To factor the expression, we perform the following steps:
Separate the expression: [tex](x ^ 3 -4x ^ 2) + (4x-16)[/tex]
We take common factor: [tex]x ^ 2(x -4) +4 (x-4)[/tex]
We take common factor (x-4): [tex](x-4)(x ^ 2 + 4)[/tex]
We zero each expression and solve:
For the first expression:
x-4 = 0
x = 4 real root
For the second expression:
[tex]x ^ 2 + 4 = 0[/tex]
[tex]x ^ 2 = -4[/tex] complex roots
[tex]x = \sqrt{-4}[/tex] and [tex]x = -\sqrt{-4}[/tex]
[tex]x = 2\sqrt{-1}[/tex] and [tex]x = -2\sqrt{-1}[/tex]
x = 2i and x = -2i
We already have the real root and the complex roots, now we factor:
[tex]f(x) = (x-4) (x-2i) (x + 2i)[/tex]
Answer: The complete factorization of the given function over the field of complex numbers is
[tex]f(x)=(x-4)(x+2i)(x-2i).[/tex]
Step-by-step explanation: We are given to find the complete factorization of the following polynomial over the set of complex numbers :
[tex]f(x)=x^3-4x^2+4x-16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following formulas :
[tex](i)ab+cb=(a+c)b\\\\(ii)~a^2-b^2=(a+b)(a-b).[/tex]
From (i), we have
[tex]f(x)\\\\=x^3-4x^2+4x-16\\\\=x^2(x-4)+4(x-4)\\\\=(x-4)(x^2+4)[/tex]
Now, to factorize completely, we will use the following value of imaginary number i(iota) :
[tex]i=\sqrt{-1}~~~~~\Rightarrow i^2=-1.[/tex]
Therefore, we get
[tex]f(x)\\\\=(x-4)(x^2+4)\\\\=(x-4)(x^2-4i^2)\\\\=(x-4)(x^2-(2i)^2)\\\\=(x-4)(x+2i)(x-2i).[/tex]
Thus, the complete factorization of the given function over the field of complex numbers is
[tex]f(x)=(x-4)(x+2i)(x-2i).[/tex]
