How do you find the number of complex zeros for the function f(x)=27x^9+8x^6-27x^3-8f(x)=27x9+8x6−27x3−8?
1 Answer
Explanation:
By the Fundamental Theorem of Algebra
We can find the zeros by factoring by grouping and using some identities:
Difference of squares:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
Difference of cubes:
a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
Sum of cubes:
a^3+b^3 = (a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
Note that in these latter two identities, the resulting quadratic factors have no linear factor with Real coefficients.
So we find:
27x^9+8x^6-27x^3-827x9+8x6−27x3−8
= (27x^9+8x^6)-(27x^3-8)=(27x9+8x6)−(27x3−8)
= x^6(27x^3+8)-1(27x^3-8)=x6(27x3+8)−1(27x3−8)
= (x^6-1)(27x^3+8)=(x6−1)(27x3+8)
= ((x^3)^2-1^2)((3x)^3+2^3)=((x3)2−12)((3x)3+23)
= (x^3-1)(x^3+1)(3x+2)(9x^2-6x+4)=(x3−1)(x3+1)(3x+2)(9x2−6x+4)
= (x-1)(x^2+x+1)(x+1)(x^2-x+1)(3x+2)(9x^2-6x+4)=(x−1)(x2+x+1)(x+1)(x2−x+1)(3x+2)(9x2−6x+4)
So there are three Real zeros:
{ (x=1),(x=-1),(x=-2/3) :}
And six non-Real Complex zeros:
{ (x = -1/2+-sqrt(3)/2i), (x = 1/2+-sqrt(3)/2i), (x = 1/3+-sqrt(3)/3i) :}
