# Question #a65c9

##### 1 Answer

#### Answer:

#### Explanation:

Since 1 zero is purely imaginary, we denote it as

The conjugate of

We let the last root be

We can then rewrite

#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

- 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

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

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