# Another polynomial problem

• March 23rd 2009, 06:45 PM
NeedHelp18
Another polynomial problem
Find all the solutions of the equation p(x) = 0 if

p(x) = 3(x)^4 + 5(x)^3 + 25(x)^2 + 45x - 18

and 3i is one of the solutions.
• March 23rd 2009, 07:00 PM
Soroban
Hello, NeedHelp18!

Quote:

Find all the solutions of: . $3x^4 + 5x^3 + 25x^2 + 45x - 18 \:=\:0$ if $3i$ is one of the solutions.

If $3i$ is a solution, then $-3i$ is also a solution.
. .
Complex roots appear in conjugate pairs.

Then $(x-3i)(x+3i) \:=\:x^2+9$ is a factor of the polynomial.

Using long division, we have: . $(x^2+9)(3x^2+5x-2) \:=\:0$

. . which factors further: . $(x^2+9)(x+2)(3x-1) \:=\:0$

Therefore, the roots are: . $3i,\:-3i,\:-2,\:\tfrac{1}{3}$