factoring differnce of squares

Question

n² -12

Answer 1

Hi David;

n²-12

This can be factored as...

(n+√12)(n-√12)

Let's FOIL...

FIRST...(n)(n)=n²

OUTER...(n)(-√12)=-(√12)n

INNER...(√12)(n)=(√12)n

LAST...(√12)(-√12)=-12

n²-(√12)n+(√12)n-12

n²-12

Answer 2

The product (x + a)*(x - a) results in a "difference of two squares", since the linear term in x "cancels" when the product is expanded;  i.e. by FOIL (first, outer, inner, last)

 

(x + a)*(x - a) = x^2 -a*x + a*x - a^2 = x^2 - a^2

 

So, in this instance,

 

a^2 = 12

 

or

 

a = √12 = √(4*3) = √4*√3 = 2√3.

 

The factorization yields

 

(x + 2√3)*(x - 2√3)

Answer 3

 Here the difference of 2 squares can be used:

 

    n ^2 - 12 = ( n - √12) ( n + √12)