## Algebra: A Combined Approach (4th Edition)

Chapter 6 - Section 6.3 - Exercise Set: 9 (Answer) $8y^2 - 17y + 9 = (y - 1)(8y - 9)$
Chapter 6 - Section 6.3 - Exercise Set: 9 (Solution) Factorize : $8y^2 - 17y + 9$ Factors of $8y^2 : 8y\cdot y or 4y\cdot 2y$ Factors of $9 : 1\cdot 9 or 3\cdot 3 or -1\cdot -9 or -3\cdot -3$ Try combination of those factors $(8y + 1)(y + 9) = 8y^2 + 73y + 9$ (Incorrect middle term) $(8y + 3)(y + 3) = 8y^2 + 27y + 9$ (Incorrect middle term) $(8y - 1)(y - 9) = 8y^2 - 73y + 9$ (Incorrect middle term) $(8y - 3)(y - 3) = 8y^2 - 27y + 9$ (Incorrect middle term) $(4y + 1)(2y + 9) = 8y^2 + 38y + 9$ (Incorrect middle term) $(4y + 3)(2y + 3) = 8y^2 + 18y + 9$ (Incorrect middle term) $(4y - 1)(2y - 9) = 8y^2 - 38y + 9$ (Incorrect middle term) $(4y - 3)(2y - 3) = 8y^2 - 18y + 9$ (Incorrect middle term) $(y + 1)(8y + 9) = 8y^2 + 17y + 9$ (Incorrect middle term) $(y + 3)(8y + 3) = 8y^2 + 27y + 9$ (Incorrect middle term) $(y - 1)(8y - 9) = 8y^2 - 17y + 9$ (Correct middle term) $(y - 3)(8y - 3) = 8y^2 - 27y + 9$ (Incorrect middle term) Thus, $8y^2 - 17y + 9 = (y - 1)(8y - 9)$