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

2 Answers
Nov 14, 2015

Answer:

Not sure about my solution. Have a look!!

Explanation:

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

Factors of #x^2# can only be #+-(x " and " x)#
Factors of 2 can only be #+-(1 " and " 2)#. You could have #sqrt(2) times sqrt(2)color(white)(x)# I suppose!

The problem is that #(x times 1)+(x times 2) != 4x#

Suppose we had instead #(x times 2)+(x times 2) = 4x#
This works for the coefficient (number in front) of #x# but fails for the constant of 2 at the end of the expression.

#2 times 2 =4# this is 2 too much. We can get over this by subtracting the amount that is too much.

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

The problem is that when asked to #color(green)("factorise completely")# I expect that I am able to do so. Consequently, #color(blue)("either I have gone wrong or the question is not quite correct.")#

Nov 14, 2015

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#, you can express it in terms of a square and a constant term as follows:

#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))#