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

[tex]x^2+1=(x+i)(x-i)[/tex]
[tex]x^3+2x=x(x^2+2)=x(x+i\sqrt2)(x-i\sqrt2)[/tex]
[tex]x^2+4x-16i-4xi=x^2+(4-4i)x-16i=(x+4)(x-4i)[/tex]

This means

[tex](x^2+1)(x^3+2x)(x^2+4x16i-4xi)=0[/tex]

has roots of [tex]x=\pm i,\pm i\sqrt2,4i,0,-4[/tex]

Answer Link

erinna erinna · Answer 2 · 2019-10-17T11:34:16+02:00

Answer:

[tex]x=0,-4 ,4i ,\pm i, \pm \sqrt{2}i[/tex]

Step-by-step explanation:

The given equation is

[tex](x^2 + 1)(x^3 + 2x)(x^2 + 4x - 16i - 4xi) = 0[/tex]

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

Using zero product property we get

Case 1: [tex]x^2 + 1=0[/tex]

[tex]x^2=-1[/tex]

Taking square root on both sides.

[tex]x=\pm \sqrt{-1}[/tex]

[tex]x=\pm i[/tex] [tex][\because \sqrt{-1}=i][/tex]

Case 2: [tex](x^3 + 2x)=0[/tex]

[tex]x(x^2 + 2)=0[/tex]

[tex]x^2 + 2=0[/tex] and [tex]x=0[/tex]

[tex]x^2=-2[/tex]

[tex]x=\pm \sqrt{-2}[/tex]

[tex]x=\pm \sqrt{2}i[/tex]

Case 3: [tex]x^2 + 4x - 16i - 4xi= 0[/tex]

[tex]x(x + 4) - 4i(4+x)= 0[/tex]

[tex](x + 4)(x - 4i)= 0[/tex]

[tex]x=-4[/tex]

[tex]x=4i[/tex]

Therefore, the roots of given equation are [tex]x=0,-4 ,4i ,\pm i, \pm \sqrt{2}i[/tex].