factorization

2 degree polynomials

we can

$a{x}^2+bx+c=0$

with quadratic formula

we can use the quadratic formula , discriminant as a last resort

completing the square

we can try to factor things by manipulating the expression so that we can use the perfect square law

$x^2+6x+5=0$

complete the square
$x^2+6x+9-9+5=0$
$x^2+6x+9-4=0$
$(x+3{)}^2-4=0$

solve
$(x+3{)}^2=4$
$x+3=\pm 2$
$x=\pm 2-3$
$x=-5,x=1$

difference of squares
$(x+3{)}^2-(2{)}^2=0$
$(x+3+2)(x+3-2)=0$
$(x+5)(x+1)=0$
$x+5=0,x-1=0$
$x=-5,x=1$

$x^2+6x-41=0$

complete the square
$x^2+6x+9-9-41=0$
$x^2+6x+9-50=0$
$(x+3{)}^2-50=0$

solve
$(x+3{)}^2=50$
$x+3=\pm \sqrt{50}$
$x=\pm 5\sqrt{2}-3$
$x=5\sqrt{2}-3,x=-5\sqrt{2}-3$

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

complete the square
$x^2+4x+4+-4+5=0$
$x^2+4x+4+1=0$
$(x+2{)}^2+1=0$

solve
$(x+2{)}^2=-1$
$x=\pm \sqrt{-1}-2$
this solution is not a real number. it is a complex number.