How do you determine the remaining zeroes for #g(x)=2x^5-3x^4-5x^3-15x^2-207x+108# if 3i is a zero?

1 Answer
Jul 4, 2015

Answer:

Use conjugate zeros, factor theorem, division of polynomials, rational zeros theorem, division (again), then solve the resulting quadratic by you favorite method for quadratics.

Explanation:

#g(x)=2x^5-3x^4-5x^3-15x^2-207x+108#

First Two Zeros
Given that #3i# is a zero, we observe that the coefficients of #g# are real, so the Complex Conjugate Zero Theorem tells us that #-3i# is also a zero.
The Factor Theorem tells us that #x-3i# and #x+3i# are factors, so their product is also a factor:
#(x-3i)(x+3i) = x^2+9#

Divide #g(x)# by #x^2+9# to get:

#2x^5-3x^4-5x^3-15x^2-207x+108 = (x^2 +9)(2x^3-3x^2-23x+12)#

Third Zero

To find another zero, we need a zero of #2x^3-3x^2-23x+12#.

Neither of the easiest choices, #+-1#, are zeros. So use the Rational Zeros Theorem to search for another zero.

Possible rational zeros are: #+-1, +-2, +-3, +-4, +-6, +-12, +-1/2, +-3/2#
Test until you find a zero. (Synthetic division is recommended for this testing.)
Depending on the order in which you test, you'll find a rational zero.
I found #-3# is a zero first. So (or really because) #x+3# is a factor.

#2x^3-3x^2-23x+12 = (x+3)(2x^2-9x+4)#.

Last Two Zeros
Solve #2x^2-9x+4 =0# by whatever method you like.

#(2x-1)(x-4) = 0#
So the final two zeros of #g# are #1/2# and #4#.

List of Zeros

Zeros: #3i, -3i, -3, 4, 1/2#