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Grace [21]
3 years ago
8

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

Mathematics
2 answers:
quester [9]3 years ago
6 0

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)                      

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

kenny6666 [7]3 years ago
5 0

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).

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