3. Multiplying Polynomials
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Chapter 7
3.
Multiplying Polynomials
This lesson offers a comprehensive guide on how to multiply polynomials using various techniques, including the distributive property and the FOIL method. It goes beyond mere definitions to explore real-world applications. For example, it discusses how these methods can be used in calculating areas and volumes in geometry, or in solving complex equations in algebra. The FOIL method, which stands for First, Outer, Inner, Last, is explained as a mnemonic to help remember the steps involved in multiplying binomials. The distributive property is also elaborated upon, showing how it can simplify complex polynomial equations. Overall, this material serves as an invaluable lesson for anyone looking to master polynomial multiplication and its applications in different fields of mathematics.
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| | 11 Theory slides |
| | 13 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Multiplying Polynomials
Slide of 11
Polynomials are useful for modeling different real-life situations. Sometimes more than one polynomial could be involved and the polynomials may need to be multiplied. For such a reason, this lesson aims to teach different methods for multiplying polynomials.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Investigating the Product of Polynomials
When two polynomials are added or subtracted, the result is also a polynomial. What about the product of two polynomials? Consider, for example, the following pair of polynomials.
P(x)Q(x)=2x3−x+5=4(x+2)(x−1)
Is it possible to calculate the product of P(x) and Q(x)? Is P(x)⋅Q(x) also a polynomial? In the affirmative case, what are the degree, the leading coefficient, and the constant term of the resulting polynomial? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Degree <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["5"]}}
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Discussion
Multiplying a Polynomial by a Constant
Consider a polynomial P(x) written in standard form.
Note that the right-hand side is a polynomial of the same degree as P(x) — the degree of both is n. On the other hand, if C=1, the leading coefficient of C⋅P(x) is different from the leading coefficient of P(x).
Consequently, when a polynomial P(x) is multiplied by a constant C, the degree of the polynomial does not change and the leading coefficient equals C times the leading coefficient of P(x).
P(x)=anxn+an−1xn−1+⋯+a1x+a0
Using the Distributive Property, multiply P(x) by a non-zero constant C.
P(x)=anxn+an−1xn−1+⋯+a1x+a0
MultEqn
LHS⋅C=RHS⋅C
C⋅P(x)=C(anxn+an−1xn−1+⋯+a1x+a0)
Distr
Distribute C
C⋅P(x)=C⋅anxn+C⋅an−1xn−1+⋯+C⋅a1x+C⋅a0
| Polynomial | Degree | Leading Coefficient |
|---|---|---|
| P(x)=anxn+an−1xn−1+⋯+a1x+a0 | n | an |
| C⋅P(x)=C⋅anxn+C⋅an−1xn−1+⋯+C⋅a1x+C⋅a0 | n | C⋅an |
Example
Finding the Area of a Poster
Ali is on his school's student council. He is in charge of decorations for an upcoming school dance. He had the following chat with Kevin.
After the talk, Kevin made the following diagram. 
A polynomial modeling the area of the poster is obtained by substituting the poster's dimensions into the formula for the area.
The area of the poster will be 112 square feet.
a Write a polynomial A(x), in standard form, that models the area of the poster.
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b What are the degree and leading coefficient of A(x)?
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c The next day, Ali called Kevin and told him that the wall is 16 feet wide. What area will the poster have?
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Hint
a The poster is 2 feet shorter and narrower than the wall. To find its area, multiply the length by the width.
b The degree of a polynomial is the greatest exponent in the polynomial. The leading coefficient is the coefficient in front of the term with the greatest exponent.
c Evaluate A(x) at x=16.
Solution
a The area of the poster equals its length multiplied by its width.
A=ℓ⋅w
From the chat and the diagram made by Kevin, the poster has to be 2 feet shorter and narrower than the wall. Since the wall is 10 feet tall, the poster will be 8 feet tall. The wall is x feet wide, so the width of the poster is x−2 feet. A(x)=(x−2)⋅8⇓A(x)=8x−16
Note that this polynomial gives the area of the poster depending on the width of the wall.
b Start by writing the polynomial found in Part A. Then, highlight the term with the greatest exponent and its coefficient.
