Help please - Page 2

**Kaiizu** · 08-15-2008, 03:51 AM

some things can't be factored and have to be solved using other methods.

**Furanshisu917** · 08-15-2008, 03:54 AM

(2n-7)(n+8)
step 1:2nxn = 2n^2
step 2:-7xn= -7n, 8x2n=16n, 16n-7n= 9n
step 3:-7x8=-56
(2n^2+9n-56)

factors of 2n^2 = 2nxn, nx2n
factors of 56=7x8, 8x7
is that not it??

**hoboX10** · 08-15-2008, 03:55 AM

Quote:

Originally Posted by Furanshisu917

theres 3 parts in the last bracket e.g (2n-0)(n-0-0) it must end up either (2n-7)(n-8) or (2n-8)(n-7) so both 0's in (n-0-0) must equal 7 or 8
example
2nx4=8n
2nx4=8n
nx-7=-7n
total= +9n
possible example answer: (2n-7)(n+4+4)

Not sure what you're getting at but that's not right.

(sorry hobo my bad i edited before u posted, no idea what i was getting at myself there tbh)

**Randmthotz** · 08-15-2008, 03:56 AM

i used the quadratic formula (thats what hobo posted) to get my answer, and i was always taught that its okay to have decimals (since you really only need to use the quadratic forumula when you cant factor normally)

**Furanshisu917** · 08-15-2008, 04:13 AM

but im sure my answer is right now.

(( this was my answer (2n-7)(n+8), steps are above))