algII, factoring Options

[Freakerr]

i'm not quite sure how to factor these..


Factor out the GCF
(m-9)(m+1)+(m-9)(m+2)
^^I know i should use the grouping method.. so would it be (m+1)+(m+2)(m-9)??
orrrrr do i need an pair of brackets? [(m+1)+(m+2)](m-9) ??

m^5(r+s)+m^5(t+u)
^^grouping in that one too.. but i sorta have an idea but it may be wrong.. answer may be (m^5)(r+s)(t+u) ??

Factoring polynomials
(a+b)^2-(a-b)^2

that's alll.. any help would be greatlyy appreciated. thank you.

[innovation]

for factoring the GCF, the answer would be the second one because the common factor is (m-9).

for factoring polynomials, what are you asked to do? simplify? because that's already in factored form.

[Levy2k6]

does m( -9-1) + m(-9+2) work too? man i need to remmember this stuff too.. lol.

[*mona lisa*]

(m-9)(m+1)+(m-9)(m+2)--> (m+9)[(m+1)+(m+2)] cause (m+9) is included in both.

[innovation]

does m( -9-1) + m(-9+2) work too? man i need to remmember this stuff too.. lol.

nope, doesn't work. you can't exchange terms that are added or subtracted like that.

[Freakerr]

ummm..

i figured out this one..
(a+b)^2-(a-b)^2
------>x=(a+b) and y=(a-b)
^^like the different of two perf squares x^2-y^2=(x+y)(x-y)

[(a+b)-(a-b)][(a+b)+(a-b)]=
[a+b-a+b)(a+b+a-b)=
(2b)(2a)
final answer=4ab

riiiiight??


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more later

[*salcha*]

^ from what i can see, the process looks correct..