Question

is this correct?

find the zeros:

g(x)=x^2+4x+5
(x+5)(x-1)
x+5=0 x-1=0
x=-5 x=1
{-5,1}

???

Answer 1

The method is perfect, but you did not factor correctly :(

Answer 2

$$\huge x^2+4x+5
\neq(x+5)(x-1)$$

Answer 3

(x-5)(x+1)
x-5=0 x+1=0
x=5 x=-1

{5,-1}

???

Answer 4

$$\huge x^2+4x+5$$ is prime

Answer 5

I don't think you can factorize this equation, It doesn't have real roots. You have to use the formula

Answer 6

how is that?

Answer 7

https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/using-the-quadratic-formula