First, we notice that the difference of 2 squares x^2 - 1
could be written as a product of 2 factors.
x^2 - 1 =
(x-1)(x+1)
If the given polynomial is divisible by x^2 - 1,
then x = 1 and x = -1 are the roots of the
polynomial.
We'll apply reminder
theorem:
P(-1) = 0 (the reminder is 0 if x = -1 is a
root)
P(1) = 0
We'll verify if
P(-1) is cancelling if we'll substitute x by -1 in the expresison of
polynomial:
P(-1) = [(-1)^2 - 1 - 1]^(4m+1) -
(-1)
P(-1) = (1 - 2)^(4m+1) +
1
P(-1) = (- 1)^(4m+1) +
1
(-1) raised to an odd power yields
-1.
P(-1) = -1 + 1 =
0
Therefore, x = -1 is a root of
P(x).
We'll verify if P(1) is cancelling if we'll
substitute x by 1 in the expresison of polynomial:
P(1) =
[(1)^2 + 1 - 1]^(4m+1) - (1)
P(1) = (1)^(4m+1) -
(1)
P(1) = 1 - 1
P(1) =
0
Since P(1) = 0, then x = 1 is also the root of
P(x).
Since x = 1 and x = -1 are the roots of
P(x), the polynomial (x^2+x-1)^(4m+1) - x is divisible by x^2 -
1.
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