How do you find all the complex roots of #x^3+6x^2+11x+6#?

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Answer 1 · 2016-01-11T02:18:26

There are no complex roots.

Try to divide the polynomial, either through polynomial long division or synthetic division.

The factors you should try are #+-1,+-2,+-3,+-6#.

The first root I encountered was #-1#, or #(x+1)#.

#(x^3+6x^2+11x+6)/(x+1)=x^2+5x+6#

This is easily factorable into #(x+2)(x+3)#, revealing the final two roots of #-2# and #-3#.

Thus, this cubic has three real roots, #-1,-2#, and #-3# and no complex roots.

graph{x^3+6x^2+11x+6 [-11.29, 8.71, -4, 6]}