\(x^3 + 3x^2 + 7x − 11 = 0\)
simplifies to:
\((x-1)(x^2+4x+11)=0\)
and
\(x^2+4x+11=0\)
which gives an imaginary number
but for the complex part,
you get the roots of x as:
\(x=\frac{-4±\sqrt{4^2-4*1*11}}{2}\)
whic becomes
\(x=2±\sqrt{-7}\)
so the roots of
\(x^3 + 3x^2 + 7x − 11 = 0\)
are
\(x = 1\)
and
\(x=2±\sqrt{-7}\)
.
