How do you factor (y+2)^2-1?

a^2 - b^2 ≡ (a + b)(a - b)

(y + 2)^2 - 1
= [(y + 2) + 1][(y + 2) - 1]
= (y + 2 + 1)(y + 2 - 1)
= (y + 3)(y + 1)

(y+2)^2-1 =
y^2 + 4y + 4 - 1 =
y^2 + 4y + 3 =
(y + 3)(y + 1)

[ ( y + 2 ) - 1 ] [ ( y + 2 ) + 1 ]

( y + 1 ) ( y + 3 )

consider this simple formula :
((a)^2)-((b)^2)=(a+b)*(a-b)
so,your problem modifies into:
((y+2)^2)-(1^2)=((y+2)+1)((y+2)-1) [since,1=(1^2)]
or,(y+3)(y+1)
which is the answer.

(y+2)^2 = y^2 + 4x + 4;
(y+2)^2 - 1 =y^2 + 4x + 3
=(y+3)(y+1)

Use a difference of squares:

(y + 2)^2 - 1
= (y + 2 - 1)(y + 2 + 1)
= (y + 1)(y + 3)

(y+2)^2-1=
(y+2+1)(y+2-1)=
(y+3)(y+1)

> factor((y+2)^2-1) = (y + 3) (y + 1)

(y+2)^2-1
= (y+2+1)(y+2-1)
= (y+3)(y+1)
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Ideas: Use the difference of two squares formula directly.