how do you factor 81a^2-100b^6

Question

hartnn · Answer

81 is the square of ?

anonymous · Answer

a

anonymous · Answer

$$81a^2-100b^2$$

hartnn · Answer

do u know the square root of 81 and 100 ?

anonymous · Answer

9 & 10

hartnn · Answer

right.
so u can write that as 
$\large (9a)^2-(10b)^2$
ok?

hartnn · Answer

now use $\huge a^2-b^2=(a+b)(a-b)$

anonymous · Answer

i understood the first part you lost me at the second message

anonymous · Answer

& being that its $$100b^6$$ would it be $$( 10b)^6$$

hartnn · Answer

is it 100b^6
because you wrote it as 100b^2 afterwards

anonymous · Answer

oh okay , right

hartnn · Answer

so what is it?

anonymous · Answer

$$100b^6$$

hartnn · Answer

ok, so 100 b^6 can be written as (10b^3)^2
ok?

anonymous · Answer

oh okay because you multiply the exponents?

hartnn · Answer

yup.

hartnn · Answer

now u have 
(9a)^2 - (10b^3)^2
use the basic identity A^2-B^2 to factor this:
put A=9a, B = 10b^3

hartnn · Answer

$\huge A^2-B^2=(A+B)(A-B)$

hartnn · Answer

so (9a)^2- (10b^3)^2 will be 
(9a+10b^3)(9a-10b^3)
did u understand ?

anonymous · Answer

not at all :(

hartnn · Answer

which part ?
upto here (9a)^2- (10b^3)^2?

anonymous · Answer

yeah

hartnn · Answer

now there is a basic identity to be used :
$\huge A^2-B^2=(A+B)(A-B)$
put A=9a and B=10b^3 here

anonymous · Answer

ok

hartnn · Answer

did u get (9a+10b^3)(9a-10b^3)  ?

anonymous · Answer

yeah

hartnn · Answer

thats your final answer then :)

anonymous · Answer

^_^ Really?

hartnn · Answer

YES!!
absolutely :)

hartnn · Answer

u have factored that successfully

hartnn · Answer

factoring means bringing in the form of (...)(...)

anonymous · Answer

okay i just tried the next one can you correct me if im wrong

hartnn · Answer

sure.

anonymous · Answer

$$64a^4-25y^4$$ thats the question .

anonymous · Answer

$$(8a^2)^2-(5y^2)^2$$

hartnn · Answer

that is correct, and what will u do to factorize ?

anonymous · Answer

that idk

hartnn · Answer

use same thing
A^2-B^2 = ??

anonymous · Answer

(a+b)(a-b)?

hartnn · Answer

yup.so what will be $(8a^2)^2-(5y^2)^2$

anonymous · Answer

that system works for them all ?

hartnn · Answer

yes, for anything of the form of difference of squares

anonymous · Answer

aslo , if one of the numbers arent a perfect square its unfactorable ?

hartnn · Answer

then check whether its a perfect cube....but it really depends on problem