How do you factor (x+2)^2-7(x+2)+12 ?

First foil and distribute

x^2 +4x +4 - 7x-14+12

Then combine like terms (4x) and (-7x) = (-3x) ... (+4) , (-14) and (+12) = (+2)

x^2 -3x + 2

Then factor

(x - 2)(x - 1)

x+2-->y

(x+2)^2-7(x+2)+12 ---> y^2-7y+12

= (y-4) * (y-3)

then y--->x+2

= (x+2-4) * (x+2-3)

= (x-2) * (x-1)

y ² - 7 y + 12
(y - 4) (y - 3)
[ (x + 2) - 4 ] [ x + 2 - 3 ]
(x - 2) (x - 1)

You already have the method of factoring after substituting (x+2) with another variable then replacing after factoring. You can also leave it, preform the functions within the equation, consolidate and then factor, thus:

(x+2)^2 -7(x+2) +12

x^2+4x+4 -7x-14 +12

x^2 -3x +2

(x-2)(x-1)

Let x + 2 = y.

(x + 2)^2 - 7(x + 2) + 12
= y^2 - 7y + 12
= y^2 - 3y - 4y + 12
= (y^2 - 3y) - (4y - 12)
= y(y - 3) - 4(y - 3)
= (y - 3)(y - 4)
= [(x + 2) - 3][(x + 2) - 4]
= [x + 2 - 3][x + 2 - 4]
= [x - 1][x - 2]

[(x+2)-4][(x+2)-3]=(x-2)(x-1)
===================================================CHECK:x^2-3x+2
CHECK:x^2+4x+4-7x-14+12=x^2-3x+2

This is broken down to where you hopefully can understand.

(x+2)^(2)-7(x+2)+12

((x+2)-3)((x+2)-4)

(x+2-3)((x+2)-4)

(x-1)((x+2)-4)

(x-1)(x+2-4)

(x-1)(x-2)

If you can factor:
u^2 - 7u +12 you can factor your problem
It factors as:
(u - 3)( u -4)
so we get:
[(x+2) - 3][(x+2) - 4]
you can combine terms and get:
(x-1)(x-2)

(x + 2) is a binomial number, just like other numbers and variables. So, think:

How would I factor:
y^2 - 7y + 12 = (y - 4)(y - 3)

Does that help...