To solve these quadratics by factoring, let's go through them step by step:

a. $5x^2 - 9x + 4$
We need two numbers whose product is $5 \cdot 4 = 20$ and whose sum is $-9$: $-5$ and $-4$.

$5x^2 - 9x + 4 = 5x^2 - 5x - 4x + 4 = (5x - 4)(x - 1)$

b. $6x^2 + 7x + 2$
We need two numbers whose product is $6 \cdot 2 = 12$ and whose sum is $7$: $4$ and $3$.

$6x^2 + 7x + 2 = 6x^2 + 4x + 3x + 2 = (2x + 1)(3x + 2)$

c. $4x^2 + 8x + 3$
We need two numbers whose product is $4 \cdot 3 = 12$ and whose sum is $8$: $6$ and $2$.

$4x^2 + 8x + 3 = 4x^2 + 6x + 2x + 3 = (2x + 1)(2x + 3)$

d. $6x^2 - 11x + 4$
We need two numbers whose product is $6 \cdot 4 = 24$ and whose sum is $-11$: $-8$ and $-3$.

$6x^2 - 11x + 4 = 6x^2 - 8x - 3x + 4 = (2x - 1)(3x - 4)$

e. $8x^2 - 14x - 15$
We need two numbers whose product is $8 \cdot (-15) = -120$ and whose sum is $-14$: $-20$ and $6$.

$8x^2 - 14x - 15 = 8x^2 - 20x + 6x - 15 = (4x + 3)(2x - 5)$

f. $12x^2 - 17x + 6$
We need two numbers whose product is $12 \cdot 6 = 72$ and whose sum is $-17$: $-9$ and $-8$.

$12x^2 - 17x + 6 = 12x^2 - 9x - 8x + 6 = (3x - 2)(4x - 3)$

g. $6x^2 + 5x + 1$
We need two numbers whose product is $6 \cdot 1 = 6$ and whose sum is $5$: $3$ and $2$.

$6x^2 + 5x + 1 = 6x^2 + 3x + 2x + 1 = (3x + 1)(2x + 1)$

h. $6x^2 - 5x - 2$
We need two numbers whose product is $6 \cdot (-2) = -12$ and whose sum is $-5$: $-6$ and $1$.

$6x^2 - 5x - 2 = 6x^2 - 6x + x - 2 = (3x - 2)(2x + 1)$

i. $6x^2 - 5x - 1$
We need two numbers whose product is $6 \cdot (-1) = -6$ and whose sum is $-5$: $-6$ and $1$.

$6x^2 - 5x - 1 = 6x^2 - 6x + x - 1 = (2x - 1)(3x + 1)$

Would you like additional clarifications or a focus on specific parts? Here are five related questions:

1. How do we determine which numbers to split the middle term into?
2. What if a quadratic cannot be factored using integers?
3. How can we confirm our factored answers are correct?
4. What are the alternatives to factoring for solving quadratics?
5. How does factoring apply to real-world problems?

Tip: Always check your factors by expanding them to ensure they multiply back to the original quadratic.