i need help dividing y^2 -5y+6 all over y^3 by y^2+3y-10 all over 4y^2?

Question

i don't know what steps i suppose to take

Answer 1

((y^2 - 5y +6) / y^3 ) / ((y^2+3y-10) / 4y^2) =

invert and multiply
(y^2 -5y +6) * 4y^2 / (y^2+3y-10)*y^3 =

eliminate y^2 top and bottom
4(y^2 -5y +6) / y(y^2+3y-10) =

factor
4(y-3)(y-2) / y (y-2)(y+5) =

eliminate (y-2) top and bottom
4(y-3) / y(y+5)
OR
(4y-12) / (y^2 +5y)

Answer 2

When dividing fractions, you multiply by the reciprocal.

[y^2 - 5y + 6] / [y^3] * [4y^2] / [y^2 + 3y - 10]

First, factor everything.

[(y - 2)(y - 3)] / [y^2(y)] * [4y^2] / [(y - 2)(y + 5)]

Now, if you see anything on the top AND bottom, cross it out.

You can cross out the (y - 2) from the numerator of the first and the denominator of the second. You can cross out the y^2 from the denominator of the first and the numerator of the second. Leave the 4 in the numerator of the second. Now, multiply like you multiply regular fractions.

(y - 3) / y * 4 / (y + 5)

[4(y - 3)] / [y(y + 5)]

You can either leave it this way or distribute.

(4y - 7) / (y^2 + 5y)

Answer 3

y^2-5y+6/y^3 divided by y^2+3y-10/4y^2
first you need to simply the equations--
(y-3)(y-2)/y^3 divided by (y-2)(y+5)/4y^2;
Second step is to make it a multiplication problem & take the reciprocal of 2nd fraction--so you now have:
(y-3)(y-2)/y^3 * 4y^2/(y-2)(y+5)
simplify by cross reducing which eliminates the (y-2) in the numerator of 1st & denominator of 2nd fraction; it also changes the y^3 to y in the denominator of 1st fraction & leaves you w/ 4 in the numerator of 2nd fraction; You are left w/
4(y-3)/y(y+5)=4y-12/y^2+5