Question #a65c9

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Bio
Mar 24, 2016

#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)#

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