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well i don't remember completing square method.but i wil tell u another way.

eg x^2 + 6x + 5 .here the constant term is 5,coefficient of x^2 term is 1,and that of x term is 6.

what u have to do is just think of two no such that product of those numbers is equal to the product of the const number and the coef of x^2 and thir sum is equal to the coef of x .here the product of constant and coef of x^2 is 5*1 which is 5.now convert x term that 6x to 5x+x .now just group x^2+5x+x+5

now (x^2+5x) + (x+5).now take x outside of 1st bracket and u get x(x+5) +(x+5). now taking x+5 common we get (X+5)(x+1).

another eg

in some eqn this won't work so use the quadratic eqn.

for a quadratic exp in the form ax^2+bx+c=0

it's roots are (-b +square root(b^2-4ac))/2a or -b -square root(b^2-4ac))/2a .

moreover if b^2-4ac is -ve then it has no real root.

x^2 - 5x - 1 =0

b^2-4ac = 25-4*-1 =29

therfore x= (5 +or -sqrt(29))/2 ie x=(5 +or - 5.38)/2 ie x=5.192 or -0.19.

if you want to learn how to factor it would help if you know how to do diamond problems. so for your example of x^2 + 6x + 5, just think 'what plus what equals 6 and what times what equals 5?' the sets of number that you add together to get 6 also have to multiply to 5. and since the variable is just a variable with no number next to it you dont have to worry about in this case. so, 5+1=6

and 5*1= 5, so, when factored it would look like this:

x^2 + 6x + 5 = (x+5)(x+1).

and about completing the square sorry i wasnt too good with that.

x^2+6x=-5

(x+3)^2=-5+9

(x+3)^2=4

x+3=+/- sqrt 4

x+3=+/- 2

x=3+/- 2

x=5, x=1

Move your five to other side, then factor your x^2+6x (make sure you add the 9 (3^2) to the other side since (x+3)(x+3)=x^2+6x+9 <----where the 9 comes from. Then solve for x.

x^2+6x+5=0

1*5=5 and 1+5=6

so we would use positive 1 and positive 5 to factor

and get

x^2+1x+5x+5=0

--> (x^2+x)+(5x+5)=0

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

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

so x is -5 and -1

You factorize by this equation: a^2 + ab + ab