A(x)=8x−16⇕A(x)=8x1−16
The greatest exponent is 1, which means that the degree of the polynomial is 1. The coefficient in front of the term with the greatest exponent is 8. Thus, the leading coefficient is 8.
c Since x represents the width of the wall, to find the area of the poster, substitute x=16 into A(x).
A(x)=8x−16
Substitute
x=16
A(16)=8(16)−16
Multiply
Multiply
A(16)=128−16
SubTerms
Subtract terms
A(16)=112
Discussion
Methods for Multiplying Polynomials
Different methods can be used to multiply two polynomials. The following three methods are based on the Distributive Property.
Method
Multiplying Polynomials Using the Distributive Property
The product of two polynomials can be calculated by using the Distributive Property. Consider, for example, the following pair of polynomials.
Note that multiplying a polynomial with n terms by a polynomial with m terms produces n⋅m products. When two polynomials are multiplied, the product is a new polynomial whose degree equals the sum of the degrees of the polynomial factors. 
P(x)Q(x)=x3+2x2−3=x2+4
To multiply these two polynomials, the following four steps can be followed.
1
Distribute One Polynomial to All the Terms of the Other
expand_more Start by writing the product P(x)⋅Q(x).
(x3+2x2−3)(x2+4)
Next, distribute P(x) to each term of Q(x).
2
Clear Parentheses by Applying the Distributive Property
expand_more 3
Apply the Product of Powers Property
expand_more Use the Product of Powers Property to rewrite the product of two or more exponents as a single power.
4
Combine Like Terms and Simplify
expand_more Finally, combine like terms and perform all the required operations to simplify the result.
(x5+2x4−3⋅x2)+(x3⋅4+2x2⋅4−3⋅4)
Multiply
Multiply
(x5+2x4−3x2)+(4x3+8x2−12)
CommutativePropAdd
Commutative Property of Addition
x5+2x4+4x3−3x2+8x2−12
AssociativePropAdd
Associative Property of Addition
x5+2x4+4x3+(-3x2+8x2)−12
AddTerms
Add terms
x5+2x4+4x3+5x2−12
Method
Multiplying Polynomials Using the Box Method
Given two polynomials, their product can be calculated by using a box or table. Consider, for example, the following pair of polynomials.
P(x)Q(x)=x3+2x2−3=x2+4
To multiply these two polynomials, the following four steps can be followed.
1
Determine the Dimensions of the Table
expand_more Start by drawing a table that has as many rows as there are terms in the first polynomial and that has as many columns as there are terms in the second polynomial.
| Polynomial | Number of Terms |
|---|---|
| P(x)=x3+2x2−3 | 3 |
| Q(x)=x2+4 | 2 |
For example, a table with 3 rows and 2 columns is needed to multiply P(x) by Q(x).
2
Write the Row and Column Labels of the Table
expand_more Now, write each term of the first polynomial at the left of each cell of the first column. Similarly, write each term of the second polynomial above each cell of the first row.
3
Multiply the Terms to Fill the Table
expand_more Next, fill in the table's cells by multiplying the terms written on the corresponding borders of the table. For example, the top-left cell corresponds to the product of x3 and x2. The remaining cells can be filled by following the same procedure. 
4
Add Terms and Simplify
expand_more Finally, add all the expressions inside the table and combine like terms, if any.
The product of these polynomials has been found to be x5+2x4+4x3+5x2−12.
Method
The FOIL Method
The FOIL method is a mnemonic for remembering how to multiply two binomials. The word FOIL is an acronym for the words First, Outer, Inner, and Last. Consider, for example, the following product.
The following applet illustrates the FOIL method using two arbitrary binomials.
Like any other polynomial multiplication, the FOIL method is based on the Distributive Property.
(x+6)(3x−2)
These two binomials can be multiplied by following five steps.
1
Multiply the First Terms
expand_more Start by multiplying the first terms of each binomial. In this case, multiply x by 3x.
