Question 492558
<pre>
3x<sup>4</sup>+5x³+x²+5x-2

-2|3  5  1  5 -2
  |  -6  2 -6  2
   3 -1  3 -1  0

So the polynomial has been factored as

(x+2)(3x³-x²+3x-1)

Next you can factor the cubic by grouping:

Factor the first two terms as x²(3x-1) and write the last two
terms as +1(3x-1), and enclose those in brackets:

(x+2)[x²(3x-1)+1(3x-1)]

Factor out (3x-1) leaving x² and +1 in another set of
parentheses:

(x+2)(3x-1)(x²+1)

x+2=0   3x-1=0   x²+1=0      
  x=-2    3x=1     x²=-1
           x={{{1/3}}}      x=±i

The other two zeros are i and -i.

Edwin</pre>