Question 165304
{{{f(x)=(x+1)(x^3-27)}}} Start with the given function



{{{0=(x+1)(x^3-27)}}} Plug in {{{f(x)=0}}}




{{{0=(x+1)(x-3)(x^2+3x+9)}}} Factor {{{x^3-27}}} to get {{{(x-3)(x^2+3x+9)}}}. Note: use the difference of cubes formula.



{{{x+1=0}}} or {{{x-3=0}}} or {{{x^2+3x+9=0}}} Set each individual factor equal to zero



Solve the first two equations to get {{{x=-1}}} or {{{x=3}}}. These are the first two zeros



Now let's solve {{{x^2+3x+9=0}}}



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=3}}}, and {{{c=9}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(3) +- sqrt( (3)^2-4(1)(9) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=9}}}



{{{x = (-3 +- sqrt( 9-4(1)(9) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9-36 ))/(2(1))}}} Multiply {{{4(1)(9)}}} to get {{{36}}}



{{{x = (-3 +- sqrt( -27 ))/(2(1))}}} Subtract {{{36}}} from {{{9}}} to get {{{-27}}}



{{{x = (-3 +- sqrt( -27 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3 +- 3i*sqrt(3))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-3+3i*sqrt(3))/(2)}}} or {{{x = (-3-3i*sqrt(3))/(2)}}} Break up the expression.  



So the zeros of {{{x^2+3x+9}}} are {{{x = (-3+3i*sqrt(3))/(2)}}} or {{{x = (-3-3i*sqrt(3))/(2)}}} 



which approximate to {{{x=-1.5+2.598i}}} or {{{x=-1.5-2.598i}}} 




=======================================================



Answer:



So in total, the 4 zeros are: {{{x=-1}}}, {{{x=3}}}, {{{x = (-3+3i*sqrt(3))/(2)}}} or {{{x = (-3-3i*sqrt(3))/(2)}}} 



Which in decimal form are  {{{x=-1}}}, {{{x=3}}}, {{{x=-1.5+2.598i}}} or {{{x=-1.5-2.598i}}}