### Video Transcript

Using the multiplication of a
binomial by a trinomial, simplify the algebraic expression three π₯ plus two π¦ all
cubed.

It seems a little strange that we
are being asked to multiply a binomial by a trinomial when there do not appear to be
any trinomials present. Donβt worry though; one will appear
shortly. Letβs start by writing out the
three factors of our cube explicitly: three π₯ plus two π¦ cubed equals three π₯
plus two π¦ times three π₯ plus two π¦ times three π₯ plus two π¦.

The way to approach an algebraic
expansion problem such as this when there are more than two factors is to multiply
the factors together in pairs until everything is expanded. Because multiplication of
polynomials is associative, it does not matter if we multiply the first two factors
together first and leave the third for later or leave the first factor and multiply
the other two first. Letβs do this and multiply the last
two factors first.

We can do this using a grid to
multiply each term of one factor by each term of the other. Three π₯ times three π₯ is nine π₯
squared. Three π₯ times two π¦ is six
π₯π¦. Two π¦ times three π₯ is six
π₯π¦. And two π¦ times two π¦ is four π¦
squared. Adding these together, we get nine
π₯ squared plus six π₯π¦ plus six π₯π¦, which is 12π₯π¦, plus four π¦ squared.

And now we have a binomial and a
trinomial, which we can multiply together, again using a grid to multiply each term
of the binomial by each term of the trinomial. Nine π₯ squared times three π₯ is
27π₯ cubed. Nine π₯ squared times two π¦ is
18π₯ squared π¦. 12π₯π¦ times three π₯ is 36π₯
squared π¦. 12π₯π¦ times two π¦ is 24π₯π¦
squared. Four π¦ squared times three π₯ is
12π₯π¦ squared. And four π¦ squared times two π¦ is
eight π¦ cubed.

Adding all this stuff together, we
get 27π₯ cubed plus 18π₯ squared π¦ plus 36π₯ squared π¦ plus 24π₯π¦ squared plus
12π₯π¦ squared plus eight π¦ cubed. Finally, we collect like terms for
27π₯ cubed plus 54π₯ squared π¦ plus 36π₯π¦ squared plus eight π¦ cubed.