Question

I'm in need of help, end of the year is soon

a^2 + 5a + 4
-----------
a^3

divided by
a^2 + 3a + 2
-----------
a^2 - 2a

Answer 1

factor everything that you can in each fraction.

Answer 2

\(\huge\color{midnightblue}{ \rm \frac{a^2 + 5a + 4}{a^3} \div \frac{a^2 + 3a + 2}{a^2 - 2a} =}\)

\(\huge\color{midnightblue}{ \rm \frac{a^2 + 5a + 4}{a^3} \times \frac{a^2 - 2a}{a^2 + 3a + 2} =}\)

\(\huge\color{midnightblue}{ \rm \frac{(a+1)(a+4)}{a(a^2)} \times \frac{a(a-2)}{(a+1)(a+2)} =}\)

\(\huge\color{midnightblue}{ \rm \frac{(a+1)(a+4)a(a-2)}{a(a^2)(a+1)(a+2)} .}\)

Answer 3

You just need to cancel everything that you can out.

Answer 4

okay, so what you do is you change to multiplication?

Answer 5

so is that the final answer at the bottom?

Answer 6

No of course not. YOU need to cancel some terms out.

Answer 7

okay so let me see one moment please...

Answer 8

\(\huge\color{midnightblue}{ \rm \frac{\cancel{ (a+1)}(a+4)\cancel{ a}(a-2)}{\cancel{ a}(a^2)\cancel{ (a+1)}(a+2)} .}\)


\(\huge\color{midnightblue}{ \rm =? }\)

Answer 9

okay I think I got it...so the answer is...give me a sec

Answer 10

\[\frac{ (a + 4)(a - 2) }{ (a ^{2})(a + 2)}\]

Answer 11

@SolomonZelman was I right?

Answer 12

I wasn't sure if there was a way to simplify further

Answer 13

wait...\[\frac{ (a + 4)(a - 2) }{ (a^{2})(a + 2)}\]

(a + 4) and (a + 2) would be (a + 6) right?

Answer 14

or would they just cancel out?

Answer 15

yes correct

Answer 16

YAY!

Answer 17

Thank you @SolomonZelman and was I correct on that (a + 4) and (a + 2) would cancel out or be (a + 6) and let me post to final answer..

Answer 18

not (a+4) and (a+2) will not cancel

Answer 19

ok, so it would be (a + 6) right?

Answer 20

\(\large\color{midnightblue}{ \rm (a+4)(a+2) }\) numerator

\(\large\color{midnightblue}{ \rm a^2(a-2) }\) denominator

Answer 21

ooohhh I get it! Thank you so much! I gave you a medal :)

Answer 22

Have a nice day @SolomonZelman and I might be posting a few more questions, so if your available, could you help me?