Let P(x)=x4−16x3+86x2−176x+105
P(1)=1−16+86−176+105=0
∴x−1 is a factor of P(x)
Using synthetic division,
1−1686−176105x=101−1571−1051−1571−1050
∴x4−16x3+86x2−176x+105=(x−1)(x3−15x2+71x−105)
Let Q(x)=x3−15x2+71x−105
By trial and error method, we can find that x=3 is a root of Q(x)=0
Now, dividing Q(x) by x−3
1−1571−105x=303−361051−12350
∴Q(x)=(x−3)(x2−12x+35)
∴P(x)=(x−1)(x−3)(x2−12x+35)=(x−1)(x−3)(x−5)(x−7)
∴ The roots of P(x)=0 are 1,3,5,7
Diifference =|7−1|=6