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

Chapter 6 - Section 6.3 - Exercise Set: 12 (Answer) $36r^2 - 5r - 24 = (4r + 3)(9r - 8)$
Chapter 6 - Section 6.3 - Exercise Set: 12 (Solution) Factorize : $36r^2 - 5r - 24$ Factors of $36r^2$ : $36r\cdot r$ or $18r\cdot 2r$ or $12r\cdot 3r$ or $9r\cdot 4r$ or $6r\cdot 6r$ Factors of -24 : $1\cdot -24$ or $2\cdot -12$ or $3\cdot -8$ or $4\cdot -6$ or $6\cdot -4$ or $8\cdot -3$ or $12\cdot -2$ or $24\cdot -1$ Try combination of those factors, ignoring those pairs with GCF that can be further factorized $(36r + 1)(r - 24) = 36r^2 - 863r - 24$ (Incorrect middle term) $(18r + 2)(2r - 12)$ (Not to be considered as it can be factorized as 4(9r + 1)(r – 6)) $(12r + 3)(3r - 8)$ (Not to be considered as it can be factorized as 3(4r + 1)(3r – 8)) $(4r + 3)(9r - 8) = 36r^2 - 5r - 24$ (Correct middle term) Thus, $36r^2 - 5r - 24 = (4r + 3)(9r - 8)$