $b^2 + 3b - 10$

For this trinomial, we need to find two numbers that multiply to $-10$ (the constant term) and add to $3$ (the coefficient of $b$).

The numbers that work are $5$ and $-2$:

$b^2 + 3b - 10 = (b + 5)(b - 2)$

Summary of Factored Forms:

1. $a^2 - 5a + 6 = (a - 2)(a - 3)$
2. $y^2 - y - 12 = (y - 4)(y + 3)$
3. $b^2 + 3b - 10 = (b + 5)(b - 2)$

Would you like more details on the factoring process, or do you have any other questions?

Here are five related questions to explore further:

1. How do you factor trinomials where the leading coefficient is not 1?
2. What are the steps to solve a quadratic equation by factoring?
3. How can you determine if a quadratic trinomial is factorable over the integers?
4. What is the relationship between the roots of the equation and the factored form?
5. How can factoring be used to solve real-world problems involving quadratic equations?

Tip: When factoring trinomials, always check if the leading coefficient (the coefficient of the squared term) is 1. If it is, the process is simpler, as you're just looking for two numbers that multiply to the constant term and add to the middle coefficient.