ANSWER
The solutions to this equation are: x = -2, x = -4 and x = -5
EXPLANATION
If (x + 2) is a factor, then x = -2 is one of the solutions to the equation. To find the other solutions we can divide the polynomial by this factor to find a quadratic equation that can be easily solved:
This means that this polynomial can be written as:
[tex]x^3+11x^2+38x+40=(x+2)(x^2+9x+20)[/tex]Now we can solve this equation to find the other two solutions:
[tex]x^2+9x+20=0[/tex]Using the following formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]For this problem:
[tex]\begin{gathered} x=\frac{-9\pm\sqrt[]{9^2-4\cdot1\cdot20}}{2\cdot1} \\ x=\frac{-9\pm\sqrt[]{81-80}}{2} \\ x=\frac{-9\pm1}{2} \\ x_1=\frac{-9+1}{2}=\frac{-8}{2}=-4 \\ x_2=\frac{-9-1}{2}=\frac{-10}{2}=-5 \end{gathered}[/tex]So finally, the polynomial has 3 factors and can be written as:
[tex]x^3+11x^2+38x+40=(x+2)(x+4)(x+5)[/tex]
