### Video Transcript

In this video, we will learn how to
multiply polynomial expressions together by expanding the parentheses.

To begin, letβs review definitions
of a few terms that we will be using in this lesson. First, a monomial is a product of
constants and variables, where the variables may contain only nonnegative integer
exponents. Second, a polynomial is an
expression that is a sum of single monomial terms. We recall that a polynomial with
two terms is called a binomial. For example, eight π₯ cubed minus
one is a binomial. We also recall that a polynomial
with three terms is called a trinomial, such as π₯ squared minus four π₯ plus
four.

In order to multiply any two
polynomials together, we should first recall how we multiply a polynomial by a
monomial. To do this, we distribute the
multiplication over each term in the polynomial. For example, letβs multiply π₯
times the polynomial π₯ minus π¦. First, we distribute π₯ over the
π₯-term then the π¦-term, resulting in π₯ times π₯ minus π₯ times π¦. Then, after expanding this product,
we simplify the expression to π₯ squared minus π₯π¦.

We can apply this exact same
process to multiply any two polynomials. Letβs consider the product π₯ plus
one times π₯ plus two. This is not quite the same form
since neither factor is a monomial. However, we can treat the factor π₯
plus one as a single term. This means we just multiply each
term of π₯ plus two by the entire factor π₯ plus one as follows. Each term is now the product of a
monomial and binomial. So, we can expand the expression by
distributing π₯ over π₯ plus one and two over π₯ plus one. So, we have π₯ times π₯ plus π₯
times one plus two times π₯ plus two times one. Then, we simplify each product and
finally collect the like terms. In particular, we add the linear
terms by combining their coefficients. So, we have shown that π₯ plus one
times π₯ plus two is equal to π₯ squared plus three π₯ plus two.

In general, we can apply this
process to find the product of any two polynomials. We distribute one factor over every
term in the second factor, expand, and then simplify. Letβs see an example of applying
this process to find the product of two binomials.

Expand two π₯ plus five
squared.

Weβre asked to expand the square of
a binomial. We can do this by first recalling
that squaring an expression means multiplying that expression by itself. So, two π₯ plus five squared is
equal to two π₯ plus five times two π₯ plus five. We can now expand this product by
distributing the first factor over every term in the second factor. So, we multiply two π₯ by two π₯
plus five and five by two π₯ plus five. This is written as two π₯ times two
π₯ plus five plus five times two π₯ plus five. Then, we expand each term by
distributing the factor over each binomial, like so: two π₯ times two π₯ plus two π₯
times five plus five times two π₯ plus five times five.

We can now simplify each term by
recalling that π₯ times π₯ is equal to π₯ squared and collecting like terms. So we have four π₯ squared plus
10π₯ plus 10π₯ plus 25, which simplifies to four π₯ squared plus 20π₯ plus 25. This is the square of two π₯ plus
five.

We can also answer this question by
recalling that we can square a binomial using the formula π plus π squared equals
π squared plus two ππ plus π squared. Substituting π equals two π₯ and
π equals five into the formula yields two π₯ squared plus two times two π₯ times
five plus five squared, which simplifies to four π₯ squared plus 20π₯ plus 25.

In our next example, we will expand
and simplify the product of a trinomial and a binomial with multiple variables and a
negative coefficient.

Expand and simplify π₯ minus two π¦
plus three times two π₯ plus π¦.

To expand the product of two
polynomials, we need to distribute one polynomial over each term in the other. So that means we could distribute
the trinomial over the binomial or the binomial over the trinomial. In this case, it seems the first
option is more straightforward. So, letβs start with distributing
the trinomial over each term in the binomial as follows. We can then expand each of the two
terms by distributing the first factor over each trinomial. For the first term, we have two π₯
times π₯ plus two π₯ times negative two π¦ plus two π₯ times three, which simplifies
to two π₯ squared minus four π₯π¦ plus six π₯. For the second term, we have π¦
times π₯ plus π¦ times negative two π¦ plus π¦ times three, which simplifies to π₯π¦
minus two π¦ squared plus three π¦.

Now we can add the expressions
together and collect like terms to simplify. We obtain two π₯ squared minus
three π₯π¦ minus two π¦ squared plus six π₯ plus three π¦. We have fully expanded and
simplified the product of π₯ minus two π¦ plus three times two π₯ plus π¦.

In our next example, we will use
this process of multiplying polynomials to find a simplified expression for a shaded
region.

A rectangle with dimensions π₯
minus π¦ and π₯ plus π¦ plus one is cut out from a larger rectangle with dimensions
two π₯ plus π¦ plus three and two π₯ plus π¦. Find a simplified expression for
the shaded area.

