How do I find all real and complex zeros of $x^3+4x^2+5x$ ?

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Answer 1 · 2014-09-25T21:41:26

First set the expression equal to 0.

$x^3+4x^2+5x=0$

Factor out an $x$

$x(x^2+4x+5)=0$

$x=0$ , this is one of the roots

Factor the polynomial $=> (x^2+4x+5) =>$ Use the quadratic formula

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$a=1, b=4, and c=5$

$x= (-(4)+-sqrt((4)^2-4(1)(5)))/(2(1))$

Simplify

$x= (-4+-sqrt(16-20))/2$

$x= (-4+-sqrt(-4))/2$

$x= (-4+-2i)/2$

$x= -2+-i =>$ 2 complex roots

This function has 3 roots. One of the roots is real and other 2 roots are complex numbers.

The roots are $0, -2+i, and -2-i.$

How do I find all real and complex zeros of x^3+4x^2+5xx^3+4x^2+5x?