2y2-3y+2+5y2-1

Factor y2-3y+2 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 2 and whose sum is -3.

-2,-1

Write the factored form using these integers.

2(y-2)(y-1)+5y2-1

2(y-2)(y-1)+5y2-1

Simplify the denominator.

Rewrite 1 as 12.

2(y-2)(y-1)+5y2-12

Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=y and b=1.

2(y-2)(y-1)+5(y+1)(y-1)

2(y-2)(y-1)+5(y+1)(y-1)

2(y-2)(y-1)+5(y+1)(y-1)

To write 2(y-2)(y-1) as a fraction with a common denominator, multiply by y+1y+1.

2(y-2)(y-1)⋅y+1y+1+5(y+1)(y-1)

To write 5(y+1)(y-1) as a fraction with a common denominator, multiply by y-2y-2.

2(y-2)(y-1)⋅y+1y+1+5(y+1)(y-1)⋅y-2y-2

Multiply 2(y-2)(y-1) and y+1y+1.

2(y+1)(y-2)(y-1)(y+1)+5(y+1)(y-1)⋅y-2y-2

Multiply 5(y+1)(y-1) and y-2y-2.

2(y+1)(y-2)(y-1)(y+1)+5(y-2)(y+1)(y-1)(y-2)

Reorder the factors of (y-2)(y-1)(y+1).

2(y+1)(y+1)(y-1)(y-2)+5(y-2)(y+1)(y-1)(y-2)

2(y+1)(y+1)(y-1)(y-2)+5(y-2)(y+1)(y-1)(y-2)

Combine the numerators over the common denominator.

2(y+1)+5(y-2)(y+1)(y-1)(y-2)

Apply the distributive property.

2y+2⋅1+5(y-2)(y+1)(y-1)(y-2)

Multiply 2 by 1.

2y+2+5(y-2)(y+1)(y-1)(y-2)

Apply the distributive property.

2y+2+5y+5⋅-2(y+1)(y-1)(y-2)

Multiply 5 by -2.

2y+2+5y-10(y+1)(y-1)(y-2)

Add 2y and 5y.

7y+2-10(y+1)(y-1)(y-2)

Subtract 10 from 2.

7y-8(y+1)(y-1)(y-2)

7y-8(y+1)(y-1)(y-2)

Simplify 2/(y^2-3y+2)+5/(y^2-1)