Why can 8(4x^2-y^2) be factored further but 8(4x^2+y^2) cannot be factored further
because the first case is a difference of squares and your second case is prime
Because the first function is a difference of squares; when you factor it out, and multiply it back together, the middle term cancels the other one out. It doesn't work like that in the second function.
Thanks
CAn you explain what prime means
Has no factors except for itself and 1
We say that 7 is prime because its only factors are 7 x 1
but cant the second one become 8(4x^2+y^2) 8(2x+y)(2x+y) ????
shouldn't that be the final factored form?
Well why can't 7 become 3 x 5?
One I said cannot be factoed any further because 8(2x+y)(2x+y) is not equavolent to 8(4x^2+y^2) I understand now
Very good. It has to multiply back.
Thanks alot :)
prime in this case, means it can not be factored
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