Respuesta :
If [tex]p(x)[/tex] is a polynomial of degree [tex]N[/tex] and [tex]x_1,\ldots x_N[/tex] are his roots, we can write
[tex]p(x)=(x-x_1)(x-x_2)\ldots(x-x_N)[/tex]
So, the exercise is basically asking you to find the three roots of the polynomial.
Cubic polynomials don't have a closed formula to solve them (well, they have, but it's very complicated), so we'd better use the rational root theorem.
This theorem states that the possible rational solutions of a polynomial with integer coefficients are of the form [tex]\frac{p}{q}[/tex], where p is a divisor of the known term, and q is a divisor of the leading term.
So, in our case, we have to try all the divisors of 243 (with both signs). We find out that
[tex]p(3) = 0[/tex]
So, x=3 is a solutions, and we can write as
[tex]p(x) = (x-3)r(x)[/tex]
And we can find the remaining polynomial by writing
[tex]r(x) = \dfrac{p(x)}{x-3} = x^2+81[/tex]
The solutions of this polynomial are
[tex]x^2+81 = 0 \iff x^2=-81 \iff x = \pm 9i[/tex]
This means that the solutions of [tex]x^3-3x^2+81x-243[/tex] are [tex]x=3[/tex], [tex]x=9i[/tex], [tex]x=-9i[/tex]
So, we can factor it as
[tex]x^3-3x^2+81x-243=(x-3)(x-9i)(x+9i)[/tex]
