Given that:
The polynomial is s( x) =x 4 - 5 x 3 + 5 x 2 + 5x - 6.
By using,
The Rational Zero Theorem:
If f (x) = a n x n + a n - 1 x n - 1 + . . . + a1 x + a0 has integer coefficients and
( where is reduced ) is a rational zero .
p = a0 and q = a n .
To find the actual zeros of the polynomial:
Degree of the given polynomial is 4 then it has 4 zeros of some kind.
By using Rational zero theorem,
Constant term of the polynomial is 6.
Leading coefficient = 1
Factors of constant term is - 6 :
-6 =
Factors of leading coefficient 1 :
1 = 1
All possible combinations of .
Some or none may be zeros of the functions.
these are the possible zeros of the given polynomial.
To determine which of the possible zeros are actual zeros by substituting these x values for x in s (x) :
s( 1) =1 4 - 5 (1) 3 + 5 (1) 2 + 5(1) - 6
= 1 - 5 + 5 + 5 - 6
= 0
s( 1) =(-1) 4 - 5 (-1) 3 + 5 (-1) 2 + 5(-1) - 6
= 1 + 5 + 5 - 5 - 6
= 0
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