## Algebra 1: Common Core (15th Edition)

Factor the equation, and then use the zero product property. (If the product is 0, one of the factors is zero.) $3y^2−17y+24=0$ (3y−?)(y−?)=0 ⟶ The 2nd term is negative and the 3rd is positive, so the second terms in the factored equation are both negative. (3y−8)(y−3)=0 3y−8=0 OR y−3=0 ⟶ solve for t using addition property of equality 3y−8+8=0+8 OR y−3+3=0+3 3y=8 OR y=3 3y÷3=8÷3 OR y=3⟶ use the multiplication property of equality y=8/3 OR y=3 y=2 2/3 OR y=3