Respuesta :
Answer:
x = 2i, x = -2i and x = 4 are the roots of given polynomial.
Step-by-step explanation:
We are given the following expression in the question:
[tex]f(x) = x^3 - 4x^2+ 4x - 16[/tex]
One of the zeroes of the above polynomial is 2i, that is :
[tex]f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0[/tex]
Thus, we can write
[tex](x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16[/tex]
Now, we check if -2i is a root of the given polynomial:
[tex]f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0[/tex]
Thus, we can write
[tex](x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16[/tex]
Therefore,
[tex](x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16[/tex]
Dividing the given polynomial:
[tex]\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4[/tex]
Thus,
[tex](x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16[/tex]
X = 4 is a root of the given polynomial.
[tex]f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0[/tex]
Thus, 2i, -2i and 4 are the roots of given polynomial.
