Radicals and binomial expansions

Tricky Brackets

You've encountered expanding brackets before, and may have even expanded a product of brackets with two terms in each. We call this latter kind of expansion binomial expansion. Binomial just means involving two terms, as you can probably guess from the bi which we see in other two-related words such as bicycle, bicentennial, or biped, meaning an animal that stands on two legs!

Let's see if we remember how to expand an expression like $\left(a+b\right)\left(c+d\right)$(a+b)(c+d).

Before we move on to the multiplication it's important to group the sign with each of the four terms, like below. It is very important to think about whether a term is negative or positive when multiplying it.

Now we know that to multiply these two brackets we must multiply everything in the first set of brackets by everything in the second. That means each term should give two terms, eg. $b$b should multiply both $c$c and $d$d.

And so we end up with these weird pretzel shape that gives us a guide on how to proceed. And so we should end up with the sum of four terms as below.

Another way of thinking about this is using what's called the rectangle method:

Think of $a+b$a+b and $c+d$c+d as two sides of a rectangle, so then the product $\left(a+b\right)\left(c+d\right)$(a+b)(c+d) is calculating the area of the rectangle. From the diagram we can see that this rectangle can also be split up into $4$4 smaller rectangles, so the area can also be found by calculating the four individual areas and then adding them together. Can you see how everything fits together?

So now we know that:

$\left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$(a+b)(c+d)=ac+ad+bc+bd

 

Special Cases

When you move through many different types of binomial expansions, you'll find some examples which will give you interesting answers!

For example, perfect square cases where both brackets are the same thing. $\left(a+b\right)^2$(a+b)2 can be rewritten as $\left(a+b\right)\left(a+b\right)=a^2+ab+ab+b^2$(a+b)(a+b)=a2+ab+ab+b2 using what we've learned above. This can be simplified to $a^2+2ab+b^2$a2+2ab+b2. Wow, this is so much simpler!

Therefore: $\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2

We can use similar techniques to also get: $\left(a-b\right)^2=a^2-2ab+b^2$(a−b)2=a2−2ab+b2

Can you explain why only the term $2ab$2ab changes sign?

 

Another case is when we have both brackets the same except for the sign in the middle. Eg. $\left(a+b\right)\left(a-b\right)$(a+b)(a−b) have the same numbers but $b$b has changed sign. Again we can use similar techniques to figure out that this equals $a^2-ab+ab-b^2$a2−ab+ab−b2 which simplifies down to $a^2-b^2$a2−b2.

So we also have: $\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2

It's fascinating how the middle terms have disappeared and we're just left with one square subtracting another! That's why this expansion is called the difference of two squares. 

 

Example

Expand and simplify $\left(\sqrt{3}-\sqrt{10}\right)\left(\sqrt{3}+\sqrt{12}\right)$(√3−√10)(√3+√12)

Think of using the pretzel or the rectangle method, and remember to keep simplifying square root expressions until you reach a radical or an integer

Do: Using the rectangle method:

$\left(\sqrt{3}-\sqrt{10}\right)\left(\sqrt{3}+\sqrt{12}\right)$(√3−√10)(√3+√12)	$=$=	$\sqrt{3}\times\sqrt{3}+\sqrt{3}\times\sqrt{12}-\sqrt{10}\times\sqrt{3}-\sqrt{10}\times\sqrt{12}$√3×√3+√3×√12−√10×√3−√10×√12
	$=$=	$\left(\sqrt{3}\right)^2+\sqrt{3\times12}-\sqrt{10\times3}-\sqrt{10\times12}$(√3)2+√3×12−√10×3−√10×12
	$=$=	$3+\sqrt{36}-\sqrt{30}-\sqrt{120}$3+√36−√30−√120
	$=$=	$3+6-\sqrt{30}-\sqrt{4\times30}$3+6−√30−√4×30
	$=$=	$9-\sqrt{30}-2\sqrt{30}$9−√30−2√30
	$=$=	$9-3\sqrt{30}$9−3√30

 

 

Here are some worked examples.

Question 1

Question 2

Question 3