# How do you factor completely #x^2+ 4x + 2#?

##### 2 Answers

Not sure about my solution. Have a look!!

#### Explanation:

Certainly at the moment I can not think of an alternative to:

Factors of

Factors of 2 can only be

The problem is that

Suppose we had instead

This works for the coefficient (number in front) of

So

The problem is that when asked to

Complete the square, then use the difference of squares identity to find:

#x^2+4x+2=(x+2-sqrt(2))(x+2+sqrt(2))#

#### Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

In our example:

#x^2+4x+2#

#=x^2+4x+4-2#

#=(x+2)^2-2#

#=(x+2)^2-(sqrt(2))^2#

#=(x+2-sqrt(2))(x+2+sqrt(2))#

In general, given

#ax^2+bx+c#

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

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

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

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

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