(x+6)(3x−2)=x(3x)−2x+18x−12
The empty box is there as a reminder that there are still missing terms. 2
Multiply the Outer Terms
expand_more Next, multiply the outer terms — that is, multiply the first term of the left-hand side binomial by the second term of the right-hand side binomial. In this case, multiply x by -2.
(x+6)(3x−2)=x(3x)+x(-2)18x−12
3
Multiply the Inner Terms
expand_more Now multiply the inner terms — that is, multiply the second term of the left-hand side binomial by the first term of the right-hand side binomial. In this case, multiply 6 by 3x.
(x+6)(3x−2)=x(3x)+x(-2)+6(3x)−12
4
Multiply the Last Terms
expand_more Next, multiply the last terms of each binomial — that is, multiply the second term of the left-hand side binomial by the second term of the right-hand side binomial. In this case, multiply 6 by -2.
(x+6)(3x−2)=x(3x)+x(-2)+6(3x)+6(-2)
5
Simplify
expand_more Finally, find each product and combine like terms, if any, to simplify the resulting expression.
(x+6)(3x−2)=x(3x)+x(-2)+6(3x)+6(-2)
CommutativePropMult
Commutative Property of Multiplication
(x+6)(3x−2)=3x⋅x+x(-2)+6(3x)+6(-2)
ProdToPowTwoFac
a⋅a=a2
(x+6)(3x−2)=3x2+x(-2)+6(3x)+6(-2)
MultPosNeg
a(-b)=-a⋅b
(x+6)(3x−2)=3x2−x(2)+6(3x)−6(2)
Multiply
Multiply
(x+6)(3x−2)=3x2−2x+18x−12
AddTerms
Add terms
(x+6)(3x−2)=3x2+16x−12
Example
Modeling the Pigpen Area
Diego's parents recently bought a piece of land where they plan to raise pigs. They need to fence off a rectangular pigpen before buying the pigs. A farmer friend told Diego that the dimensions of the pigpen, in yards, vary according to the number of pigs x that are being raised in it.
a Write a polynomial A(x), in standard form, that models the pigpen area. Then, state the degree and the leading coefficient of A(x).
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b If Diego plans to raise 15 pigs, what should be the area of the pigpen?
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Hint
a The area of a rectangle equals the length times the width.
b Evaluate A(x) at x=15.
Solution
a The area of a rectangle is obtained by multiplying its length by its width. Start by writing the dimensions suggested by Diego's friend.
Length:Width:5x+23x+1
By multiplying the two binomials, a polynomial modeling the pigpen area will be obtained.
A(x)=(5x+2)(3x+1)
To write A(x) in standard form, the product on the right-hand side can be performed by using the FOIL method.
b To determine the area of a pigpen large enough to raise 15 pigs, substitute 15 for x into A(x)=15x2+11x+2 and simplify.
A(x)=15x2+11x+2
Substitute
x=15
A(15)=15(15)2+11(15)+2
CalcPowProd
Calculate power and product
A(15)=15(225)+165+2
Multiply
Multiply
A(15)=3375+165+2
AddTerms
Add terms
A(15)=3542
Example
Determining the Area of a Triangle
Consider a right triangle on the coordinate plane with vertices J(-10,0), K(x,0), and L(x,P(x)), where P(x)=x3+3x2−18x+54.
The next step is to fill the table by multiplying the terms written on the corresponding borders.
Now, add all the terms inside the table and simplify by combining like terms.
Having computed JK⋅KL, a polynomial modeling the area of △JKL can be found.
Consequently, the degree of A(x) is 4 and its leading coefficient is 21.
To find the area of the above triangle, substitute 3 for x into the polynomial A(x) found in Part A.
In consequence, when K has coordinates (3,0), the triangle JKL has an area of 351.
To determine which triangle has the larger area, find A(-4) and A(-1) and compare them. Start by finding A(-4).
Next, find the value of A(-1).
As can be seen, A(-1) is greater than A(-4). Therefore, the triangle generated by K2(-1,0) has the larger area.
a Write a polynomial A(x), in standard form, that represents the area of △JKL. Then, state the degree and leading coefficient of A(x).
