How do you solve #(y+5)(y-1)=27#?
multiply the parentheses together
combine like terms
subtract 27 from both sides
figure out what numbers add to 4 and -32. It's -4 and 8
for the left side of the equation to equal zero, either y-4 or y+8 must equal zero (because zero times anything makes zero). y-4 is equal to zero when y = 4, and y+8 is equal to zero when y = -8. therefore,