What is the quotient?

startfraction 2 y squared minus 6 y minus 20 over 4 y + 12 endfraction divided by startfraction y squared + 5 y + 6 over 3 y squared + 18 y + 27 endfraction
startfraction 2 over 3 (y minus 5) endfraction
startfraction 3 (y minus 5) over 2 endfraction
startfraction (y minus 5) (y + 2) squared over 6 (y + 3) squared endfraction
startfraction 3 (y + 5) (y minus 2) over 2 (y + 2) endfraction

StartFraction 3 (y minus 5) Over 2 EndFraction

Step-by-step explanation:

First we need to factor all the polynomials here using quadratic equation (-b +/- √(b^2 - 4ac)) / 2a :

* 2y^2 - 6y - 20 = 2(y^2 -3y - 10)

From quadratic equation, solutions are 5 and -2, which means that we can factor this to;

2(y+2)(y-5)

* y^2 + 5y + 6, in the same manner, is found to have solutions -3 and -2, so it can be factored to:

(y+3)(y+2)

* 3y^2 + 18y + 27 = 3(y^2 + 6y + 9)

Again, solving quadratic equation, we find solution to be -3, so we factor this to:

3(y+3)(y+3)

* 4y + 12 equals to 4(y+3)

Now, let's rewrite our polynomials:

2(y+2)(y-5) • 3(y+3)(y+3) / 4(y+3) • (y+3)(y+2)

We see that we can cancel out some factors here (x+3) and (x+2), so we are left with 6(y-5) / 4 which is 3(y-5)/2.

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.