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b If K has coordinates (3,0), what is the area of △JKL?
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c Which one of the following triangles has the larger area?
{"type":"choice","form":{"alts":["The triangle generated by <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">K<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:-0.07153em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">4<\/span><span class=\"mpunct\">,<\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span>","The triangle generated by <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\" style=\"margin-right:0.07153em;\">K<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:-0.07153em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">1<\/span><span class=\"mpunct\">,<\/span><span class=\"mspace\" style=\"margin-right:0.16666666666666666em;\"><\/span><span class=\"mord\">0<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
Hint
a The area of a triangle equals half the product of its base and its height. Note that the height of △JKL is P(x) and the base is x+10.
b Evaluate A(x) at x=3.
c Find and compare A(-4) and A(-1).
Solution
a The area of a triangle is half the product of its base and its height. Since the given triangle is a right triangle, its area equals half the product of its legs.
A=21⋅JK⋅KL
The length of the legs can be written using the given expressions.
JKKL=x+10=x3+3x2−18x+54
Next, substitute the previous expressions into the formula for the area.
A(x)=21(x+10)(x3+3x2−18x+54)
For simplicity, the product of the two polynomials will be computed first. Then, the resulting polynomial will be multiplied by 21. To perform the polynomial multiplication, the Box Method can be used. To do so, start by drawing a 2×4 table and writing the row and column labels. A(x)=21JK⋅KL
Substitute
JK⋅KL=x4+13x3+12x2−126x+540
A(x)=21(x4+13x3+12x2−126x+540)
Distr
Distribute 21
A(x)=21x4+213x3+6x2−63x+270
b If the coordinates of K are (3,0), it implies that the coordinates of L are (3,P(3)).
A(x)=21x4+213x3+6x2−63x+270
▼
Substitute 3 for x and simplify
Substitute
x=3
A(3)=21(3)4+213(3)3+6(3)2−63(3)+270
CalcPowProd
Calculate power and product
A(3)=281+2351+54−189+270
AddFrac
Add fractions
A(3)=281+351+54−189+270
AddSubTerms
Add and subtract terms
A(3)=2432+135
CalcQuot
Calculate quotient
A(3)=216+135
AddTerms
Add terms
A(3)=351
c The triangles generated by K1 and K2 look as follows.
A(x)=21x4+213x3+6x2−63x+270
▼
Substitute -4 for x and simplify
Substitute
x=-4
A(-4)=21(-4)4+213(-4)3+6(-4)2−63(-4)+270
CalcPowProd
Calculate power and product
A(-4)=2256−2832+96+252+270
CalcQuot
Calculate quotient
A(-4)=128−416+96+252+270
AddSubTerms
Add and subtract terms
A(-4)=330
A(x)=21x4+213x3+6x2−63x+270
▼
Substitute -1 for x and simplify
Substitute
x=-1
A(-1)=21(-1)4+213(-1)3+6(-1)2−63(-1)+270
CalcPowProd
Calculate power and product
A(-1)=21−213+6+63+270
AddFrac
Add fractions
A(-1)=21−13+6+63+270
SubTerms
Subtract terms
A(-1)=-212+6+63+270
CalcQuot
Calculate quotient
A(-1)=-6+6+63+270
AddSubTerms
Add and subtract terms
A(-1)=333
Example
Determining a Toy's Volume
Izabella bought her nephew a bag of magic grow toys for his birthday. Among the toys, there was a trailer and its rectangular container. Initially, the container was 5 centimeters long, 2 centimeters wide, and 3 centimeters high.
ℓ(t)w(t)h(t)=41t2+5=-91t2+2t+2=t+3
Here, t represents the number of hours the toy has been underwater. a Write a polynomial V(t), in standard form, that represents the volume of the container after being underwater for t hours.
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b Izabella's nephew placed the toy underwater at 10:00 AM and took it out at 4:00 PM. What is the volume of the container now?
