SOLUTION: Hello I am currently trying to figure out this problem. Find all real and imaginary zeros for each polynomial function f(x)=x^3-27 This is how I started working out the

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello I am currently trying to figure out this problem. Find all real and imaginary zeros for each polynomial function f(x)=x^3-27 This is how I started working out the       Log On


   

Question 88390: Hello I am currently trying to figure out this problem.
Find all real and imaginary zeros for each polynomial function
f(x)=x^3-27
This is how I started working out the problem, I have no examples to really work with on this style of problem so I am not to sure if I have even started it correctly.
f(x)=x^3-27
x^3-27=0
+27=+27
x^3=27
I am stuck and not sure where to go next. Your help at solving this problem will be greatly appreciated.
Thank you
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3-27=0

%28x-3%29%28x%5E2%2B3x%2B9%29=0 Factor using the difference of cubes

x-3=0 Set the first factor equal to zero

x=3 Add 3 to both sides

x%5E2%2B3x%2B9=0 Set the second factor equal to zero


Now let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve x%5E2%2B3%2Ax%2B9=0 (notice a=1, b=3, and c=9)

x+=+%28-3+%2B-+sqrt%28+%283%29%5E2-4%2A1%2A9+%29%29%2F%282%2A1%29 Plug in a=1, b=3, and c=9



x+=+%28-3+%2B-+sqrt%28+9-4%2A1%2A9+%29%29%2F%282%2A1%29 Square 3 to get 9



x+=+%28-3+%2B-+sqrt%28+9%2B-36+%29%29%2F%282%2A1%29 Multiply -4%2A9%2A1 to get -36



x+=+%28-3+%2B-+sqrt%28+-27+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-3+%2B-+3%2Ai%2Asqrt%283%29%29%2F%282%2A1%29 Simplify the square root (note: since we cannot take the square root of a negative value, we must factor out i=sqrt%28-1%29 to make the radicand positive)



x+=+%28-3+%2B-+3%2Ai%2Asqrt%283%29%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=-1.5+%2B+2.59807621135332i or x=-1.5+-+2.59807621135332i


So the polynomial has one real zero:

x=3

and 2 imaginary zeros

x=-1.5+%2B+2.59807621135332i or x=-1.5+-+2.59807621135332i


Notice if we graph the polynomial y=x%5E3-27, we get

+graph%28+900%2C+900%2C+-35%2C+10%2C+-35%2C+10%2C+x%5E3-27%29+ graph of y=x%5E3-27

and we can see that the polynomial has one x-intercept (which is x=3). Since this polynomial has at most 3 complex roots, there must be 2 imaginary roots (because one is real). So our answer is verified.