Question

HOW TO SOLVE (a-m)(b-m)..(x-m)(y-m)(z-m)

Answer 1

is anyone there\

Answer 2

Distribute.

Answer 3

how

Answer 4

(abcdefghijklmnopqrstuvwxyz)\(\times\)(26m)=?

Answer 5

0

Answer 6

?

Answer 7

I meant (abcdefghijklmnopqrstuvwxyz)-(26m)=?

Answer 8

No. Not zero.

Answer 9

doesnt it stay the same?

Answer 10

You're not saying (26m)-(26m). Just (abcdefghijklmnopqrstuvwxyz)-(26m)=?. Yes, it stays the same.

Answer 11

no it doesnt\

Answer 12

I'm fairly sure it does. There is nothing for you to go on but the alphabet.

Answer 13

what about m-m

Answer 14

It's not m+n+o, it's m\(\times\)n\(\times\)o. You can't take away from the product.

Answer 15

hmmm

Answer 16

Oh! I see where you're coming from. Gah, I'm so dense. That would leave you with 0 \(\times\) any random number, and commutative property wouldn't apply to that, so you have to leave it be.

Answer 17

so the answer is 0?

Answer 18

lol im still very confused

Answer 19

it is zero...because of the m-m term

Answer 20

elaborate i still dont understand...

Answer 21

(a-m)(b-m)...(l-m)(m-m)(n-m)...(y-m)(z-m)
=(a-m)(b-m)...(l-m)(0)(n-m)...(y-m)(z-m)=0

Answer 22

(a-m)(b-m) so -m-m

Answer 23

(a-m)(b-m)=ab-m^2????????????????????????????????????/

Answer 24

if you want (a-m)(b-m) you get

\[ab-am-mb+m^2\]

Answer 25

GAAAAAAAAAAAAAAAHHHHHH...loll

Answer 26

that would give you m^2 no?

Answer 27

?

Answer 28

ok when you look at the operation do you see -m-m
?

Answer 29

no

Answer 30

i see an ....

m-m

Answer 31

which is zero

Answer 32

but before the st m is a -

Answer 33

1st*

Answer 34

this is just a bunch of multiplication...one of the terms is (m-m)=0. Therefore the entire product is zero