Answer:
StartFraction 3 (y minus 5) Over 2 EndFraction
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.