Question

Question #a65c9

Answer 1

`k=2/3`

`P(x) = 3(x + 2/3)(x - sqrt15 i)(x + sqrt15 i)`

Explanation:

Since 1 zero is purely imaginary, we denote it as `iy`, where `y` is the imaginary part of the zero.

The conjugate of `iy` is `-iy`. We know from the complex conjugate root theorem that roots of a polynomial with real coefficients exists as conjugate pairs. Therefore, `-iy` is also a root of the cubic equation.

We let the last root be `z`. `z` is purely real.

We can then rewrite `P(x)` in product form. We know that the leading coefficient is 3.

`P(x) = 3(x - z)(x - iy)(x + iy)`

`= 3(x - z)(x^2 + y^2)`

Upon expanding, we get

`P(x) = 3x^3 + (-z) x^2 + (y^2) x + (-z y^2)`

Comparing the coefficients with the original `P(x)`

• Coefficient of `x^0` term

`-z y^2 = 10`

• Coefficient of `x^1` term

`y^2 = 15`

• Coefficient of `x^2` term

`-z = k`

Solving the 3 equations, we have

`y = +-sqrt15`
`z = -2/3`
`k = 2/3`

Hence, the roots of P(x) are `-2/3`, `sqrt15 i`, `-sqrt15 i`.

`P(x)` in linear product form

`P(x) = 3(x + 2/3)(x - sqrt15 i)(x + sqrt15 i)`

`=(3x+2)(x^2+15)`