### Video Transcript

Use the distributive property to
fully expand two π₯ plus π¦, π₯π¦ minus two π§.

In this question, the phrase βto
use the distributive propertyβ means that we can multiply the sum by multiplying
each addend separately and then adding the products. We could use a number of methods to
expand the brackets. Here, weβre going to look at two of
those. The first method is the FOIL
method, which is an acronym. We can recall that it stands for
the First, the Outer, the Inner, and the Last terms of our binomials.

So with our two binomials set, we
start by multiplying the first two terms, which is two π₯ times π₯π¦. Next, our outer terms will be two
π₯ times negative two π§. Itβs always worth taking extra care
when we have the variable π§, because we donβt want to confuse it with the digit
two. Next, our inner terms will be π¦
multiplied by π₯π¦. And finally, we add on the product
of our last terms, which is π¦ times negative two π§.

We now need to simplify our
products. The coefficient of our first term
is two. π₯ multiplied by π₯ will be π₯
squared. And we have our π¦. So the first term simplifies to two
π₯ squared π¦. The coefficient of our second term
is found by multiplying plus two by negative two, giving us negative four. And we then have π₯ times π§. Our third term simplified will be
π₯π¦ squared since we have π¦ times π¦. And our final term will be negative
two π¦π§. At this stage, we always check if
we can simplify our answer by collecting any like terms. And as there are none here, this
would be our final answer.

As an alternative method, we could
use the area or grid method to expand our binomials. To set up our grid, we split up our
binomials into their component terms with one on the row and one on the column. And it doesnβt matter which way
round we put them. To fill in each cell in our grid,
we multiply the row value by the column value.

So taking our first cell, we have
π₯π¦ times two π₯, which would be two π₯ squared π¦. For our next cell, we multiply π₯π¦
by π¦ to give us π₯π¦ squared. On our next row then, we have
negative two π§ times two π₯, which would give us negative four π₯π§. And our final grid value would be
negative two π¦π§. To find our answer from this grid,
we add our four products, giving us two π₯ squared π¦ minus four π₯π§ plus π₯π¦
squared minus two π¦π§, which is the same as we achieved using the FOIL method.