How do you factor #x^2 + 5x + 6#?

Because the #x^2# coefficient is #1# we know the coefficient for the #x# terms in the factor will also be #1#:

#(x )(x )#

Because the constant is a positive and the coefficient for the #x# term is also positive we know the sign for the constants in the factors will both be positive.

#(x + )(x + )#

Now we need to determine the factors which multiply to 6 and also add to 5:

#1 xx 6 = 6#; #1 + 6 = 7# <- this is not the factor

#2 xx 3 = 6#; #2 + 3 = 5# <- this IS the factor

#(x + 2)(x + 3)#

#x^2 + (a+b)x + ab = (x+a)*(x+b)#

#x^2+5x+6 = (x+a)*(x+b)#

Equation system:
#1. a+b = 5#
#2. ab = 6#
Then #a = 2# and #b=3#

#color(purple)("First Answer:")#

Using prime factorization of #6#

#+6=2xx3=1xx6=-2xx-3=-1xx-6#

since the coefficient of #x# is positive

#color(red)("the answers "-2xx-3 " and " -1xx-6 " are refused ")#

#color(blue)("and the answers "2xx3 " and " 1xx6 " are accepted ")#

and then we try adding the two numbers to get the coefficient of #x#

#1+6=7##rarr##color(red)("refused")#

#2+3=5##rarr##color(blue)("accepted")#

so it will be

#x^2+5x+6=(x+2)(x+3)#

#color(purple)("The Second Answer")#

if you have a polynomial function
#color(blue)(ax^2+bx+c=0#

the solution set will be

#color(green)(x=(-bcolor(red)(+-)sqrt(b^2-4ac))/(2a)#

and using it like this

#a(x-(-bcolor(red)(+)sqrt(b^2-4ac))/(2a))(x-(-bcolor(red)(-)sqrt(b^2-4ac))/(2a))#

so for your question

#x^2+5x+6#

#a=1,b=5,c=6#

#x=(-5+-sqrt(25-4xx1xx6))/(2xx1)=(-5+-1)/2#

and by using it in the formula

#1*(x-(-5+1)/2)(x-(-5-1)/2)#

#=(x+2)(x+3)#

The key realization is that we need to think of two numbers that add up to the middle term, and have a product of the middle term.

What two numbers sum up to #5# and have a product of #6#?

After some trial and error, we arrive at #2# and #3#, because

#2+3=5# (Middle term) and

#2*3=6# (Last term)

These will be our factors. So we can write our expression as

#(x+2)(x+3)#

Hope this helps!