# Thread: Find the solutions of this polynomial

1. ## Find the solutions of this polynomial

p(x) = 3x^4 - 11x^3 + 35x^2 + 21x - 26

if one of the solutions is 2 + 3i.

ok, so its conjugate is also a solution? 2 - 3i?

i multiplied the conjugates and then applied long division

and got 3x^2 + x + 2

but im stuck..

how do i get the other two solutions? is the above answers even right?

thankz

2. Originally Posted by NeedHelp18
p(x) = 3x^4 - 11x^3 + 35x^2 + 21x - 26

if one of the solutions is 2 + 3i.
Are you sure you have written the problem correctly? $2+3i$ is not a root of $p.$

3. Hello, NeedHelp18!

There seems to be a typo . . . the middle term is $33x^2$, isn't it?

$p(x)\;=\;3x^4 - 11x^3 + {\color{red}33}x^2 + 21x - 26 \:=\:0$ .if one of the roots is $2 + 3i.$

If $2 + 3i$ is a root, then so is $2 - 3i$

Then: $[x - (2+3i)]$ and $[x - (2-3i)]$ are factors.

Dividing by: $[x-(2i+3)][x-(2i-3)] \:=\:x^2 - 4x + 13$, we get:

. . $p(x) \;=\;(x^2-4x+13)\,(3x^2+x-2)\:=\:0 \quad\Rightarrow\quad (x^2-4x+13)(x+1)(3x-2) \:=\:0$

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