Question

Solve the following equation by identifying all of its roots including any imaginary numbers. In your final answer, include the necessary steps and calculations. Hint: Use your knowledge of factoring polynomials. (x 2 + 1)(x 3 + 2x)(x 2 + 4x - 16i - 4xi) = 0

Answer 1

`x^2+1=(x+i)(x-i)`

`x^3+2x=x(x^2+2)=x(x+i\sqrt2)(x-i\sqrt2)`

`x^2+4x-16i-4xi=x^2+(4-4i)x-16i=(x+4)(x-4i)`

This means


`(x^2+1)(x^3+2x)(x^2+4x16i-4xi)=0`

has roots of
`x=\pm i,\pm i\sqrt2,4i,0,-4`

Answer 2

`x=0,-4 ,4i ,\pm i, \pm √(2)i`

Explanation:

The given equation is

`(x^2 + 1)(x^3 + 2x)(x^2 + 4x - 16i - 4xi) = 0`

According to zero product of property, if ab=0, then either a=0 or b=0.

Using zero product property we get

Case 1:
`x^2 + 1=0`

`x^2=-1`

Taking square root on both sides.

`x=\pm √(-1)`

`x=\pm i`
`[\because √(-1)=i]`

Case 2:
`(x^3 + 2x)=0`

`x(x^2 + 2)=0`

`x^2 + 2=0` and
`x=0`

`x^2=-2`

`x=\pm √(-2)`

`x=\pm √(2)i`

Case 3:
`x^2 + 4x - 16i - 4xi= 0`

`x(x + 4) - 4i(4+x)= 0`

`(x + 4)(x - 4i)= 0`

`x=-4`

`x=4i`

Therefore, the roots of given equation are
`x=0,-4 ,4i ,\pm i, \pm √(2)i`.