Question

What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27​

Answer 1

B. 3(y-5)/2 on edge

Explanation:

Answer 2

Answer with explanation:

`\rightarrow ((2y^2-6 y-20)/(4 y+12))/((y^2+5 y+6)/(3 y^2+28 y+27))\\\\\rightarrow ((y^2-3y-10)/(2 y+6))/(((y+2)(y+3))/(3 y^2+28 y+27))\\\\\rightarrow (((y-5)(y+2))/(2 (y+3)))/(((y+2)(y+3))/(3 y^2+28 y+27))\\\\\rightarrow ((y-5)(y+2))/(2 (y+3))} * {(3 y^2+28 y+27)/((y+2)(y+3))}\\\\ \rightarrow((y-5)*(3 y^2+28 y+27))/(2 (y+3)^2)}`

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

=(y+2)(y+3)

→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)