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Hint
a The volume of a rectangular container is obtained by multiplying its three dimensions.
b How many hours was the container underwater? Evaluate V(t) at this value.
Solution
a The volume of a rectangular container is found by multiplying its three dimensions.
V=w⋅ℓ⋅h
Therefore, to find a polynomial representing the container's volume after passing t hours underwater, multiply the expressions that model the dimensions of the container.
w(t)⋅ℓ(t)⋅h(t)⇓(41t2+5)(-91t2+2t+2)(t+3)
By applying the Commutative Property of Multiplication, the previous product can be rewritten to have the two binomials next to each other.
(-91t2+2t+2)(41t2+5)(t+3)
Then, these two binomials can be multiplied using the FOIL method. For simplicity, the trinomial will not be written during this multiplication. - Start by multiplying the first terms.(41t2+5)(t+3)=41t2⋅t+⋯
- Continue by multiplying the outer terms.(41t2+5)(t+3)=41t2⋅t+41t2⋅3+⋯
- Next, multiply the inner terms.(41t2+5)(t+3)=41t2⋅t+41t2⋅3+5⋅t+⋯
- Now, multiply the last terms.(41t2+5)(t+3)=41t2⋅t+41t2⋅3+5⋅t+5⋅3
- Finally, simplify the resulting expression by performing the required multiplications.(41t2+5)(t+3)=41t3+43t2+5t+15
V(t)=(-91t2+2t+2)(41t3+43t2+5t+15)
▼
Multiply (-91t2+2t+2) by (41t3+43t2+5t+15)
Distr
Distribute -91t2+2t+2
V(t)=(-91t2+2t+2)41t3+(-91t2+2t+2)43t2+(-91t2+2t+2)5t+(-91t2+2t+2)15
Distr
Distribute 41t3,43t2,5t,&15
V(t)=(-91t2⋅41t3+2t⋅41t3+2⋅41t3)+(-91t2⋅43t2+2t⋅43t2+2⋅43t2)+(-91t2⋅5t+2t⋅5t+2⋅5t)+(-91t2⋅15+2t⋅15+2⋅15)
CommutativePropMult
Commutative Property of Multiplication
V(t)=(-91⋅41t2⋅t3+2⋅41t⋅t3+2⋅41t3)+(-91⋅43t2⋅t2+2⋅43t⋅t2+2⋅43t2)+(-91⋅5t2⋅t+2⋅5t⋅t+2⋅5t)+(-91⋅15t2+2⋅15t+2⋅15)
▼
Simplify right-hand side
MultPow
am⋅an=am+n
V(t)=(-91⋅41t5+2⋅41t4+2⋅41t3)+(-91⋅43t4+2⋅43t3+2⋅43t2)+(-91⋅5t3+2⋅5t2+2⋅5t)+(-91⋅15t2+2⋅15t+2⋅15)
MultFrac
Multiply fractions
V(t)=(-361t5+42t4+42t3)+(-363t4+46t3+46t2)+(-95t3+2⋅5t2+2⋅5t)+(-915t2+2⋅15t+2⋅15)
SimpQuotProd
Simplify quotient and product
V(t)=(-361t5+21t4+21t3)+(-121t4+23t3+23t2)+(-95t3+10t2+10t)+(-35t2+30t+30)
CommutativePropAdd
Commutative Property of Addition
V(t)=-361t5+(21t4−121t4)+(21t3−95t3+23t3)+(23t2−35t2+10t2)+(10t+30t)+30
AddSubTerms
Add and subtract terms
V(t)=-361t5+125t4+913t3+659t2+40t+30
b The polynomial V(t) found in Part A gives the volume of the container after passing t hours underwater. It is given that Izabella's nephew put the container underwater at 10:00 AM and took it out at 4:00 PM. Therefore, the container was underwater for 6 hours.
6hours10:00 AM⟶4:00 PM
Consequently, to find the current volume of the container, evaluate V(t) at t=6.
