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]}

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