How to factor 5(x+2)^2 -4(x+2) - 1

i got this far:

substitute (x+2) for a to get
5a^2 - 4a - 1

the answer says you then get:

(5a +1)(a-1)

however, I am confused as to how the 1, -1 get there? don't you have to find two numbers that multiply to -1 and add to -4?

Thank you!

there's only 1 way to factor the 5 in front of a², and that gives you
(5a ....) (a ....)
there's only 1 way to factor the 1 at the end, and that gives you
(5a .. 1) (a .. 1)
since the 1 at the end is negative, one of the factors must be negative, and since the 4 in the middle gets its sign from the larger of the 2 numbers adding up to it, the negative goes with the 1 that multiplies the 5, giving you
(5a + 1) (a – 1)
and when you check to see you got the signs right, the partial products are
first : 5a²
outside -5a
inside: +a
last: -1
and the middle two add to -4a.

then, of course, you put the x+2 back in place of a and get
[ 5(x+2) + 1] [ (x+2) – 1] =
(5x + 11) (x + 1)

1 * -1 = -1
5 * -1 + 1 * 1 = -4
Therefore,
5a^2 - 4a - 1 =
(5a + 1)(a - 1) =
(5x + 10 + 1)(x + 2 - 1) =
(5x + 11)(x + 1)

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

= 5a^2 -4a -1 where a = x+2
= (5a +1)( a -1)
= ( 5x +10+1)(x+2 -1) ///5a = 5(x+2) = 5x +10
= (5x +11)(x +1)

If you cross multiply you get

5a x a = 5a^2

5a x -1 = -5a

1 x a = a

1 x -1 = -1

adding -5a + a gives you your -4a

:-)