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Question

The absolute difference of the maximum and the minimum value of x satisfying x416x3+86x2176x+105=0 is 


Solution

Let P(x)=x416x3+86x2176x+105
P(1)=116+86176+105=0
x1 is a factor of P(x)

Using synthetic division,
11686176105x=1011571105115711050
x416x3+86x2176x+105=(x1)(x315x2+71x105)
Let Q(x)=x315x2+71x105
By trial and error method, we can find that x=3 is a root of Q(x)=0

Now, dividing Q(x) by x3
11571105x=30336105112350
Q(x)=(x3)(x212x+35)
P(x)=(x1)(x3)(x212x+35)=(x1)(x3)(x5)(x7)
The roots of P(x)=0 are 1,3,5,7

Diifference =|71|=6

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