ORCCA Multiplying Polynomials

Section 6.5 Multiplying Polynomials

Previously, we have learned to multiply monomials in Section 6.2 (such as $(4xy)\left(3x^2\right)$) and to add and subtract polynomials in Section 6.4 (such as $(4x^2-3x)+(5x^2+x-2)$). In this section, we will learn how to multiply polynomials.

Example 6.5.1. Revenue.

Avery owns a local organic jam company that currently sells about $1500$ jars a month at a price of $\$13$ per jar. Avery has found that every time they raise the price by $25$ cents a jar, 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.

If we let $x$ represent the number of $25$-cent increases in the price, then the price per jar will be the current price of thirteen dollars/jar plus $x$ times $0.25$ dollars/jar, or $13+0.25x\text{.}$

Continuing with $x$ representing the number of $25$-cent increases in the price, we know the company will sell $50$ fewer jars each time the price increases by $25$ 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 $x$ $25$-cent price increases.

Combining this, we can now write a formula for our revenue model:

\begin{align*} \text{revenue} \amp= \left(\text{price per item}\right)\left(\text{number of items sold}\right)\\ R \amp= \left(13+0.25x\right)\left(1500-50x\right) \end{align*}

To simplify the expression $\left(13+0.25x\right)\left(1500-50x\right)\text{,}$ we'll need to multiply $13+0.25x$ by $1500-50x\text{.}$ In this section, we'll learn how to multiply these two expressions that each have multiple terms.

Subsection 6.5.1 Review of the Distributive Property

The first step in almost every polynomial multiplication exercise will be a step of distribution. Let's quickly review the distributive property from Section 2.9, which states that $a(b+c)=ab+ac$ where $a, b\text{,}$ and $c$ are real numbers or variable expressions.

When we multiply a monomial with a binomial, we apply this property by distributing the monomial to each term in the binomial. For example,

\begin{align*} \highlight{-4x}(3x^2+5) \amp= \multiplyleft{(-4x)}\left(3x^2\right)+\multiplyleft{(-4x)}(5)\\ \amp=-12x^3-20x \end{align*}

A visual approach to the distributive property is to treat the product as finding a rectangle's area. Such rectangles are referred to as generic rectangles and they can be used to model polynomial multiplication.

a rectangle with two regions; the region on the left indicates 2x*3x=6x^2; the region on the right indicates that 2x*4=8x; together they add up to 6x^2+8x — Figure 6.5.2. A Generic Rectangle Modeling $2x(3x+4)$

The big rectangle consists of two smaller rectangles. The big rectangle's area is $2x(3x+4)\text{,}$ and the sum of those two smaller rectangles is $2x\cdot3x+2x\cdot4\text{.}$ Since the sum of the areas of those two smaller rectangles is the same as the bigger rectangle's area, we have:

\begin{align*} 2x(3x+4) \amp= 2x\cdot3x+2x\cdot4\\ \amp= 6x^2+8x \end{align*}

Generic rectangles are frequently used to visualize the distributive property.

Multiplying a monomial with a polynomial involves two steps: distribution and monomial multiplication. We also need to rely on the rules of exponents 6.2.14 when simplifying.

Checkpoint 6.5.3.

Checkpoint 6.5.4.

Checkpoint 6.5.5.

Remark 6.5.6.

We can use the distributive property when multiplying on either the left or the right. This means that we can state $a(b+c)=ab+ac\text{,}$ or that $(b+c)a=ba+ca\text{,}$ which is equivalent to $ab+ac\text{.}$ As an example,

\begin{align*} (3x^2+5)\highlight{(-4x)} \amp= (3x^2)\multiplyright{(-4x)}+(5)\multiplyright{(-4x)}\\ \amp=-12x^3-20x \end{align*}

