Answer:
The factors are (x+6),
and ![x+2i\sqrt{2}](https://tex.z-dn.net/?f=x%2B2i%5Csqrt%7B2%7D)
Therefore the given polynomial
![x^3+6x^2+8x+48=(x+6)(x-2i\sqrt{2})(x+2i\sqrt{2})](https://tex.z-dn.net/?f=x%5E3%2B6x%5E2%2B8x%2B48%3D%28x%2B6%29%28x-2i%5Csqrt%7B2%7D%29%28x%2B2i%5Csqrt%7B2%7D%29)
Step-by-step explanation:
Given polynomial is ![x^3+6x^2+8x+48](https://tex.z-dn.net/?f=x%5E3%2B6x%5E2%2B8x%2B48)
To factorise the polynomial equate the given polynomial to zero
![x^3+6x^2+8x+48=0](https://tex.z-dn.net/?f=x%5E3%2B6x%5E2%2B8x%2B48%3D0)
By using synthetic division we can find the factors :
![x^3+6x^2+8x+48=0](https://tex.z-dn.net/?f=x%5E3%2B6x%5E2%2B8x%2B48%3D0)
_-6| 1 6 8 48
0 -6 0 -48
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1 0 8 0
Therefore (x+6) is a factor
The quadratic equation is ![x^2+8=0](https://tex.z-dn.net/?f=x%5E2%2B8%3D0)
To solve it we can use ![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Here a=1 b=0 and c=8
![x=\frac{-(0)\pm\sqrt{0^2-4(1)(8)}}{2(1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%280%29%5Cpm%5Csqrt%7B0%5E2-4%281%29%288%29%7D%7D%7B2%281%29%7D)
![=\frac{\pm\sqrt{-32}}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cpm%5Csqrt%7B-32%7D%7D%7B2%7D)
where ![i^2=-1](https://tex.z-dn.net/?f=i%5E2%3D-1)
![=\frac{\pm4i\sqrt{2}}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cpm4i%5Csqrt%7B2%7D%7D%7B2%7D)
![=\pm2i\sqrt{2}](https://tex.z-dn.net/?f=%3D%5Cpm2i%5Csqrt%7B2%7D)
![x=\pm2i\sqrt{2}](https://tex.z-dn.net/?f=x%3D%5Cpm2i%5Csqrt%7B2%7D)
Therefore
and ![x=-2i\sqrt{2}](https://tex.z-dn.net/?f=x%3D-2i%5Csqrt%7B2%7D)
Therefore the factors are (x+6),
and ![x+2i\sqrt{2}](https://tex.z-dn.net/?f=x%2B2i%5Csqrt%7B2%7D)
Therefore the given polynomial
![x^3+6x^2+8x+48=(x+6)(x-2i\sqrt{2})(x+2i\sqrt{2})](https://tex.z-dn.net/?f=x%5E3%2B6x%5E2%2B8x%2B48%3D%28x%2B6%29%28x-2i%5Csqrt%7B2%7D%29%28x%2B2i%5Csqrt%7B2%7D%29)