Solve x^2+6x+8 by completing the square.

(x - 4) (x-2)

hope this help

you have to find factors that when you do the foil method ends up being the original equation

"Completing the square" is how the quadratic equation is derived.  If the quadratic ax^2+bx+c=0, for unknown values of a,b, and c, you arrive at:

x=(-b±√(b^2-4ac))/(2a)

You proceed in this manner:

x^2+6x+8=0  Since the leading coefficient is already one, no division is necessary.  You then move the constant to the other side of the equation by subtracting 8 from both sides...

x^2+6x=-8  Then you halve the linear coefficient, square it, and add that value to both sides.  (6/2)^2=9 so

x^2+6x+9=1  Now the left side is a "perfect square" equal to:

(x+3)^2=1  Now take the square root of both sides

x+3=±√1  Subtract 3 from both sides

x=-3±√1 So x is equal to:

x=-3-1 and -3+1

x=-4 and -2 or in factored form

(x+4)(x+2)

The same process for unknown a,b, and c:

ax^2+bx+c=0

x^2+bx/a+c/a=0

x^2+bx/a=-c/a

x^2+bx/a+b^2/(4a^2)=b^2/(4a^2)-c/a

x^2+bx/a+b^2/(4a^2)=(b^2-4ac)/(4a^2)

(x+b/(2a))^2=(b^2-4ac)/(4a^2)

x+b/(2a)=±√(b^2-4ac)/(4a^2)

x+b/(2a)=±√(b^2-4ac)/(2a)

x=(-b±√(b^2-4ac))/(2a)