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

$x^{2}$ + bx + 15 Here, the numeric factors have to be factors of 15. Thus factors of 15 are: 1,3,5 and 15 Factors grouped such that their product is 15 $\rightarrow$ (1 $\times$ 15 and 3 $\times$ 5) Thus $x^{2}$ + bx + 15 could be = $x^{2}$ + x + 15x + 15 = $x^{2}$+ 16x + 15 , or = $x^{2}$ + 3x + 5x +15 = $x^{2}$ + 8x + 15 Thus b could be 16 or 8