Question

What is the complete factorization of the polynomial function over the set of complex numbers? f(x)=x^3+3x^2+16x+48
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Answer 1

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).