Subsection 6.5.2 Approaches to Multiplying 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 know we can distribute the $3$ in $(x+2)3$ to obtain $(x+2)\multiplyright{3}=x\multiplyright{3}+2\multiplyright{3}\text{.}$ We can actually distribute anything across $(x+2)\text{.}$ For example:

\begin{equation*} (x+2)\cat=x\cdot \cat + 2\cdot \cat \end{equation*}

With this in mind, we can begin multiplying $(x+2)(x+3)$ by distributing the $(x+3)$ across $(x+2)\text{:}$

\begin{equation*} (x+2)\highlight{(x+3)} = x\highlight{(x+3)} + 2\highlight{(x+3)} \end{equation*}

To finish multiplying, we'll continue by distributing again, but this time across $(x+3)\text{:}$

\begin{align*} (x+2)\highlight{(x+3)} \amp= x\highlight{(x+3)} + 2\highlight{(x+3)}\\ \amp= x \cdot x + x \cdot 3 + 2 \cdot x + 2 \cdot 3\\ \amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}

To multiply a binomial by another binomial, we simply had to repeat the step of distribution and simplify the resulting terms. In fact, multiplying any two polynomials will rely upon these same steps.

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".

\begin{align*} (x+2)(x+3)\amp= (\overbrace{{x} \stackrel{}{\cdot} {x}}^{\text{F}}) + (\overbrace{{3} \stackrel{}{\cdot} {x}}^{\text{O}}) + (\overbrace{{2} \stackrel{}{\cdot} {x}}^{\text{I}}) + (\overbrace{{2} \stackrel{}{\cdot} {3}}^{\text{L}})\\ \amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}

F: $x^2$: The $x^2$ term was the result of the product of first terms from each binomial.
O: $3x$: 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.
I: $2x$: 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.
L: $6$: The constant term $6$ was the result of the product of the last terms of each binomial.

a diagram that shows how to multiply using FOIL; x*x + x*3 + 2*x + 6 — Figure 6.5.7. Using 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 $(x+2)(x+3)\text{.}$

a two by two rectangle with the terms x and 2 above the columns and the terms x and 3 on the left side of the rows — Figure 6.5.8. Setting 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 $\text{area}=\text{base}\cdot\text{height}\text{.}$

the previous generic rectangles with the areas computed; x*x=x^2, x*2=2x, 3*x=3x and 3*2=6 — Figure 6.5.9. Using Generic Rectangles to Multiply $(x+2)(x+3)$

To finish finding this product, we need to add the areas of the four smaller rectangles:

\begin{align*} (x+2)(x+3)\amp=x^2+3x+2x+6\\ \amp=x^2+5x+6 \end{align*}

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. Both the FOIL method and generic rectangles approach are different ways to represent the distribution that is occurring.

Example 6.5.10.

Multiply $(2x-3y)(4x-5y)$ using distribution.

Explanation

To use the distributive property to multiply those two binomials, we'll first distribute the second binomial across $(2x-3y)\text{.}$ Then we'll distribute again, and simplify the terms that result.

\begin{align*} (2x-3y)\highlight{(4x-5y)}\amp=2x\highlight{(4x-5y)}-3y\highlight{(4x-5y)}\\ \amp=8x^2-10xy-12xy+15y^2\\ \amp=8x^2-22xy+15y^2 \end{align*}

Example 6.5.11.

Multiply $(2x-3y)(4x-5y)$ using FOIL.

Explanation

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.

\begin{align*} (2x-3y)(4x-5y)\amp= (\overbrace{{\stackrel{}{2x}}\cdot{4x}}^{\large\text{F}})+ (\overbrace{{\stackrel{}{2x}}\cdot{(-5y)}}^{\large\text{O}})+ (\overbrace{{\stackrel{}{-3y}}\cdot{4x}}^{\large\text{I}})+ (\overbrace{{\stackrel{}{-3y}}\cdot{(-5y}}^{\large\text{L}})\\ \amp=8x^2-10xy-12xy+15y^2\\ \amp=8x^2-22xy+15y^2 \end{align*}

Example 6.5.12.

Multiply $(2x-3y)(4x-5y)$ using generic rectangles.

