Browse · MATH Print → jmc algebra intermediate Problem The expression 3y2−y−24 can be written as (3y+a)(y+b), where a and b are integers. What is a−b? Solution — click to reveal We see that 3y2−y−24=(3y+8)(y−3), thus a=8 and b=−3. Hence, a−b=11. Final answer 11 ← Previous problem Next problem →