Solve the following equation by identifying all of its roots including all real and complex numbers. In your final answer, include the necessary steps and calculations. Hint: Use your knowledge of factoring polynomials.

(x2 + 1)(x3 + 2x)(x2 - 64) = 0

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kudzordzifrancis kudzordzifrancis · Answer 1 · 2020-01-26T07:31:33+01:00

Answer:

[tex]{x}= \pm \: i\: or \: x = 0 \: or \: {x}= \pm \sqrt{2}i \: or \: {x}= \pm8[/tex]

Step-by-step explanation:

The given polynomial is

[tex]( {x}^{2} + 1)( {x}^{3} + 2x)( {x}^{2} - 64) = 0[/tex]

By the zero product principle,

[tex] {x}^{2} + 1 = 0 \: or \: {x}^{3} + 2x = 0 \: or \: {x}^{2} - 64 = 0[/tex]

Or

[tex]{x}^{2} + 1 = 0 \: or \: x( {x}^{2} + 2)= 0 \: or \: {x}^{2} - 64 = 0[/tex]

This implies that;

[tex]{x}^{2}= - 1 \: or \: x = 0 \: or \: ( {x}^{2} + 2)= 0 \: or \: {x}^{2} = 64 [/tex]

[tex]{x}= \pm \sqrt{ - 1} \: or \: x = 0 \: or \: {x} = \pm \sqrt{ - 2} \: or \: {x}= \pm \sqrt{64} [/tex]

Hence the roots are:

[tex]{x}= \pm \: i\: or \: x = 0 \: or \: {x}= \pm \sqrt{2}i \: or \: {x}= \pm8[/tex]