V(t)=-361t5+125t4+913t3+659t2+40t+30
▼
Substitute 6 for t and evaluate
Substitute
t=6
V(6)=-361(6)5+125(6)4+913(6)3+659(6)2+40(6)+30
CalcPow
Calculate power
V(6)=-361⋅7776+125⋅1296+913⋅216+659⋅36+40(6)+30
MoveRightFacToNum
ca⋅b=ca⋅b
V(6)=-367776+125⋅1296+913⋅216+659⋅36+40(6)+30
Multiply
Multiply
V(6)=-367776+126480+92808+62124+240+30
CalcQuot
Calculate quotient
V(6)=-216+540+312+354+240+30
AddSubTerms
Add and subtract terms
V(6)=1260
Discussion
Analyzing the Product of Two Polynomials
In all the examples seen throughout this lesson, the product of two or more polynomials has resulted in a new polynomial. This is not a coincidence. In fact, the following property guarantees that multiplying polynomials always produces a polynomial.
Rule
Closure Property of Polynomial Multiplication
Given two polynomials P(x) and Q(x), the product P(x)⋅Q(x) is always a polynomial.
Multiplying two polynomials produces a new polynomial.
In other words, the polynomials are closed under multiplication.
Proof
Consider two arbitrary polynomials P(x) and Q(x) written in standard form.
Since P(x) and Q(x) are polynomials, all the exponents are whole numbers. Furthermore, because the whole numbers are closed under addition, the exponents of the resulting expression are whole numbers. Then, the new expression can be rewritten as follows.
P(x)Q(x)=anxd1+⋯+a1x+a0=bnxd2+⋯+b1x+b0
These two polynomials can be multiplied by using the Distributive Property. The Product of Powers Property can also be applied to simplify the resulting expression.
P(x)Q(x)=ckxD+⋯+c1x+c0
Consequently, the new expression is a polynomial. Therefore, the product of two polynomials produces a polynomial, which proves that the polynomials are closed under multiplication.
Pop Quiz
Practicing Multiplication of Polynomials
Given two polynomials P(x) and Q(x), compute their product and find the required information.
Closure
Computing a Polynomials Product
Before finishing this lesson, take another look at the challenge presented at the beginning. There it was given the following pair of polynomials.
Having both polynomials written in standard form, the required product can be found by using the Distributive Property.
As expected, the resulting expression is a polynomial. The degree of the resulting polynomial is 5, its leading coefficient is 8, and its constant term is -40.
P(x)Q(x)=2x3−x+5=4(x+2)(x−1)
In the challenge, it was asked whether it was possible to find the product of P(x) and Q(x). Now, with all the knowledge learned, it can be said that it is possible to multiply these two polynomials, and the resulting expression will also be a polynomial. To compute P(x)⋅Q(x), first, rewrite Q(x) in standard form. To do so, the FOIL Method can be applied.
Q(x)=4(x+2)(x−1)
MultII
Multiply (x+2) by (x−1)
Q(x)=4(x(x)+x(-1)+2x+2(-1))
▼
Simplify right-hand side
MultPow
am⋅an=am+n
Q(x)=4(x2+x(-1)+2x+2(-1))
MultPosNeg
a(-b)=-a⋅b
Q(x)=4(x2−x⋅1+2x−2⋅1)
MultByOne
a⋅1=a
Q(x)=4(x2−x+2x−2)
AddTerms
Add terms
Q(x)=4(x2+x−2)
Distr
Distribute 4
Q(x)=4x2+4x−8
P(x)⋅Q(x)
SubstituteExpressions
Substitute expressions
(2x3−x+5)(4x2+4x−8)
(2x3⋅4x2−x⋅4x2+5⋅4x2)+(2x3⋅4x−x⋅4x+5⋅4x)+(2x3(-8)−x(-8)+5(-8))
▼
Simplify
MultPow
am⋅an=am+n
(2⋅4x5−4x3+5⋅4x2)+(2⋅4x4−4x2+5⋅4x)+(2x3(-8)−x(-8)+5(-8))
Multiply
Multiply
(8x5−4x3+20x2)+(8x4−4x2+20x)+(-16x3+8x−40)
CommutativePropAdd
Commutative Property of Addition
8x5+8x4+(-4x3−16x3)+(20x2−4x2)+(20x+8x)−40
AddSubTerms
Add and subtract terms
8x5+8x4−20x3+16x2+28x−40
Tadeo invited his friends to watch the Champions League final next Sunday. He plans to meet in the backyard and wants to project the game on a wall of the house using a video beam. However, on Saturday night, Tadeo made a test and the image did not look so clear, so he decided to buy a projector screen.
