Previously, we have learned to add and subtract polynomials. In this section, we will learn how to multiply polynomials, which will make use of our polynomial addition and subtraction skills.
Figure6.3.1Alternative Video Lesson
Subsection6.3.1Multiplying Monomials
When we multiply the monomials \(2\) and \(xy\text{,}\) note that we have only a single term being multiplied by a single term. Simply multiply the coefficients with the coefficients and combine variables using the product rule (2.7.1). Thus we have that \(2(xy)=2xy\text{.}\) Please remember that you only use <<xref without ref, first/last, or provisional attribute (check spelling)>> across addition or subtraction, never multiplication or division. This is why the \(2\) doesn't “distribute” to both the \(x\) and \(y\) separately.
To multiply these two monomials, we have to recognize which factors in either polynomial can interact. Numbers can interact with numbers, \(x\)'s can interact with \(x\)'s, \(y\)'s can interact with \(y\)'s, and so on.
Formally, we'd start by recognizing the only operation involved is multiplication, we'd then use the commutative property to move related factors next to each other, and then we'd then regroup those related factors using the associative property.
\begin{align*}
\left( 2x \right)\left( 4xy \right) \amp= \left(2 \cdot x\right) \cdot \left( 4 \cdot x \cdot y \right)\\
\amp= 2 \cdot x \cdot 4 \cdot x \cdot y \\
\amp= 2 \highlight{{}\cdot 4 \cdot x } \cdot x \cdot y \\
\amp= \left(2 \cdot 4\right) \cdot \left(x \cdot x\right) \cdot \left(y\right) \\
\amp= 8 \cdot x^2 \cdot y \\
\amp= 8 x^2 y
\end{align*}
The fact is that much of what we just did on paper can be done in our minds. Normally the previous example would not take six steps of work on paper. It would normally be done as a one-step problem. Even if this only takes one step of work on paper, all of that work will still be done mentally.
Subsection6.3.2Multiplying a Monomial and a Polynomial
Recall the Distributive property. What properties are involved with multiplying \(2(x+y)\text{?}\) How is this different than what we were just discussing?
In this example, we have addition inside the parentheses and by having addition (or subtraction) inside the parentheses. We are now in a situation that will require distribution.
When we multiply a monomial with a binomial, we need to first distribute the monomial to each term in the binomial.
The first step in almost every polynomial multiplication exercise will be a step of distribution. Let's quickly review the distributive property with an area model. Let's look at the product \(2(3+4)\text{.}\) We can treat the product as finding a rectangle's area, where the rectangle's height is \(2\) units, and its base is \(3+4\) units. We call such rectangles Generic Rectangles and they can be used to model polynomial multiplication.
The big rectangle consists of two smaller rectangles. The big rectangle's area is \(2(3+4)\text{,}\) and the sum of those two smaller rectangles is \(2\cdot3+2\cdot4\text{.}\) Since the sum of the areas of those two smaller rectangle is the same as the bigger rectangle's area, we have:
In these examples, it turned out that there weren't any like terms to combine at the end of our work.
Example6.3.10
A rectangle's length is \(4\) meters longer than its width. Assume its width is \(w\) meters. Use an expanded polynomial to model the rectangle's area.
The rectangle's area can be modeled by \(w^2+4w\) square meters.
In the second line of work above, we should recognize that \((w+4)(w)\) is equivalent to \(w(w+4)\text{.}\) Whether the \(w\) is written before or after the binomial, we still are able to use distribution to sipmlify the product.
Multiplying a monomial with a polynomial involves two steps: distribution and monomial multiplication.
Exercise6.3.11
Exercise6.3.12
Exercise6.3.13
Exercise6.3.14
Subsection6.3.3Multiplying Binomials
Multiplying Binomials Using Distribution
Whether we're multiplying a monomial with a polynomial or two larger polynomials together, the first step to carrying out the multiplication is a step of distribution. We'll start with multiplying binomials and then move to working with larger polynomials.
We just stated that multiplying \((x+2)(x+3)\) will require us to use the distributive property.
