Question

What is the complete factorization of the polynomial function over the set of complex numbers?

f(x)=x3−4x2+4x−16



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Answer 1

To factor the expression, we perform the following steps:

Separate the expression:
`(x ^ 3 -4x ^ 2) + (4x-16)`

We take common factor:
`x ^ 2(x -4) +4 (x-4)`

We take common factor (x-4):
`(x-4)(x ^ 2 + 4)`

We zero each expression and solve:

For the first expression:

x-4 = 0

x = 4 real root

For the second expression:

`x ^ 2 + 4 = 0`

`x ^ 2 = -4` complex roots

`x = √(-4)` and
`x = -√(-4)`

`x = 2√(-1)` and
`x = -2√(-1)`

x = 2i and x = -2i

We already have the real root and the complex roots, now we factor:

`f(x) = (x-4) (x-2i) (x + 2i)`

Answer 2

Answer: The complete factorization of the given function over the field of complex numbers is

`f(x)=(x-4)(x+2i)(x-2i).`

Step-by-step explanation: We are given to find the complete factorization of the following polynomial over the set of complex numbers :

`f(x)=x^3-4x^2+4x-16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

We will be using the following formulas :

`(i)ab+cb=(a+c)b\\\\(ii)~a^2-b^2=(a+b)(a-b).`

From (i), we have

`f(x)\\\\=x^3-4x^2+4x-16\\\\=x^2(x-4)+4(x-4)\\\\=(x-4)(x^2+4)`

Now, to factorize completely, we will use the following value of imaginary number i(iota) :

`i=√(-1)~~~~~\Rightarrow i^2=-1.`

Therefore, we get

`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).`

Thus, the complete factorization of the given function over the field of complex numbers is

`f(x)=(x-4)(x+2i)(x-2i).`