a^2+4a+4 2. 6x^2+22x+12 3. 8a^2-2a-1 u b^2+4b-5 square 16-49a^2

Let's factor each expression step by step:

1. $a^2 + 4a + 4$

This is a perfect square trinomial. It can be factored as:
$$(a + 2)^2$$

2. $6x^2 + 22x + 12$

First, factor out the greatest common factor (GCF), which is 2:
$$6x^2 + 22x + 12 = 2(3x^2 + 11x + 6)$$

Next, factor the quadratic inside the parentheses:
$$3x^2 + 11x + 6$$

We need to find two numbers that multiply to $3 \times 6 = 18$ and add up to 11. These numbers are 9 and 2.

So, we can rewrite the quadratic as:
$$3x^2 + 9x + 2x + 6$$

Now, factor by grouping:
$$= 3x(x + 3) + 2(x + 3)$$
$$= (3x + 2)(x + 3)$$

Thus, the factored form is:
$$2(3x + 2)(x + 3)$$

3. $8a^2 - 2a - 1$

This quadratic does not factor nicely with integer coefficients. We can use the quadratic formula to find the roots if needed:
$$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 8$, $b = -2$, and $c = -1$.

4. $b^2 + 4b - 5$

We need to find two numbers that multiply to $-5$ and add up to $4$. These numbers are $5$ and $-1$.

So, we can rewrite the quadratic as:
$$b^2 + 5b - b - 5$$

Now, factor by grouping:
$$= b(b + 5) - 1(b + 5)$$
$$= (b - 1)(b + 5)$$

Thus, the factored form is:
$$(b - 1)(b + 5)$$

5. $16 - 49a^2$

This is a difference of squares. It can be factored as:
$$16 - 49a^2 = (4)^2 - (7a)^2$$
$$= (4 - 7a)(4 + 7a)$$

So, the factored form is:
$$(4 - 7a)(4 + 7a)$$

In summary:
1. $(a + 2)^2$
2. $2(3x + 2)(x + 3)$
3. $8a^2 - 2a - 1$ (no simple factorization)
4. $(b - 1)(b + 5)$
5. $(4 - 7a)(4 + 7a)$