Why can 8(4x^2-y^2) be factored further but 8(4x^2+y^2) cannot be factored further
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OpenStudy (precal):
because the first case is a difference of squares and your second case is prime
OpenStudy (sw050399):
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.
OpenStudy (waheguru):
Thanks
OpenStudy (waheguru):
CAn you explain what prime means
OpenStudy (mertsj):
Has no factors except for itself and 1
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OpenStudy (mertsj):
We say that 7 is prime because its only factors are 7 x 1
OpenStudy (waheguru):
but cant the second one become
8(4x^2+y^2)
8(2x+y)(2x+y) ????
OpenStudy (waheguru):
shouldn't that be the final factored form?
OpenStudy (mertsj):
Well why can't 7 become 3 x 5?
OpenStudy (waheguru):
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
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