Answer:
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)](https://img.qammunity.org/2019/formulas/mathematics/high-school/q22mv1t0f7af8q57657lwmzo6y4dbwwv2g.png)
Taking
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)](https://img.qammunity.org/2019/formulas/mathematics/high-school/ieur2rmu7zk6h2rvjwxv8lpd0y9pckfnlp.png)
Now, to factor in complex from we have to break term
![x^2+16](https://img.qammunity.org/2019/formulas/mathematics/high-school/lo3nzn5zcrvv6i79f5oekw6enxuh5sq0rn.png)
![f(x)= (x+3){x^2-(-4i)^2}](https://img.qammunity.org/2019/formulas/mathematics/high-school/ybixrf6zi1co2vle5qfnt2n7zuanaf4woy.png)
As,
![i^2 = -1 , therefore (-4i)^2 = 16i^2 =-16](https://img.qammunity.org/2019/formulas/mathematics/high-school/9gowptabkvcggcr1zohf1lmfmrwmf37n74.png)
Also using identity
![a^2-b^2 =(a+b)(a-b)](https://img.qammunity.org/2019/formulas/mathematics/high-school/jjnfxe5l1z0dzafw7ly8z0x4vcdptiye5o.png)
On solving
![f(x) = (x+3)(x+4i)(x-4i)](https://img.qammunity.org/2019/formulas/mathematics/high-school/vr50nhf2wv5ldndfob0iv6nnxiwv8492vd.png)
is the required factorization of f(x).