3. The Binomial Theorem and Other Polynomial Identities
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Chapter 2
3.
The Binomial Theorem and Other Polynomial Identities
This lesson explores the Binomial Theorem, a mathematical principle that allows for the expansion of expressions raised to a power. It introduces Pascal's Triangle as a tool for finding the coefficients in these expansions. The focus is not just on theory but also on practical applications. For example, understanding the Binomial Theorem can simplify complex calculations in fields like engineering, physics, and data science. The lesson also delves into polynomial expansion, explaining how coefficients play a crucial role in these mathematical expressions. Overall, the information aims to provide a thorough understanding of these topics, making complex calculations more approachable.
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Lesson Settings & Tools
| | 9 Theory slides |
| | 15 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The Binomial Theorem and Other Polynomial Identities
Slide of 9
Expanding binomials can be difficult if the exponent is large. Considering that there is a formula for the square of a binomial, it might be possible to find similar expressions for greater exponents. This lesson aims to show an array of numbers that can be used to expand binomials.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Recognizing Patterns
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Example
Difference of Squares
Two good friends, Diego and Maya, like to challenge each other on math related topics. One summer afternoon, Diego came up with an interesting problem.
Help Maya prove that the difference of the squares of any two consecutive integers is odd.
Answer
See solution.
Hint
Let n be an arbitrary integer. Find the difference of the squares of n and its consecutive integer.
Remember that the division of an even number by two always result on an integer.
Solution
To prove that the difference of the squares of any two consecutive integers is odd, let n be an integer number. The second of the consecutive integers is obtained by adding 1 to n. Therefore, the consecutive integer to n is n+1. Each number will be squared.
cc
Numbers & Squares of the Numbers n andn+1 & n^2 and (n+1)^2
Next, the difference of the squared numbers can be calculated. For simplicity the smaller integer is subtracted from the larger, but it should be noted that either way works.
(n+1)^2 - n^2
▼
Simplify
ExpandPosPerfectSquare
(a+b)^2=a^2+2ab+b^2
n^2 + 2n(1) + 1^2 - n^2
IdPropMult
Identity Property of Multiplication
n^2 + 2n + 1^2 - n^2
BaseOne
1^a=1
n^2 + 2n + 1 - n^2
SubTerm
Subtract term
2n + 1
To determine whether this difference is an odd number, the expression 2n+1 can be divided by 2. If the result is not an integer, the difference is odd.
2n+1/2
▼
Simplify
WriteSumFrac
Write as a sum of fractions
2n/2 + 1/2
MovePartNumRight
a* b/c=a/c* b
2/2 n+ 1/2
QuotOne
a/a=1
1n+ 1/2
IdPropMult
Identity Property of Multiplication
n + 1/2
Since n is an integer number and 12 is not, the number n+ 12 is not an integer. This means that the number 2n+1 is odd for any integer n. Therefore, the difference of the squares of any two consecutive integers odd.
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Example
Difference of Cubes
After solving Diego's challenge, Maya now creates one for Diego. She asks him to prove that the difference of the cubes of any two consecutive integers is also an odd number.
Help Diego prove this statement!
Answer
See solution.
Hint
To prove this statement, consider an even number 2n, where n is any integer. Its consecutive number is 2n+1 and its previous number is 2n-1. Consider the difference of the cubes of 2n+1 and n, and the difference of n and 2n-1.
Solution
Any even number can be written as 2n, where n is an integer. Consider the cube of this number.
(2n)^3 = 8n^3
Now, when considering two consecutive integers, one is even and the other is odd. The odd number can be before or after 2n. These cases can be expressed by subtracting 1 from or adding 1 to 2n, respectively. Number Beforen: & 2n -1 Number Aftern: & 2n + 1
Recall the formula for the cube of a binomial.
(a± b)^3 = a^3± 3a^2b+3ab^2± b^3
By using this formula, the cubes of 2n-1 and 2n+1 can be found. These formulas are obtained by substituting 2n for a and 1 for b.
