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

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Answer 1 · 2014-09-25T21:41:26

First set the expression equal to 0.

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

Factor out an $x$ $x$

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

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

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x = \frac{- (4) \pm \sqrt{{(4)}^{2} - 4 (1) (5)}}{2 (1)}$ $x= (-(4)+-sqrt((4)^2-4(1)(5)))/(2(1))$

Simplify

$x = \frac{- 4 \pm \sqrt{16 - 20}}{2}$ $x= (-4+-sqrt(16-20))/2$

$x = \frac{- 4 \pm \sqrt{- 4}}{2}$ $x= (-4+-sqrt(-4))/2$

$x = \frac{- 4 \pm 2 i}{2}$ $x= (-4+-2i)/2$

$x = - 2 \pm i \Rightarrow$ $x= -2+-i =>$ 2 complex roots

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

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

How do I find all real and complex zeros of x^3+4x^2+5xx3+4x2+5xx^3+4x^2+5x?