9514 1404 393
Answer:
x ∈ {1/2, 1, 3, i, -i}
Explanation:
We assume your function is supposed to be ...
f(x) = 2x^5 -9x^4 +12x^3 -12x^2 +10x -3=0
I find it convenient to use a graphing calculator to find the real roots of higher-degree polynomials. The attached shows the real roots are ...
x = 1/2, 1, 3
Factoring those out leaves the quadratic factor ...
x^2 +1 = 0
That one has roots ±i.
So, the full complement of roots is ...
x ∈ {1/2, 1, 3, i, -i}
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Descartes' rule of signs tells you the real roots are all positive. The rational root theorem tells you that rational roots are in the set {1/2, 1, 3/2, 3}. The sum of coefficients is 0, telling you that x=1 is a real root.
Factoring that from the polynomial leaves the 4th-degree factor ...
2x^4 −7x^3 +5x^2 −7x +3 . . . . . division shown in 2nd attachment
For x=1, this has a negative value (-4), so you know there is a root between f(0) =3 and f(1) = -4. The only rational choice is 1/2.
Similarly, factoring that from the polynomial (along with a factor of 2) leaves the cubic ...
x^3 -3x^2 +x -3 . . . . . division shown in 3rd attachment
This is clearly factorable by grouping to (x -3)(x^2 +1), so we have our third real root and the quadratic that yields the complex roots.