Algebra 1 please help?

Question

4r^2 st^3 - 2 r^3 s^2t / 2rst

- im suppose to simplify - and im so confused i keep forgetting how to do this . please explain .

Answer 1

4r^2st^3-2r^3s^2t/ 2rst

2r^2st(2t^2-rs)/2rst

r(2t^2-rs)

Answer 2

4r^2 st^3 - 2 r^3 s^2t / 2rst

factor / get the common:

4r^2 st^3 - 2 r^3 s^2t = 2r^2st ( 2t^2 - rs)

2r^2st ( 2t^2 - rs) / 2rst

[2r^2st / 2rst] ( 2t^2 - rs) = r ( 2t^2 - rs)

r ( 2t^2 - rs) or (2rt^2 - r^2s) -----> ans

Answer 3

(4r^2st^3 - 2r^3s^2t)/2rst

= 2r^2st(2t^2 - rs)/2rst

= (2r^2st/2rst)(2t^2 - rs)

= (2/2)(r^2/r)(s/s)(t/t)(2t^2 - rs)

= [r^(2 - 1)][s^(1 - 1)][2t^2 - rs]

= [r][s^0][2t^2 - rs]

= r(2t^2 - rs)

Answer 4

here..

first,  you factorise for 2r^2st for both numerators

                                

                                  2r^2st ( 2t^2 - rs )

then divide all by 2rst

                                  2r^2st( 2t^2 - rs ) / 2rst = r(2t^2 - rs).

done.

Answer 5

4r^2 st^3 - 2 r^3 s^2t / 2rst

First we work with the integer coefficients

4, 2 and 2 are all factors of 2

so divide the nominator and the denom. by 2

2r^2 st^3 -  r^3 s^2t / rst

Now divide the two terms of the numerator by rst

that means subtract one from their exponents

and if the expon. is one to start with just cancel that variable

2rt² - r²s =====this is the answer

Answer 6

First factor the numerator:

2rst (2rt^2 - r^2s)

Then divide

2rst cancels 2 rst

you the have 2rt^2-r^2s

factor out r

r(2t^2-rs)