I find no rational zeros, but the real zero is very near -1.3, so
f(x) = (x+1.3)(2x^2+6.4x-6.12)
I assume you have some graphical or numeric tools for solving general polynomials.
what are the real and imaginary zeros of the equation 2x^3 + 5x^2 - 3x + 7
2 answers
oops - the real root is at -3.282