Either there is a mistake in
or I don't believe it is possible to find the exact real roots for this polynomial.
If what you posted is correct, then there are no rational roots. The only possible rational roots are 1 and -1 and neither of them work out to be roots. So the only real roots of this function would be irrational. The best one could do, I believe, is to use a graphing calculator:- Graph
- Then use the trace function to find decimal approximations for the x-coordinates of all the places where the graph intersects the x-axis. (If the graph never intersects the x-axis then there are no real roots.) I'll show you a graph at then end.
I suspect that the function is supposed to be:
Again, the only possible rational roots are 1 and -1. But this time 1 is actually a root. Here's the synthetic division:
1 | 1 -8 28 -56 70 -56 28 -8 1
---- 1 -7 21 -35 35 -21 7 -1
--------------------------------------------
1 -7 21 -35 35 -21 7 -1 0 <== a root!
Trying 1 again:
1 | 1 -7 21 -35 35 -21 7 -1
---- 1 -6 15 -20 15 -6 1
----------------------------------------
1 -6 15 -20 15 -6 1 0 <== a root!
Trying 1 again:
1 | 1 -6 15 -20 15 -6 1
---- 1 -5 10 -10 5 1
-----------------------------------
1 -5 10 -10 5 -1 0 <== a root!
Trying 1 again:
1 | 1 -5 10 -10 5 -1
---- 1 -4 6 -4 1
------------------------------
1 -4 6 -4 1 0 <== a root!
I think by now you can see where this is going. 1 is a root 8 times. In other words it is a root of multiplicity 8. And f(x) in factored form is:
FWIW, The following is the graph of the function you posted. It shows 5 points where the graph intersects the x-axis. But there are more such points outside of what we see on this graph. There will be 1, 2 or 3 additional points where the graph intersects the x-axis. We know there is at least 1 because an 8th degree polynomial which has a positive leading coefficient will go toward infinity for large positive and negative x's. So that steep "dive" we see at about x = 1.2 has to eventually come back up and cross the x-axis somewhere out to the right.