Answer:
The solutions are [tex]x=-3,\:x=3[/tex]
Step-by-step explanation:
To factor this cubic polynomial [tex]x^3+3x^2-9x-27[/tex] you must:
[tex]x^3+3x^2-9x-27=\left(x^3+3x^2\right)+\left(-9x-27\right)[/tex]
[tex](-9x-27)=-9\left(x+3\right)[/tex]
[tex](x^3+3x^2)=x^2\left(x+3\right)[/tex]
[tex]x^3+3x^2-9x-27=-9\left(x+3\right)+x^2\left(x+3\right)[/tex]
[tex]-9\left(x+3\right)+x^2\left(x+3\right)=\left(x+3\right)\left(x^2-9\right)[/tex]
[tex]x^3+3x^2-9x-27=\left(x+3\right)\left(x^2-9\right)[/tex]
[tex]x^2-9=\left(x+3\right)\left(x-3\right)[/tex]
[tex]x^3+3x^2-9x-27= \left(x+3\right)\left(x+3\right)\left(x-3\right)[/tex]
[tex]x^3+3x^2-9x-27=\left(x+3\right)^2\left(x-3\right)=0[/tex]
[tex]x+3=0, \quad{x=-3}\\x-3=0, \quad{x=3}[/tex]
The solutions are
[tex]x=-3,\:x=3[/tex]