# Thread: More Factoring

1. ## More Factoring

I'm having some more trouble: Please help me, a little is very appreciated and a lot of help would be a God-send. Thanks, I love this community and I wish I would have found this place many years ago.
Problem 11:
a^2(b+c)(2b+c)+a(b+c)
I got this far: (b+c)[a^2(2b+c)+a]
Same as problem 12. Doesn't come out with same answer when I plug in values to check.

Problem: 12:
t^3(2r+s)(3r+2s)+t^2(3r+2s)
I got this far: (3r+2s)[t^3(2r+s)+t^2]
If I factor out t, I can check to see if it works. by doing t=5, r=4, s=3 and I get the same answer as the problem originally. However, if I factor out t^2 it makes the answer different...any reason or explanation why?

Problem 13:
(2a+1)(a+3)-(2a+1)(a-3)
I got this far: (2a+1)[(a+3)-(a-3)]
I know this is a difference of squares, but what do I do with it?

Problem 15: 3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)
I got this far: (5a+2b)[3a(3a-5b)-9(a^2)(4a-b)]
No idea where to go with this one.

and Problem 16: 6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)]
I got this far: 3n-2m[(6m)(n+m)+8(m^2)(n+m)]
And tried : n+m(3n-2m)[6m+8(m^2)]
Same as before, no idea where to go either.

Any help would be appreciated.
Thanks

2. Originally Posted by andy011789
I'm having some more trouble: Please help me, a little is very appreciated and a lot of help would be a God-send. Thanks, I love this community and I wish I would have found this place many years ago.
Problem 11:
a^2(b+c)(2b+c)+a(b+c)
I got this far: (b+c)[a^2(2b+c)+a]
Same as problem 12. Doesn't come out with same answer when I plug in values to check.
$\displaystyle a^2(b + c)(2b + c) + a(b + c)$

Pull the $\displaystyle a(b + c)$ out from both terms:
$\displaystyle a(b + c)[a(2b + c) + 1] = a(b + c)(2ab + ac + 1)$

Originally Posted by andy011789
Problem: 12:
t^3(2r+s)(3r+2s)+t^2(3r+2s)
I got this far: (3r+2s)[t^3(2r+s)+t^2]
If I factor out t, I can check to see if it works. by doing t=5, r=4, s=3 and I get the same answer as the problem originally. However, if I factor out t^2 it makes the answer different...any reason or explanation why?
The odds are you did something wrong with your simplification. Since you didn't show us your work, I can only speculate.

$\displaystyle t^3(2r + s)(3r + 2s) + t^2(3r + 2s)$

You can factor a $\displaystyle t^2(3r + 2s)$:
$\displaystyle t^2(3r + 2s)[t(2r + s) + 1] = t^2(3r + 2s)(2rt + st + 1)$

Originally Posted by andy011789
Problem 13:
(2a+1)(a+3)-(2a+1)(a-3)
I got this far: (2a+1)[(a+3)-(a-3)]
I know this is a difference of squares, but what do I do with it?
Difference of what squares? I don't see any.

You factored correctly, now simplify:
$\displaystyle (2a+1)[(a+3)-(a-3)] = (2a + 1)[a + 3 - a + 3] = 6(2a + 1)$

Originally Posted by andy011789
Problem 15: 3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)
I got this far: (5a+2b)[3a(3a-5b)-9(a^2)(4a-b)]
No idea where to go with this one.
$\displaystyle 3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)$

You can factor a $\displaystyle 3a(5a + 2b)$ from each term:
$\displaystyle 3a(5a + 2b)[(3a - 5b) - 3a(4a - b)]$

Now simplify.

Originally Posted by andy011789
Problem 16: 6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)]
I got this far: 3n-2m[(6m)(n+m)+8(m^2)(n+m)]
And tried : n+m(3n-2m)[6m+8(m^2)]
Same as before, no idea where to go either.
Remember the parenthesis!!

$\displaystyle 6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)$

Again you can factor more than you did: you can take a $\displaystyle 2m(3n - 2m)(n + m)$ out of both terms.

$\displaystyle 2m(3n - 2m)(n + m)[3 + 4m]$

If your terms have more than one thing in common you can factor them all out. If you like, for simplicity and until you get used to the idea, you can take them out one by one. No harm there and it will make checking your work easier. With this last one your only flaw (aside from the parenthesis(!)) was that you didn't go far enough. Both terms still had something you could factor out.

-Dan