ORCCA Multiplying Polynomials

Section 5.4 Multiplying Polynomials

Objectives: PCC Course Content and Outcome Guide

MTH 65 CCOG 1.b

Previously in Section 5.2, we learned to multiply two monomials together (such as $4xy\cdot3x^2$). And in Section 5.1, we learned how to add and subtract polynomials even when there is more than one term (such as $(4x^2-3x)+(5x^2+x-2)$). In this section, we will learn how to multiply polynomials with more than one term.

Figure 5.4.1. Alternative Video Lesson

Example 5.4.2. 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 for each time they would raise the price of a jar by $25$ cents, they will sell $50$ fewer jars of jam per 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 times the price was raised by $25$ cents, then the price will be $13+0.25x\text{.}$

Conversely, the number of jars the company will sell will be the $1500$ they currently sell each month, minus $50$ times $x\text{.}$ This gives us the expression $1500-50x$ to represent how many jars the company will sell after raising the price $x$ times.

Combining these expressions, we can write a formula for the revenue model:

\begin{align*} \text{revenue} \amp= \left(\text{price per item}\right)\times\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 learn how to do that.

Subsection 5.4.1 Review of the Distributive Property

Polynomial multiplication relies on the distributive property , and may also rely on the rules of exponents. 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*}

Remark 5.4.3.

We can use the distributive property when multiplying on either the left or the right. This means that $a(b+c)=ab+ac\text{,}$ but also $(b+c)a=ba+ca\text{.}$

Example 5.4.4.

A rectangle's length is $4$ meters longer than its width. Assume its width is $w$ meters. Use a simplified polynomial to model the rectangle's area in terms of $w$ as the only variable.

Explanation

Since the rectangle's length is $4$ meters longer than its width, we can model its length by $w+4$ meters.

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The rectangle's area would be:

\begin{align*} A\amp=\ell w\\ \amp=(w+4)w\\ \amp=w^2+4w \end{align*}

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 are still able to use distribution to simplify the product.

Checkpoint 5.4.5.

The distributive property can be understood visually with a generic rectangle.

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 5.4.6. 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 can be used to visualize multiplying polynomials.

Subsection 5.4.2 Multiplying Binomials

Multiplying Binomials Using Distribution.

Whether we're multiplying a monomial with a polynomial or two larger polynomials together, the first step is still based on the distributive property . We'll start with multiplying two binomials and then move on to 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 $(x+2)$ if it is multiplied. For example:

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

With this in mind, we can multiply $(x+2)(x+3)$ 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 \highlight{x} + x \cdot \highlight{3} + 2 \cdot \highlight{x} + 2 \cdot \highlight{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 acronym 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 5.4.7. 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 5.4.8. 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 5.4.9. 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 5.4.10.

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 5.4.11.

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 5.4.12.

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 5.4.13. 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 5.4.14. 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*}

Example 5.4.15.

Multiply and simplify the formula for Avery's organic jam revenue, $R$ (in dollars), from Example 5.4.2 where $R= (13+0.25x)(1500-50x)$ and $x$ represents the number of times they raised the price by 25 cents.

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 5.4.16.

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

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

Explanation

Currently, each painting costs $200. After raising the price $x$ times, each time by $20, 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 revenue can be calculated by multiplying each painting's price by the number of paintings he would sell:

\begin{align*} \text{annual revenue}\amp=(\text{price})(\text{number of sales})\\ \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, Tyrone's annual income from paintings would be $-100x^2+1000x+20000$ dollars.

Subsection 5.4.3 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 5.4.17.

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.

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Figure 5.4.18. Multiply Each Term by Each Term

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 \highlight{x^2} - x\cdot \highlight{4x} +x\cdot \highlight{6}+5\cdot \highlight{x^2} - 5\cdot \highlight{4x} +5\cdot \highlight{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.

Checkpoint 5.4.19.

Reading Questions 5.4.4 Reading Questions

1.

Describe three ways you can go about multiplying $(x+3)(2x+5)\text{.}$

2.

If you multiplied out $(a+b+c)(d+e+f+g)\text{,}$ how many terms would there be? (Try to answer without actually writing them all down.)