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


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).


User Brian Carper
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