first factor the denominator:
[tex]8 {x}^{2} - 14x - 15[/tex]
factors of 8: 1, 2, 4, 8
factors of 15: 1, 3, 5, 15
2×3 = 6 and 4×5 = 20, and 20-6 = 14
So now we have:
(2x-5)(4x+3)
let's make that our common denominator, so now it becomes:
[tex] \frac{3(2x - 5)}{(2x - 5)(4x + 3)} + \frac{21}{(2x - 5)(4x + 3)} [/tex]
add the numerators, then place that over the common denominator
[3(2x - 5) + 21] / [(2x-5)(4x+3)]
6x-15+21 = 6x+6
= 6(x+1) / [(2x-5)(4x+3)]
Therefore C), or the third answer, is correct
