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.#