how would you factor (a+b)^2 - 64
64 is perfect square
\(x^2-y^2=(x+y)(x-y)\) you will need this difference of two perfect squares
\((a+b)^2\) this quantity is a perfect square by it self
any clue? what is the number if we multiply it by it self we get 64
yeah the number would be 8
Good so you agree that \(64=8*8=8^2\)
yes i do agree so I would break it apart by (a+8)(a-8) because its negative 64
no no! just follow since you agreed now we can write \((a+b)^2-64=(a+b)^2-8^2\)
yes?
ok yes i see that
now treat a+b as one entity let's call it z a+b=z alright so \((a+b)^2-8^2=z^2-8^2\)
i don't need to always call it some name i just want to explain it to you so i becomes easy to deal with okay
yep no problem
now we have \(z^2-8^2\) difference of two square like is said before \(x^2-y^2=(x-y)(x+y)\) we do the same with \(z^2-8^2=(z-8)(z+8)\)
yes?!
yes
now z=a+b because i just called z it was a+b so i have \(z^2-8^2=(z-8)(z+8)=(a+b-8)(a+b+8)\)
So \((a+b)^2-64=(a+b-8)(a+b+8)\)
ok that makes so much more sense thank you for your help
My pleasure! once you know how to do it you don't need all those steps you go directly to write you factored form because we know that's the difference of two perfect squares
yeah i understood the difference of perfect squares i was just confusing myself when i had to difference of perfect squares to combine
the only step you need is \((a+b)^2-64=(a+b)^2-8^2\) then you write what it is directly
if you foil this \((x+y)(x-y)\) you will get \(x^2-y^2\) which is called difference of perfect squares
yeah i understand that now thank you do you think you can help with other problem too?
welcome! actually i have to go out now sorry
ok no problem
oh if my help satisfied you there is best response thing click on it^_^
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