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.

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- Mar 23rd 2009, 05:45 PMNeedHelp18Another 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. - Mar 23rd 2009, 06:00 PMSoroban
Hello, NeedHelp18!

Quote:

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

If $\displaystyle 3i$ is a solution, then $\displaystyle -3i$ is also a solution.

. . Complex roots appear in conjugate pairs.

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

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

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

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