In Algebra 2 {3b^2+3[2b-(5-b)]}=?

Question

ok what do i do first I know the pemdas but come on lol

Answer 1

if you really know what pemdas is, you know what things go first.

Answer 2

looks like a load of distribution to clear parenthesis to me, start on the inside and work your way out

so i think that first you would distribute the - to 5-b giving you...

2(3b^2+3(2b-5+b))

next the 3 to the 2b-5+b gives you

2(3b^2+6b-15+3b)

next distribute the 2...

6b^2+12b-30+6b

then combine.....6b^2+18b-30

then you could brake that down into factors looks like you got a GCF of 6 so

6(b^2+3b-5)

and it doesn't look like it brakes down any more than that

Answer 3

{3b^2+3[2b-5+b]}

{3b^2+3[3b-5]}

{3b^2+9b-15}

Answer 4

2[3b^2+3(3b-5)]

2[3b^2+(9b-15)]

= 6b^2+18b-30

= 6(b^2+3b-5)

=6(b^2+3b-5)

idk how to solve it from there.

(that's if the 2 is in front of the equation, if you meant to just write Algebra 2 and then the equation, it would be this:

[3b^2+3(3b-5)]

[3b^2+(9b-15)]

= 3b^2+9b-15

= 3(b^2+3b-5)

=3(b^2+3b-5)