It is often difficult to factor a polynomial, but notice that the ratio of the coefficient of the first term to that of the second term is 1:3, and that the ratio of the coefficient of the third term to that of the fourth is also 1:3. Therefore, we can factor this cubic by grouping.
[tex]x^3-3x^2+5x-15=0\\x^2(x-3)+5(x-3)=0\\(x^2+5)(x-3)=0\\x^2+5=0,\ x-3=0[/tex]
We have three solutions in all.
Two imaginary:
[tex]x^2=-5\\x=\pm i \sqrt{5} [/tex]
One real:
[tex]x=3[/tex]
