(y^3+4y^2-32y)÷(y+8)=

Question

(y^3+4y^2-32y)÷(y+8)=

Answer 1

1) y^3 + 4y^2 - 32y = y( y^2 + 4y - 32 ) = y( y - 4 )( y + 8 )
D = 16 + 128 = 144 = 12^2 
y1 = ( - 4 + 12 ) : 2 = 4 
y2 = ( - 4 - 12 ) : 2 = - 8
2) ( y( y - 4 )( y + 8 )) : ( y + 8 ) = y( y - 4 )