Write a polynomial equation of degree 3 such that two of its roots are 2 and an imaginary number.

Solution:

A polynomial equation of degree 3 such that two of its roots are 2 and an imaginary number [Given]

We know that the equation has 3 roots one is 2 and the other two are imaginary

If one root is 2 then the equation must have (x - 2) as its factor.

Consider i, -i as the other two imaginary numbers

(x - i)(x + i)

Roots of the polynomial of degree 3 is (x - 2)(x - i)(x + i)

Solve the roots to get the equation

(x - 2)(x² + xi - xi - i²)

We know that i² = 1

= (x - 2)(x² - i²)

So we get

= (x³ + x - 2x² - 2)

= x³ - 2x² + x - 2

Therefore, the polynomial equation of degree 3 is (x³ - 2x² + x - 2).

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Write a polynomial equation of degree 3 such that two of its roots are 2 and an imaginary number.

Summary:

The polynomial equation of degree 3 such that two of its roots are 2 and an imaginary number is (x³ - 2x² + x - 2).