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Because the projector screen has a rectangular shape, its area equals its length multiplied by its width. A = l * w From the diagram, we can see that the projector screen is 1.25 feet shorter and narrower from each side of the wall. The wall is 12 feet tall, so we can find the height of the screen by subtracting 1.25 feet from each side. 12-1.25-1.25=9.5 We can find the width of the screen in a similar way. x-1.25-1.25=x-2.5 Let's look at these dimensions in the diagram!
A polynomial modeling the area of the projector screen is obtained by substituting its dimensions into the formula for the area. P(x) = (x-2.5)(9.5) ⇕ P(x) = 9.5x-23.75 Note that this polynomial gives the area of the projector screen depending on the width of the wall. The polynomial can also be written using fractions instead of decimal numbers. To do this, we expand the corresponding terms.
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Use any multiplication method to find the following products. Write each result in standard form.
(x−4)(x+2)
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(y−3)(y+3)
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(z+2)(1−z)
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Let's find the required product using the Distributive Property. To start, we distribute the left-hand side binomial to each term of the right-hand side binomial. ( x-4)( x+2) ⇓ ( x-4) x+( x-4) 2 Next, we apply the Distributive Property once more to clear the parentheses and continue combining like terms to simplify the resulting expression.
To find the second product, we will use the FOIL Method. Remember, the word FOIL stands for First, Outer, Inner, and Last.
With this in mind, we can proceed with the required multiplication. Let's start by multiplying the first terms. ( y-3)( y+3) ⇓ y* y The empty box is there as a reminder that there are still missing terms. Next, we multiply the outer terms. ( y-3)(y+ 3) ⇓ y* y + y* 3 Now, it is the turn of the inner terms. (y- 3)( y+3) ⇓ y* y + y* 3 - 3* y Then, we multiply the last terms. (y- 3)(y+ 3) ⇓ y* y + y* 3 - 3* y - 3* 3 Finally, we perform each of the products and combine like terms to simplify the resulting polynomial.
For the third product, let's use the Box Method. First, we determine the dimensions of the table. To do this, let's count the number of terms in each polynomial. ccc Polynomial & & Number of Terms [0.1cm] (z+2) & & 2 [0.1cm] (1-z) & & 2 Since each polynomial has two terms, we will need a table with two rows and two columns. However, we will add an extra row and column to write the terms of each polynomial. Let's write the terms of the first polynomial in the first column and the terms of the second polynomial in the first row.
| * | 1 | - z |
|---|---|---|
| z | ||
| 2 |
Next, we fill in the table's cells by multiplying the terms written on the corresponding borders of the table.
| * | 1 | - z |
|---|---|---|
| z | z(1) | z(- z) |
| 2 | 2(1) | 2(- z) |
Finally, let's add all the expressions inside the table and combine like terms.
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Write a polynomial, in standard form, that represents the area of the following shaded regions.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["g"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["18g^2 + \\dfrac{1}{2}g - 1"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["z"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.04398em;\">z<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["47z^2 + 10z - 1"]}}
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To start, let's write the formula for finding the area of a rectangle. A = l * w In our case, the rectangle has dimensions 3x+1 and x-4. Let's substitute them into the formula. A(x) = (3x+1)(x-4) Finally, we write the previous polynomial in standard form by multiplying the two binomials.
The last polynomial represents the area of the given rectangle.
We are interested in finding the area of the given triangle. Therefore, let's begin by writing the formula that we will use.
A = 1/2bh
Since the given triangle is a right triangle, we can use the legs to find the area — that is, b=4g+1 and h=9g-2. Let's substitute these expressions into the formula.
