## A3.3 Difference of two squares

What is the DOTS rule? If you have a squared expression subtracted from another squared expression, you can factorise this quickly according to the DOTS rule.

The difference of two squares formula is commonly used in mathematics. It allows us to factorise terms such as \[\begin{align*} x^{2}-36 & ,\\ 5x^{2}-20y^{2} & ,\\ a^{2}-b^{2} & . \end{align*}\] This module explains the formula and shows how it may be used to factorise algebraic expressions.

It is an important rule that you should commit to memory.

Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on factorisation, the difference of two squares or dots.

Let’s first of all review our knowledge of removing brackets, X plus five X minus five in this example. And if you go down three lines we end up with an answer of X squared minus 25. Notice that is the same as X squared minus five squared, in other words what we have here is terms in brackets differ only in the sign in each bracket, on the left hand side, and the expansion on the right hand side is the difference of two terms both of which are perfect squares, eight squared and five squared. So in general we can state that the dots rule says A plus B times A minus B is the same as A squared minus B squared, so the terms in brackets on the left hand side differ only in the sign, plus or minus, in each bracket and the expansion on the right hand side is the difference of two terms, both of which are perfect squares.

Let’s look at an example. Here we have X squared minus 36, and the first thing to do is to see if both these terms are perfect squares. Well they are because X squared is the same as X times X, X squared, and 36 is the same as six times six which is 36. So we can now go on and state the rule which says that X squared minus 36 is the same as X plus six X minus six, the only difference is the sign in each brackets.

Let’s look at another one. Four A squared minus Y squared. The first question are each of the terms perfect squares, well they are, four A squared is the same as two A all squared or two A times two A and Y squared is the same as Y times Y, so the expression is the difference of two squared or dots. So we can use the rule which says that four A squared minus Y squared is the same as two A minus Y two A plus Y, the only difference is the sign in each brackets.

Now let’s look at a slightly more difficult question. We have three X squared minus 48. So the first question is, is the expression a dots, well it’s not because neither of those terms are perfect squares, however when we factorise this expression by taking three out of the brackets we have three brackets X squared minus 16, but notice this time that X squared minus 16 is a perfect square, X times X and 16 is four times four, in other words three X squared minus 48 is three, open up big sets of brackets in blue, then we have X squared in red minus four squared in red which gives us three brackets X plus four X minus four.

Now, in this final example we have X plus two all squared minus nine. First question is are these two terms perfect squares. Well the first term must be because it’s X plus two all squared, so that is. Nine is a perfect square because nine is three times three, so we now have the expression X plus two all squared minus three all squared. And the rule, that gives us X plus two plus three, notice the blue plus, minus X plus two minus three, notice the blue minus, which simplifies to X plus five brackets X minus one and that’s our answer.

Now try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.

### Factorisation: Difference of Two Squares (DOTS)

Consider the following expansion:

\[\begin{align*} (x+5)(x-5) & =x^{2}-5x+5x-25\\ & =x^{2}-25\\ & =x^{2}-5^{2} \end{align*}\] In general,

\[\begin{alignat*}{1} (x+a)(x-a) & =x^{2}-ax+ax-a^{2}\\ & =x^{2}-a^{2}. \end{alignat*}\]

Note that:

The terms in the brackets differ only in the sign of the second term

The expansion is the difference of two terms, both of which are perfect squares More generally:

\[\begin{alignat*}{1} (a+b)(a-b) & =a^{2}-b^{2} \end{alignat*}\]

so

\[\begin{alignat*}{1} a^{2}-b^{2} & =(a+b)(a-b), \end{alignat*}\] and is known as the DOTS (Difference Of Two Squares) rule.

This can be used to factorise expressions of the form \(a^{2}-b^{2}\).

Watch a short video on factorising using " Difference of two squares"

Download a transcript of the video on " Difference of two squares"

#### Examples:

- Factorise \(a^{2}-36\).

Solution: \[\begin{align*} a^{2}-36 & =a^{2}-\left(6\right)^{2}\ \text{expression is the difference of two squares}\\ & =\left(a+6\right)\left(a-6\right)\ \text{using the DOTS rule}\\ & =\left(a-6\right)\left(a+6\right)\ \text{order doesn't matter, either is correct.} \end{align*}\]

- Factorise \(4^{2}-y^{2}\).

Solution: \[\begin{align*} 4^{2}-y^{2} & =\left(2\right)^{2}-y^{2}\ \text{expression is the difference of two squares}\\ & =\left(2+y\right)\left(2-y\right)\ \text{using the DOTS rule}. \end{align*}\]

- Factorise \(3x^{2}-48\).

Solution:

At first sight, we cannot use DOTS. But taking out a common factor of \(3\) we have \[\begin{align*} 3x^{2}-48 & =3\left(x^{2}-16\right)\\ & =3\left(x^{2}-4^{2}\right)\\ & =3\left(x+4\right)\left(x-4\right)\ \text{using the DOTS rule}. \end{align*}\]

- Factorise \(\left(x+2\right)^{2}-9\).

Solution:

\[\begin{align*} \left(x+2\right)^{2}-9 & =\left(x+2\right)^{2}-3^{2}\ \text{expression is the difference of two squares}\\ & =\left(x+2+3\right)\left(x+2-3\right)\ \text{using the DOTS rule}\\ & =\left(x+5\right)\left(x-1\right). \end{align*}\]

- Factorise \(y^{2}+36\).

Solution:

This expression is the sum of two squares, not the difference hence the DOTS rule cannot be applied. There are no real factors for this expression. 1 There are complex factors but we do not consider them in this module.

### Exercise

Factorise the following expressions using the DOTS rule (if possible):

\(\text{a.$\ x^{2}-4\qquad\text{b.$\ a^{2}-100\qquad\text{c.$\ 49-x^{2}\qquad\text{d.$\ 64x^{2}-1$ }$ }$ }$ }\)

\(\text{e.$\ 121x^{2}-49y^{2}\qquad\text{f.$\ a^{2}b^{2}-25\qquad\text{g.$\ 5x^{2}-20\qquad\text{h.$\ a^{2}+100$ }$ }$ }$ }\)

\(\text{i.$\ x^{2}y^{3}-36y\qquad\text{j.$\ \left(x+2\right)^{2}-y^{2}\qquad\text{$\text{k.$\ \left(x-5\right)^{2}-36\qquad$ $\text{l.$\ \left(a+1\right)^{2}-\left(b-2\right)^{2}$ }$ }$ }$ }$ }\)

\(\text{a.$\ (x+2)(x-2)\qquad\text{b.$\ (a+10)(a-10)\qquad\text{c.$\ (7+x)(7-x)$ }$ }$ }\)

\(\text{d.$\ $ $\left(8x+1\right)\left(8x-1\right)$ $\qquad\text{e.$\ \left(11x+7y\right)\left(11x-7y\right)\qquad\text{$\text{f.$\ \left(ab+5\right)\left(ab-5\right)$ }$ }$ }$ }\)

\(\text{g.$\ 5\left(x+2\right)\left(x-2\right)\qquad\text{h. Does not factorise.$\qquad\text{i.$\ y\left(xy+6\right)\left(xy-6\right)$ }$ }$ }\)

\(\text{j.$\ \left(x+2+y\right)\left(x+2-y\right)\qquad\text{k.$\ \left(x-11\right)\left(x+1\right)\qquad\text{l.$\ \left(a+b-1\right)\left(a-b+3\right).$ }$ }$ }\)

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