## Calculus with Applications (10th Edition)

$\frac{m^{2}+4}{4}$
Step 1: First, you have to factor the numerator and the denominator so they are completely simplified. To simplify the numerator, the only method you can use is the difference of squares since $m^{4}$ and 16 are both squares and they one is being subtracted from another. For difference of squares, you take the square root of both terms separately and both add and subtract them from each other in 2 separate parenthesis that are being multiplied together. This is what it would simplify to: $(m^{2}+4)$$(m^{2}-4). However, that is still not completely simplified because there is another difference of squares since m^{2} and 4 are both perfect squares. So you repeat the difference of squares process to that term only. You CANNOT factor (m^{2}+4) because they are not being subtracted, they are being added and it is the difference of squares, not the sum. So after applying the same process, you should get: (m^{2}+4)$$(m+2)$$(m-2)$ which is completely simplified. To factor the denominator is close to the same process. However, there is a common factor in both $4m^2$ and 16, and that is 4. You can factor 4 out of the binomial by dividing it from the coefficients. $4\div4=1$ and $16\div4=4$, therefore you are left with: $4(m^{2}-4)$ which is again a difference of squares. You can apply the same process as above and you will end up with: 4(m+2)(m-2) in the denominator. Be careful to not forget the 4! It is still apart of the equation! Step 2: Once you have everything factored in your fraction you will have: $\frac{(m^{2}+4)(m+2)(m-2)}{4(m+2)(m-2)}$. Now you can finish simplifying the fraction by canceling out any terms that are in both the numerator and the denominator. You can do this because any number divided by itself equals 1. The (m+2) and the (m-2) both cancel out and the remaining terms that cannot be simplified any more is your answer which is: $\frac{m^{2}+4}{4}$