A(g) = 1/2(4g+1)(9g-2)
To write the polynomial A(g) in standard form, we can multiply the two binomials using the FOIL Method.
Note that the area of the shaded region is the area of the rectangle minus the area of the square. A_(Shaded) = A_R - A_S Let's start by finding the area of the rectangle. To do this, we multiply its dimensions.
Next, let's find the area of the square.
Finally, let's subtract the square's area from the rectangle's area.
The obtained polynomial represents the area of the given shaded region.
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The surface area of a three-dimensional solid is the sum of the areas of all its faces. Knowing this, write a polynomial representing the surface area of the following solids.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["p"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">p<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["102p^2 - 28p - 10"]}}
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Let's recall the formula for the surface area of a cube. SA =6l ^2 Here l is the side length of the cube. In our case l= 2k+1. Let's substitute this expression into the formula. A(k)=6( 2k+1)^2 Now, let's simplify the obtained expression and write it in standard form.
We can multiply the binomials using the FOIL Method. Recall that the word FOIL stands for First, Outer, Inner, and Last.
Finally, we combine like terms to simplify the polynomial.
The surface area of the given cube is represented by A(k)=24k^2 + 24k + 6.
Let's disassemble the given rectangular prism to see the faces individually.
With this in mind, we can write an expression representing the surface area of the prism. SA = 2A_1 + 2A_2 + 2A_3 ⇓ SA = 2(A_1 + A_2 + A_3) Let's start by finding the area of the bases, whose dimensions are 2p and 3p+1.
In a similar way, let's find the area of the lateral face with dimensions 3p+1 and 9p-5.
Next, let's find the area of the lateral face with dimensions 2p and 9p-5.
Now we are ready to find the polynomial representing the surface area of the rectangular prism. To do this, we substitute the expressions we found for A_1, A_2, and A_3.
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Last class, Paulina learned how to use the box method to multiply polynomials. However, while trying to take notes, her teacher erased the board and she could not finish writing all the calculations. Paulina missed two expressions inside the table and the final result!
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mord mathdefault\">Q<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["2x^4-4x^3+15x^2-6x+18"]}}
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According to the box method, the top row labels are the terms of one of the polynomials being multiplied. Similarly, the elements at the left of the table form the other polynomial. Let's start by writing the polynomials being multiplied. P(x) &= x^2-2x+6 Q(x) &= 2x^2+3 Now, the cells of the table are filled by multiplying the terms written on the corresponding borders of the table. For example, the top-left cell corresponds to the product of 2x^2 and x^2.
In a similar fashion, the second cell of the second row corresponds to the product of 3 and -2x.
Knowing this, we can find A and B. A &= 2x^2* x^2 = 2x^4 B &= 3(-2x) = -6x Finally, we add the two previous expressions. A+B = 2x^4 - 6x
Now that we know the expressions for A and B, let's begin by writing the in the table.
The result of the polynomial multiplication is the sum of all the expressions inside the table. Therefore, let's add them and combine like terms, if any.
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P(x)Q(x)=2x3−5=4x6+2
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["14"]}}
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The degree of a polynomial is the highest exponent in the polynomial. Therefore, let's start by finding R(x).
From the resulting polynomial, we can see that the highest exponent is 9. Therefore, the degree of R(x) is 9.
The leading coefficient of a polynomial is the coefficient in front of the term with the highest exponent. Knowing this, let's highlight the leading coefficients of P(x), Q(x), and R(x).
P(x) &= 2x^3-5
Q(x) &= 4x^6+2
R(x) &= 8x^9 - 20x^6 + 4x^3 - 10
The leading coefficients of P(x), Q(x), and R(x) are 2, 4, and 8, respectively. Let's find the required sum using this information!
2 + 4 + 8 = 14
The constant term of a polynomial is the term with no variables including its sign.
R(x) = 8x^9 - 20x^6 + 4x^3 - 10
Therefore, the constant term of R(x) is -10.
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Multiplying Polynomials
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