While we know we can distribute \(2(x+3)=2\cdot x+2\cdot3\text{,}\) it's important to understand that what we distribute doesn't need to be a number. We can actually distribute anything into \((x+3)\text{.}\) For example:
\begin{equation*}
😀(x+3)=x😀+3😀
\end{equation*}
Note that on the right hand side of the \(=\) we chose to place the 😀 behind each of the terms of the binomial to motivate what is below. You can place the 😀 either in front or behind the terms according to the commutative property of multiplication (2.9.1).
Similarly, we can begin multiplying \((x+2)(x+3)\) by distributing the \((x+2)\) into \((x+3)\text{:}\)
While we have done a step of distribution and removed one set of parentheses, we need to use the distributive property again to fully remove all the parentheses. Once this is done, we can multiply the individual monomial pairs as before and then combine like terms. The full solution is:
\begin{align*}
\highlight{(x+2)}(x+3) \amp= x\highlight{(x+2)} + 3\highlight{(x+2)}\\
\amp= x \cdot x + x \cdot 2 + 3 \cdot x + 3 \cdot 2\\
\amp=x^2+2x+3x+6\\
\amp=x^2+5x+6
\end{align*}
We had previously said that multiplying a monomial with a polynomial involves two steps: distribution and monomial multiplication. The truth is that multiplying any two polynomials will rely upon the same two steps. The difference with larger polynomials is that they simply require more distribution.
Multiplying Binomials Using FOIL
While multiplying two binomials requires two applications of the distributive property, people often remember this distribution process using the mnemonic FOIL. FOIL refers to the pairs of terms from each binomial that end up distributed to each other.
If we take another look at the example we just completed, \((x+2)(x+3)\text{,}\) we can highlight how the FOIL process works. FOIL is the acronym for "First, Outer, Inner, Last".
The \(x^2\) term was the result of the product of first terms from each binomial.
\(3x\)
The \(3x\) was the result of the product of the outer terms from each binomial. This was from the \(x\) in the front of the first binomial and the \(3\) in the back of the second binomial.
\(2x\)
The \(2x\) was the result of the product of the inner terms from each binomial. This was from the \(2\) in the back of the first binomial and the \(x\) in the front of the second binomial.
\(6\)
The constant term \(6\) was the result of the product of the last terms of each binomial.
To make the spelling of FOIL fit more cleanly, we used the commutative property to switch the order of the two middle terms \(2x\) and \(3x\text{.}\)
Figure6.3.15Using FOIL Method to multiply \((x+2)(x+3)\)
Multiplying Binomials Using Generic Rectangles
We can also approach this same example using the generic rectangle method. To use generic rectangles, we treat \(x+2\) as the base of a rectangle, and \(x+3\) as the height. Their product, \((x+2)(x+3)\text{,}\) represents the rectangle's area. The next diagram shows how to set up generic rectangles to multiply \((x+2)(x+3)\text{.}\)
Figure6.3.16Setting up Generic Rectangles to Multiply \((x+2)(x+3)\)
The big rectangle consists of four smaller rectangles. We will find each small rectangle's area in the next diagram by the formula \(\text{area}=\text{base}\cdot\text{height}\text{.}\)
Figure6.3.17Using Generic Rectangles to Multiply \((x+2)(x+3)\)
To finish finding this product, we need to add the areas of the four smaller rectangles:
You should notice that the areas of the four smaller rectangles are exactly the same as the four terms we obtained using distribution, which are also the same four terms that came from the FOIL method. This is not a coincidence. Both the FOIL method and generic rectangles approach are different ways to represent the distribution that is occurring.
To use the distributive property to multiply those two binomials, we'll first distribute \((2x-3y)\) through the second binomial. Then we'll distribute again, to distribute the \(4x\) and \(-5y\) through the remaining parentheses.
First, Outer, Inner, Last: Either with arrows on paper or mentally in our heads, we'll pair up the four pairs of monomials and multiply those pairs together.
The local organic jam company currently sells about 1500 jars a month at a price of thirteen dollars/jar. They have also realized that for every 20 cent increase in the selling price of a jar of jam, they will sell 50 fewer jars of jam each month.