Explanation

We begin by drawing four rectangles and marking their bases and heights with terms in the given binomials:

a two by two rectangle with the terms 2x and -3y above the columns and the terms 4x and -5y on the left of the rows — Figure 6.5.13. Setting up Generic Rectangles to Multiply $(2x-3y)(4x-5y)$

Next, we calculate each rectangle's area by multiplying its base with its height:

the previous generic rectangles with the areas multiplied; 2x*4x=8x^2, 2x*-5y=-10xy, -3y*4x=-12xy and -3y*-5y=15y^2 — Figure 6.5.14. Using Generic Rectangles to Multiply $(2x-3y)(4x-5y)$

Finally, we add up all rectangles' area to find the product:

\begin{align*} (2x-3y)(4x-5y)\amp=8x^2-10xy-12xy+15y^2\\ \amp=8x^2-22xy+15y^2 \end{align*}

Subsection 6.5.3 More Examples of Multiplying Binomials

When multiplying binomials, all of the approaches shown in Subsection 6.5.2 will have the same result. The FOIL method is the most direct and will be used in the examples that follow.

Checkpoint 6.5.15.

Checkpoint 6.5.16.

Example 6.5.17.

Multiply and simplify the formula for Avery's jam company's revenue, $R$ (in dollars), from Example 6.5.1 where $R= (13+0.25x)(1500-50x)$ and $x$ represents the number of 25-cent price increases to the selling price of a jar of jam.

Explanation

To multiply this, we'll use FOIL:

\begin{align*} R \amp= \left(13+0.25x\right)\left(1500-50x\right)\\ \amp= \left(13\cdot1500\right) + \left(13 \cdot (-50x) \right) + \left( 0.25x \cdot 1500 \right) + \left( 0.25x \cdot (-50x) \right)\\ \amp= 19500 - 650x + 375x - 12.5x^2\\ \amp= -12.5x^2 - 275x + 19500 \end{align*}

Example 6.5.18.

Tyrone is an artist and he sells each of his paintings for $\$200\text{.}$ Currently, he can sell $100$ paintings per year. Thus, his annual income from paintings is $200\cdot100=20000$ dollars. He plans to raise the price. However, for each $\$20$ price increase per painting, his customers would buy $5$ fewer paintings annually.

Assume Tyrone would raise the price of his paintings $x$ times, each time by $\$20\text{.}$ Use an expanded polynomial to represent his new income per year.

Explanation

Currently, each painting costs $\$200\text{.}$ After raising the price $x$ times, each time by $\$20\text{,}$ each painting’s new price would be $200+20x$ dollars.

Currently, Tyrone 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 by the number of paintings he would sell:

\begin{align*} \text{annual income}\amp=(200+20x)(100-5x)\\ \amp=200(100)+200(-5x)+20x(100)+20x(-5x)\\ \amp=20000-1000x+2000x-100x^2\\ \amp=-100x^2+1000x+20000 \end{align*}

After raising the price $x$ times, each time by $\$20\text{,}$ Tyrone’s annual income from paintings would be $-100x^2+1000x+20000$ dollars.

Subsection 6.5.4 Multiplying Polynomials Larger Than Binomials

The foundation for multiplying any pair of polynomials is distribution and monomial multiplication. Whether we are working with binomials, trinomials, or larger polynomials, the process is fundamentally the same.

Example 6.5.19.

Multiply $\left( x+5 \right)\left( x^2-4x+6 \right)\text{.}$

We can approach this product using either distribution generic rectangles. We cannot directly use the FOIL method, although it can be helpful to draw arrows to the six pairs of products that will occur.

Using the distributive property, we begin by distributing across $\left( x^2-4x+6 \right)\text{,}$ perform a second step of distribution, and then combine like terms.

\begin{align*} \left(x+5\right)\highlight{\left( x^2-4x+6 \right)} \amp= x\highlight{\left( x^2-4x+6 \right)}+5\highlight{\left( x^2-4x+6 \right)}\\ \amp= x\cdot x^2 - x\cdot 4x +x\cdot 6 +5\cdot x^2 - 5\cdot 4x +5\cdot 6\\ \amp= x^3 -4x^2 +6x +5x^2 -20x +30\\ \amp= x^3+x^2-14x+30 \end{align*}

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.