The difference of two squares

Algebra: Notable Products

The difference of two squares

Difference of two squares

With the difference between two squares, we can factorize with the following rule:

\[\blue a^2-\green b^2=(\blue a+\green b) (\blue a-\green b)\]

\[\begin{array}{rcl} x^2-16&=& \blue{x}^2-\green{4}^2 \\ &=& (\blue{x}+\green{4}) (\blue{x}-\green{4})
\end{array}\]

We can also use this formula, the other way around, to eliminate brackets:

\[(\blue a+\green b) (\blue a-\green b) = \blue a^2-\green b^2\]

\[\begin{array}{rcl} (\blue{x}+\green{5}) (\blue{x}-\green{5}) &=& \blue{x}^2-\green{5}^2 \\ &=& x^2-25 \\
\end{array}\]

The derivation of this rule follows from the banana method: \[\begin{array}{rclcl}(\blue a+\green b)(\blue a-\green b)&=&\blue a^2-\blue a\green b +\green b\blue a -\green b^2&\phantom{x}&\blue{\text{banana method }}\\ &=&\blue a^2-\blue a \green b+\blue a \green b-\green b^2 &&\blue{\text{rule } ab=ba}\\ &=&\blue a^2-\green b^2&&\blue{\text{terms added }}\end{array}\]

Not for the sum of two squares

Please note that this rule does not apply for the sum of two squares:

\[(a+b)^2 \;\red\ne \; a^2 +b^2\]

But, according to the square of a sum, we have:

\[(a+b)^2 = a^2 + 2 a b+b^2\]

Hence, do not forget the term #2 a b#.

Factorize \({36}{c}^2-289\).

#(6c+17)(6c-17)#

#\begin{array}{rcl}{36}{c}^2-289&=&(6c)^2-17^2\\&&\phantom{xxx}\blue{\text{recognize the square}}\\&=&(6c+17)(6c-17)\\&&\phantom{xxx}\blue{\text{factorize}}\end{array}#

New example