OpenStudy (anonymous):

Show all work in dividing ((y^2-y-12)/(y^2-6y+8)) / ((y^2+2y-15)/(y^2-7y+10)). Use complete sentences to explain how to simplify this expression.

OpenStudy (anonymous):

alright so this is what we are going to do i am going to define some variables to make our life a little bit easier alright so we will let a= y^2-y-12, b =y^2-6y+8, c =y^2+2y-15, and d=y^2-7y+10 and now our expressions looks like (a/b)/(c/d) or in other words we have (a/b)*(d/c) and notice we had to flip our c to d because we are multiplying okay so now we will factor a, b,c, and d so for a = y^2 - y- 12 our product is -12 and sum is -1 our two numbers are -4 , 3 so we got (y-4)(y+3) now to factor b = y^2-6y+8 our product is 8 and sum is -6 our two numbers are -4, -2 so we got (y-4)(y-2). for c=y^2+2y-15 our product is -15 and sum is 2 our two numbers are 5, -3 so (y+5)(y-3) and finally d=y^2-7y+10 our product is 10 and sum is -7 our two numbers are -5, -2 so we got (y-5)(y-2) and now let us sub all our factorization into our expression and then we can maybe cancel some expressions out so now we got [(y-4)(y+3)/ (y-4)(y-2)]*[(y-5)(y-2)/(y+5)(y-3) ] notice y-4 cancels out and now we have [(y-5)(y-2)(y+3)/(y+5)(y-3)(y-4) ] and as far as i can see nothing else cancels out :)

OpenStudy (anonymous):

Thanks, you rock!

OpenStudy (anonymous):

@hamza_b23 the final thing you wrote which was [(y-5)(y-2)(y+3)/(y+5)(y-3)(y-4) ], this is the final answer? simplified and all?

OpenStudy (anonymous):