help with Alg 2 ( factor by grouping)?

Question

please explain how to do it:

1. y^2 - a^2 - 6a - 9

2. a^2 - a -ac + c

3. 2nr + 3p^2 + 6pr + pn

Answer 1

Ok, let's see if I remember, what you have to do is that you group two similar expressions in the equation, take for example the problem number one

y^2 - a^2 - 6a - 9

ok, in this case you have to group the parts that contain an a and the 9, so you get

y^2 - (a^2 + 6a + 9)

now, when you factor the expression into the parenthesis you get

y^2 - (a + 3)^2

and you can factor it even more like this

(y-a-3)*(y+a+3)

and there you have it :D it's kind of the same with the rest

2. a^2 - a -ac + c

you group the terms this way

(a^2 - a) -(ac - c)

and you take out the common term

a*(a-1)-c*(a-1)

and regroup

(a-c)*(a-1)

and that's it, :D

3. 2nr + 3p^2 + 6pr + pn

you group them this way

(3p^2 + 6pr) + (pn + 2nr)

you take our the common term

3p(p+2r) + n(p+2r)

and you regroup

(3p+n)*(p+2r)

easy ;o), hope this helps.

Answer 2

1) y² - a² - 6a - 9 =
y² - ( a² + 6a + 9 ) =
y² - [(a + 3)(a+3)] =
y² - (a+3)² =
(y + (a+3)) (y - (a+3) ) =
(y + a + 3 ) (y - a -3 )

2) a² - a - ac + c =
(a (a -1)) + (c ( -a +1) )=
(a (a-1) ) + ( -c (a - 1)) =
(a-1)(a-c)

3) 2nr + 3p² + 6pr + pn =
2r(n + 3p) + p( 3p + n) =
(n +3p) (2r+p)

Answer 3

1. y^2 - a^2 - 6a - 9
=y² - ( a² + 6a + 9 )
=y² - (a+3)²
=(y + (a+3)) (y - (a+3) )
=(y + a + 3 ) (y - a -3 )

2. a^2 - a -ac + c
=a(a-1)-c(a-1)
=(a-c)(a-1)

3. 2nr + 3p^2 + 6pr + pn
=3p^2 + 6pr + pn +2nr
=3p(p+2r)+n(p+2r)
=(3p+n)(p+2r)

Answer 4

(y+a-3)(y-a+3)
(a-c)(a-1)
(n+3p)(2r+p)