Question

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

Answer 1

Answer:

f(x)=(x-2)(x+2)(x-5)

Step-by-step explanation:

f(x)=x³-5x²+4x-20                 1. Just group randomly

f(x)=x²(x-5)+4(x-5)                2. Factor the groupings

f(x)=(x²+4)(x-5)                      

f(x)=(x-2)(x+2)(x-5)                3. Factor the difference of two squares

Answer 2

Answer:$(x-5)(x+2i)(x-2i)$  is the required factorization of f(x).

Step-by-step explanation:

To factor the expression we must first group the terms and then take out common from these groups

$f(x)=x^3-5x^2+4x-20=(x^3-5x^2)+(4x-20)$

Taking $x^2$ common from first group and the 4 from second group we get:

$f(x) = x^2(x-5)+4(x-5) = (x-5)(x^2+4)$

Now, to factor in complex from we have to break term $x^2+4$

$f(x)= (x-5){x^2-(-2i)^2}$

As, $i^2 = -1 , therefore (-2i)^2 = 4i^2 =-4$

Also using identity $a^2-b^2 =(a+b)(a-b)$

On solving

$f(x) = (x-5)(x+2i)(x-2i)$

$(x+5)(x+2i)(x-2i)$  is the required factorization of f(x).