### Video Transcript

Expand the product two π₯ plus one,
3π₯ minus two.

In this question, the word product
indicates that weβll be multiplying our two binomials. Weβre going to use two different
methods to demonstrate the expansion of these binomials. The first is the FOIL method. And the second is the area or grid
method. So starting with the FOIL method,
this means that we take our first, then the outer, then the inner, and then the last
terms and multiply. Looking at our binomials, the first
terms are the two π₯ and the three π₯. So we multiply those, writing two
π₯ times three π₯.

The outer terms will be our two π₯
and our negative two. So we add two π₯ times negative
two. The inner terms will be our plus
one and our three π₯. Therefore, we add our one times our
three π₯. And finally, the last two terms
will be the plus one and the negative two. So we add plus one times negative
two. We may not always need to write
this line of workings. But it can be helpful to illustrate
our method. And now, we work out the
products. Two π₯ times three π₯ is six π₯
squared. Two π₯ times negative two is
negative or minus four π₯ plus one times three π₯ is equivalent to plus three
π₯. And finally, plus one times
negative two is negative two.

The next thing to do at this stage
of the question is to simplify by collecting any like terms. Here, we have two terms in π₯. So we can write this as six π₯
squared minus π₯ minus two, because our negative four π₯ plus three π₯ is equivalent
to a negative π₯. Letβs see how we would get the same
answer using the grid method. To set up our grid, we put one
binomial horizontally and one binomial vertically. The first binomial will give us two
π₯ and one horizontally. And the second binomial of three π₯
minus two can be split into the terms three π₯ and negative two.

We then calculate the product of
each row and column, starting with three π₯ times two π₯. This will give us six π₯
squared. Next, three π₯ times one is three
π₯. In our second row, negative two
times two π₯ is negative four π₯. And finally, negative two times one
is negative two. We then simplify by adding all four
terms and seeing if we can collect any like terms. We start with six π₯ squared. We then have a plus three π₯ and a
negative four π₯, which simplifies to negative π₯ and then our final term of
negative two. So using either of our methods
would give us the answer of six π₯ squared minus π₯ minus two.