Question

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

Answer 1

(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

Answer 2

"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)