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

##### 2 Answers

#### Answer:

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

#### Answer:

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))#