Factor completely, can someone do this

(x+4)^2-7(x+4)+6


please?

x^2 + 8x + 16 - 7x - 28 + 6
x^2 + x - 6
(x+3)(x-2)

First, make this simpler for yourself by replacing the (x+4) with a different variable, such as "a"
a^2 - 7a + 6

This might look like a more familiar factoring problem now.

Find two numbers that add to -7 and multiply to 6. Using trial and error, you can figure out:
-6 + -1 = -7
-6 * -1 = 6

So the numbers -6 and -1 work out. Write that as:
(a - 6)(a - 1)

Now, remember that "a" used to be (x+4). Put that back in, in place of a.
(x + 4 - 6)(x + 4 - 1)

And simplify by doing the subtraction of the numbers within each set of parentheses:
(x - 2)(x + 3)

x^2 + 8x + 16 - 7x - 28 + 6
x^2 + x - 6
(x + 3)(x - 2)

x^2 + 8x + 16 - 7x - 28 + 6
x^2 + x - 6
(x + 3)(x - 2)

u = x+4

= u^2 - 7u + 6
(u - 6) (u - 1),

(x+4-6)(x+4-1) = (x - 2)(x + 3)

to multiply (x+4)^2 it's x^2+2ab+b^2

x^2+8x+16-7x-28+6
x^2+x-6
(x-2)(x+3)

x^2+x-6