In general, this company's revenue can be calculated by multiplying the cost per jar by the total number of jars of jam sold. What expression can model the revenue, \(R\) (in dollars), for the jam company?
If we let \(x\) represent the number of 20 cent increases in the price, then the price per jar will be the current price of thirteen dollars/jar plus \(x\) times 0.20 dollars/jar, or \(13+0.2x\text{.}\)
Continuing with \(x\) representing the number of 20-cent increases in the price, we know the company will sell 50 fewer jars each time the price increases by 20 cents. The number of jars the company will sell will be the 1500 they currently sell each month, minus 50 jars times \(x\text{,}\) the number of price increases. This gives us the expression \(1500-50x\) to represent how many jars the company will sell after \(x\) price increases.
Combining this, we can now write a formula for our revenue model:
\begin{align*}
\text{revenue} \amp= \left(\text{cost per item}\right)\left(\text{number of items sold}\right)\\
R \amp= \left(13+0.2x\right)\left(1500-50x\right)
\end{align*}
Multiply and simplify the formula for the company's revenue, \(R\) (in dollars), \(R= (13+0.2x)(1500-50x)\text{,}\) where \(x\) represents the number of 20 cent price increases to the selling price of a jar of jam.
An artist sells his paintings at $\(10.00\) per piece. Currently, he can sell \(100\) paintings per year. Thus, his annual income from paintings is \(10\cdot100=1000\) dollars. He plans to raise the price. However, for each $\(2.00\) of price increase per painting, his customers would buy \(5\) fewer paintings annually.
Assume the artist would raise the price of his painting \(x\) times, each time by $\(2.00\text{.}\) Use an expanded polynomial to represent his new income per year.
Currently, each painting costs $\(10.00\text{.}\) After raising the price \(x\) times, each time by $\(2.00\text{,}\) each painting’s new price would be \(10+2x\) dollars.
Currently, the artist sells \(100\) paintings per year. After raising the price \(x\) times, each time selling \(5\) fewer paintings, he would sell \(100-5x\) paintings per year.
His annual income can be calculated by multiplying each painting’s price with the number of paintings he would sell:
Solution: After raising the price x times, each time by $\(2.00\text{,}\) the artist’s annual income from paintings would be \(-10x^2+150x+1000\) dollars.
Exercise6.3.25
Exercise6.3.26
Subsection6.3.4Multiplying Polynomials Larger Than Binomials
The foundation for multiplying any polynomials is distribution and monomial multiplication. Whether we are working binomial, trinomials, or larger polynomials, the process is fundamentally the same.
We can approach this product using any of the three methods from before: distribution, arrows, or generic rectangles.
Using the distributive property, we'd begin by distributing \((x+3)\) through \(\left( x^2-4x+6 \right)\) and then finish with a second step of distribution.
We finished, as we often will, with a step of combing like terms.
If we had approached this using the arrows, similar to what we did with FOIL, our work would look like this.
Figure6.3.28Using Arrows to Distribute \(\left( x+3 \right)\left( 2x^2-4x+6 \right)\)
We should again note that the six arrows draw perfectly match with the six pairs of monomial products we obtained in the second and third lines of work in the previous approach. Based on the pairs of products identified by the arrows, we'd have:
It's also important to recognize calling this approach "FOIL" in this context doesn't make sense. There are more than just the first, outer, inner, and last products in this example. While the name isn't appropriate, the idea of the approach does still work.
We can also approach this using the generic rectangles approach. We'll treat \(x+3\) as the height of the large rectangle and \(x^2-4x+6\) as the width:
Figure6.3.29Setting up Generic Rectangles to Multiply \((x+3)(x^2-4x+6)\)
Next, we calculate each rectangle's area by multiplying its base with its height:
Figure6.3.30Using Generic Rectangles to Multiply \((x+3)(x^2-4x+6)\)
As before, each smaller rectangle's area represents one of the pairs obtained in our distribution process. We would finish in a very similar manner:
With the foundation of monomial multiplication and understanding how distribution applies in this context, we are able to find the product of any two polynomials.