We can find an expression for the
shaded area by finding the area of the larger rectangle and subtracting the area of
the smaller rectangle from the area of the larger rectangle. We recall the area of a rectangle
is found by multiplying its length times its width. Letβs start by expanding and
simplifying the expression for the area of the larger rectangle, two π₯ plus π¦ plus
three times two π₯ plus π¦. We can do this by distributing the
trinomial over each term in the binomial. So, we have two π₯ times two π₯
plus π¦ plus three plus π¦ times two π₯ plus π¦ plus three.

Then, we can distribute the
monomials over the trinomials. Distributing two π₯ over two π₯
plus π¦ plus three gives four π₯ squared plus two π₯π¦ plus six π₯. And distributing π¦ over the same
trinomial gives two π₯π¦ plus π¦ squared plus three π¦. We can add these expressions
together and collect like terms to find the area of the larger rectangle, four π₯
squared plus four π₯π¦ plus π¦ squared plus six π₯ plus three π¦.

Now, letβs clear some space to find
the area of the smaller rectangle. For the smaller rectangle, we have
dimensions π₯ plus π¦ plus one and π₯ minus π¦. So, the expression for its area
comes from the product of these two expressions. We can do this, once again, by
distributing the trinomial over each term in the binomial. So, we have π₯ times π₯ plus π¦
plus one minus π¦ times π₯ plus π¦ plus one. Then, we can distribute the
monomials over the trinomials. Distributing π₯ over π₯ plus π¦
plus one gives π₯ squared plus π₯π¦ plus π₯. And distributing negative π¦ over
π₯ plus π¦ plus one gives negative π₯π¦ minus π¦ squared minus π¦. We can then add these expressions
together and collect like terms to find the area of the smaller rectangle, resulting
in the expression π₯ squared minus π¦ squared plus π₯ minus π¦.

Now we need to subtract the area of
the smaller rectangle from the area of the larger rectangle to find the area of the
shaded region. We must be careful on this step to
subtract all the terms of the second polynomial, not just the first term. We can do this by distributing the
negative through each term and then collecting the like terms. Thus, we subtract π₯ squared, add
π¦ squared, subtract π₯, and add π¦ to the first expression. We can use a vertical method to
stack the like terms, resulting in three π₯ squared plus four π₯π¦ plus two π¦
squared plus five π₯ plus four π¦. This is the simplified expression
for the shaded region, which we found by subtracting the area of the smaller
rectangle from the area of the larger rectangle.

In our final example, we will
expand the product of three binomials to find the values of unknown
coefficients.

If two π₯ minus π¦ times two π₯
minus five π¦ times three π₯ plus two π¦ is equal to ππ₯ cubed plus ππ₯ squared π¦
plus ππ₯π¦ squared plus ππ¦ cubed, what are the values of π, π, π, and π?

On the left-hand side of the
equation, we have the product of three binomials. And on the right-hand side of the
equation, we have a polynomial. We can find a polynomial expression
equivalent to the left-hand side of the equation by expanding the product. Letβs start off by finding the
product of the first two factors. To do this, we can take the product
of each pair of terms from the first parentheses with the second parentheses and add
the results.

So, we can distribute the first
term from the first binomial through both terms in the second binomial, resulting in
two π₯ times two π₯ plus two π₯ times negative five π¦. Then, we can distribute the second
term from the first binomial through both terms in the second binomial, resulting in
negative π¦ times two π₯ plus negative π¦ times negative five π¦. Simplifying the results gives four
π₯ squared minus 10π₯π¦ minus two π₯π¦ plus five π¦ squared. Then, we can combine like terms to
get four π₯ squared minus 12π₯π¦ plus five π¦ squared. Now we can substitute this
expression into our product of three binomials. Then, we can expand the product by
distributing the trinomial over each term in the binomial.

For the first term, we have 12π₯
cubed minus 36π₯ squared π¦ plus 15π₯π¦ squared. And for the second term, we have
eight π₯ squared π¦ minus 24π₯π¦ squared plus 10π¦ cubed. Combining the like terms and
simplifying gives 12π₯ cubed minus 28π₯ squared π¦ minus nine π₯π¦ squared plus 10π¦
cubed. We are told this is equal to ππ₯
cubed plus ππ₯ squared π¦ plus ππ₯π¦ squared plus ππ¦ cubed. For the polynomials to be equal,
their coefficients must be equal. Thus, π equals 12, π equals
negative 28, π equals negative nine, and π equals 10.

Letβs finish by recapping some of
the important points from this video. We can multiply two polynomials by
first multiplying every term of one polynomial by the other entire polynomial then
summing their results. We can finally simplify by
combining like terms. We also saw in our last example
that the product of any two polynomials can also be found by multiplying every pair
of terms from each of the polynomials and summing the results. So for two binomials π plus π and
π plus π, this means their product can be found by summing π times π plus π
times π plus π times π plus π times π.