Answer:
3 roots are:
4, 4i, -4i
Explanation:
This is a cubic equation that has 3 roots. One root is given, we got to find the other two.
Let's group first 2 terms and last two terms and factor and solve:
![x^3 - 4x^2 + 16x - 64 = 0\\x^2(x-4) + 16(x-4)=0\\(x^2+16)(x-4)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gfvctdvf1rr38vu2efg1mmi3hbganvm9dp.png)
From here we can say:
x^2 + 16 = 0
and
x - 4 = 0 [x = 4, we already know this solution]
Let's find the other 2 roots from the 1st equation:
![x^2 + 16 = 0\\x^2=-16\\x=+-4i](https://img.qammunity.org/2020/formulas/mathematics/middle-school/btuhfyqzubie1wls2wwdnojddq9z09p8v0.png)
Note:
![√(-1)=i](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pwqqox8scny6g915yv82f9wgb2sceo8ngm.png)
So the 3 roots are:
4, 4i, -4i