1. ## Divide the expression

Can someone help me divide the following expression: 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!

2. ## misunderstood first initial post

Originally Posted by jay1
Can someone help me divide the following : 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!
$\displaystyle \frac{112a^0b^3c^{-4}}{48a^3b^0c^3} \Rightarrow \frac{112}{48}\ \frac{a^0}{a^3}\ \frac{b^3}{b^0}\ \frac{c^{-4}}{c^3} \Rightarrow \frac{7}{3}\frac{1}{a^3}\frac{b^3}{1}\frac{1}{c^7} \Rightarrow \frac{7b^3}{3a^3c^7}$

3. I think that it may be:

$\displaystyle \frac{112a^0b^3c^{-4}}{48a^3b^0c^3}$

If this is correct, then remember that any value to the power of zero equals one, and $\displaystyle \frac{c^{-4}}{c^3}=\frac{1}{c^7}$

4. Originally Posted by jay1
Can someone help me divide the following expression: 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!
$\displaystyle \frac{112a^0b^3c^-4}{48a^3b^0c^3}$.

So, what we should do here is subtract all powers of like bases in the denominator from those that are in the numerator...

So....

$\displaystyle \frac{112a^0b^3c^{-4}}{48a^3b^0c^3}=\frac{112a^{0-3}b^{3-0}c^{-4-3}}{48}=\frac{112a^{-3}b^3c^{-7}}{48}$

So, note that the negative exponenets go into the denominator and become positive leaving us with

$\displaystyle \frac{112b^3}{48a^3c^7}$

Now, reduce the numerical coefficients

$\displaystyle \frac{7b^3}{3a^3c^7}$

5. That makes sense now Von Nemo..Thanks! Now, could someone help me verify my answer on performing the indicated operations and simplfying: 5a+3/a^2 +13a +12 - 2a-3/a^2 + 13a + 12 I have: -3(a+2)/(a+12)(a+1) ?

6. You should really use some parentheses.

Do you mean: $\displaystyle \frac{5a+3}{a^2}+13a+12-\frac{2a-3}{a^2}+13a+12$ ??

7. Originally Posted by Stroodle
You should really use some parentheses.

Do you mean: $\displaystyle 5a+\frac{3}{a^2}+13a+12-2a-\frac{3}{a^2}+13a+12$ ??
I'm sorry.. It's (5a +3 /a^2 +13a +12) - (2a - 3/a^2 +13a +12) My answer was 3(a+2)/(a+12)(a+1) ?? Thanks!

8. If you mean to simplify $\displaystyle \frac{5a+3}{a^2+13a+12}-\frac{2a-3}{a^2+13a+12}$

then $\displaystyle \frac{3(a+2)}{(a+12)(a+1)}$ is correct.