5.Possible rational solutions are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30 Try 2 1 | 1 -1 -19 49 -30 | 1 0 -19 30 1 0 -19 30 0 So we have factored P(x) as Possible rational solutions to the second parenthetical expression set = 0 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30 Try 2 2 | 1 0 -19 30 | 2 4 -30 1 2 -15 0 So we have factored P(x) further as as Now just factor the trinomial: x-1=0; x-2=0; x+5=0; x-3=0 x=1; x=2 x=-5 x=3 Edwin