How do you factor completely x^2+ 4x + 2x2+4x+2?

2 Answers
Nov 14, 2015

Not sure about my solution. Have a look!!

Explanation:

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

Factors of x^2x2 can only be +-(x " and " x)±(x and x)
Factors of 2 can only be +-(1 " and " 2)±(1 and 2). You could have sqrt(2) times sqrt(2)color(white)(x)2×2x I suppose!

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

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

2 times 2 =42×2=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)^2-2x2+4x+2=(x+2)(x+2)2=(x+2)22

The problem is that when asked to color(green)("factorise completely")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.")either I have gone wrong or the question is not quite correct.

Nov 14, 2015

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

x^2+4x+2=(x+2-sqrt(2))(x+2+sqrt(2))x2+4x+2=(x+22)(x+2+2)

Explanation:

The difference of squares identity can be written:

a^2-b^2 = (a-b)(a+b)a2b2=(ab)(a+b)

In our example:

x^2+4x+2x2+4x+2

=x^2+4x+4-2=x2+4x+42

=(x+2)^2-2=(x+2)22

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

=(x+2-sqrt(2))(x+2+sqrt(2))=(x+22)(x+2+2)

In general, given ax^2+bx+cax2+bx+c, you can express it in terms of a square and a constant term as follows:

ax^2+bx+cax2+bx+c

= a(x+b/(2a))^2+(c-b^2/(4a))=a(x+b2a)2+(cb24a)

= a((x+b/(2a))^2-(b^2/(4a^2)-c/a))=a((x+b2a)2(b24a2ca))

= a((x+b/(2a))^2-(sqrt(b^2/(4a^2)-c/a))^2)=a(x+b2a)2(b24a2ca)2

=a(x+b/(2a)- sqrt(b^2/(4a^2)-c/a))(x+b/(2a)+sqrt(b^2/(4a^2)-c/a))=a(x+b2ab24a2ca)(x+b2a+b24a2ca)

=a(x+(b-sqrt(b^2-4ac))/(2a))(x+(b+sqrt(b^2-4ac))/(2a))=a(x+bb24ac2a)(x+b+b24ac2a)