Question

Find the zeros (roots) of the following equations.

Answer 1

9514 1404 393

Answer:

  x ∈ {1/2, 1, 3, i, -i}

Step-by-step 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}

_____

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.