SOLUTION: Did I also subtract the below equations correctly? 3a a+4 − (2a2−16a a2−16)= a3+8a2+16a a3+4a2−16a−64= a(a+4)(a+4) (a+4)(a+4)(a−4)= a a−4 Th

Question 1168890: Did I also subtract the below equations correctly?

3a
a+4
−
(2a2−16a
a2−16)=
a3+8a2+16a
a3+4a2−16a−64=

a(a+4)(a+4)
(a+4)(a+4)(a−4)=
a
a−4
Thank you

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
too hard to read
write it better
Tell exactly what you are given and what you want to know how to do or to understand.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

Did I also subtract the below equations correctly?

3a
a+4
−
(2a2−16a
a2−16)=
a3+8a2+16a
a3+4a2−16a−64=

a(a+4)(a+4)
(a+4)(a+4)(a−4)=
a
a−4
Thank you

Your simplification doesn't seem to make sense.
However, your final answer is CORRECT, considering that what you have is:
By the way, you need to present these problems better. This one should be written like this: 3a/(a + 4) - (2a^2 - 16a)/(a^2 - 16)
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