Algebra II

Hong Kong

Stage 1 - Stage 3

Lesson

We've already looked at how to expand brackets when we were multiplying a binomial by a single number, as well as how to expand binomial products. Now we are going to look at a special case of expanding binomial products, called the *difference of two squares*.

When we have a difference of two squares in a factorised form, it looks something like this:

$\left(a+b\right)\left(a-b\right)$(`a`+`b`)(`a`−`b`)

To expand this binomial product, will still use the distributive law, making sure we multiply both terms in the first set of brackets by both terms in the second set of brackets, as shown in the picture below.

By doing this we get the expanded expression:

$a^2-ab+ab-b^2$`a`2−`a``b`+`a``b`−`b`2

If we then collect any like terms and simplify the expression we are left with the difference of two squares:

$a^2-b^2$`a`2−`b`2

Difference of Two Squares

$\left(x+y\right)\left(x-y\right)=x^2-y^2$(`x`+`y`)(`x`−`y`)=`x`2−`y`2

So in general, if we see something of the form $\left(x+y\right)\left(x-y\right)$(`x`+`y`)(`x`−`y`) (or equivalently $\left(x-y\right)\left(x+y\right)$(`x`−`y`)(`x`+`y`)) we know it's expansion will be $x^2-y^2$`x`2−`y`2. Lets confirm this result by expanding using our rectangle method.

Set up the rectangle with the binomial expressions on each side

Multiply each term and write the answers in the relevant boxes.

When collecting like terms we can see the ab will cancel out with the $ba$`b``a`. leaving us just with $a^2-b^2$`a`2−`b`2. A difference of two squares.

Expand the following:

$\left(m+3\right)\left(m-3\right)$(`m`+3)(`m`−3)

Expand the following:

$\left(3x-8\right)\left(3x+8\right)$(3`x`−8)(3`x`+8)

Expand the following:

$6\left(8x-9y\right)\left(8x+9y\right)$6(8`x`−9`y`)(8`x`+9`y`)