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attashe74 [19]
3 years ago
8

What is the complete factorization of the polynomial function over the set of complex numbers? f(x)=x^3+3x^2+16x+48

Mathematics
1 answer:
Tamiku [17]3 years ago
8 0

Answer:(x+3)(x+4i)(x-4i)  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+3x^2+16x+48=(x^3+3x^2)+(16x+48)

Taking x^2 common from first group and the 16 from second group we get:


f(x) = x^2(x+3)+16(x+3) = (x+3)(x^2+16)


Now, to factor in complex from we have to break term x^2+16


f(x)= (x+3){x^2-(-4i)^2}

As, i^2 = -1 , therefore (-4i)^2 = 16i^2 =-16

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

On solving

f(x) = (x+3)(x+4i)(x-4i)

(x+3)(x+4i)(x-4i)  is the required factorization of f(x).


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