(a± b)^3 = a^3± 3a^2b+3ab^2± b^3
SubstituteII
a= 2n, b= 1
( 2n± 1)^3 = ( 2n)^3± 3( 2n)^2( 1)+3( 2n)( 1)^2± ( 1)^3
(2n± 1)^3 = 8n^3± 12n^2+6n± 1
There are two possible situations. The consecutive numbers can be 2n and 2n+1, or 2n-1 and 2n. These two situations will be considered one at a time.
2n and 2n-1
If the consecutive integers are 2n and 2n-1, then their cubes are 8n^3 and 8n^3-12n^2+6n-1, respectively. Next, find their difference. For the result to be positive, the cube of 2n-1 is subtracted from the cube of 2n.
8n^3 - (8n^3 -12n^2 + 6n - 1)
12n^2 - 6n + 1
Now, divide this expression by 2. An even number is always divisible by two. Therefore, if dividing the above expression by 2 results in an integer, the expression represents an even number.
12n^2 - 6n + 1/2
▼
Simplify
WriteSumFrac
Write as a sum of fractions
12n^2/2 - 6n/2 + 1/2
MovePartNumRight
a* b/c=a/c* b
12/2n^2 - 6/2 n+ 1/2
CalcQuot
Calculate quotient
6n^2 - 3n + 1/2
This expression will be examined closely to determine if it is an integer or not. Each term will be examined individually, gradually building the expression of 12n^2 - 6n + 12.
| Expression | Is it an Integer? | Reason |
|---|---|---|
| n | Yes | It is defined this way. |
| n^2 | Yes | Closure Property of Multiplication |
| 6n^2 | Yes | Closure Property of Multiplication |
| 3n | Yes | Closure Property of Multiplication |
| 6n^2-3n | Yes | Closure Property of Subtraction |
| 6n^2-3n+1/2 | No | 12 is not an integer |
Since dividing 12n^2-6n+1 by 2 did not result in an integer number, the expression 12n^2-6n+1, which is the difference of the cubes of 2n and 2n-1, is odd.
2n+1 and 2n
If the consecutive integers are 2n+1 and 2n, then their cubes are 8n^3+12n^2+6n+1 and 8n^3, respectively. Their difference can be calculated.
(8n^3 + 12n^2 + 6n + 1) - 8n^3
12n^2 + 6n + 1
Just like in the previous case, the expression that represents the difference will be divided by 2 to determine if the result is even.
12n^2 + 6n + 1/2
▼
Simplify
WriteSumFrac
Write as a sum of fractions
12n^2/2 + 6n/2 + 1/2
MovePartNumRight
a* b/c=a/c* b
12/2n^2 + 6/2 n+ 1/2
CalcQuot
Calculate quotient
6n^2 + 3n + 1/2
Similar to the previous case, 6n^2+3n is an integer. Because of the 12, the sum 6n^2+3n+ 12 is not an integer. Therefore, the difference of the cubes is odd. Now that the two cases are considered, the statement has been proved.
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Example
Square of the Sum of Two Consecutive Integers
Diego, engulfed by Maya's last challenge, asks if she could create another one for him. This time, Maya asks Diego to prove that the square of the sum of two consecutive integers is odd.
Help Diego solve Maya's challenge!
Answer
See solution.
Hint
Let n and n+1 be two consecutive integers. Find their sum, calculate its square, and prove that the obtained result is odd.
Remember that the division of an even number by two always result on an integer.
Solution
Let n and n+1 be two consecutive integer numbers. To prove the given statement, the numbers first must be added.
n + (n + 1) = 2n + 1
The sum resulted in a binomial. Then, the square of the sum is the square of a binomial.
(2n + 1)^2
ExpandPosPerfectSquare
(a+b)^2=a^2+2ab+b^2
(2n)^2 + 2(2n)(1) + 1^2
ProdPowII
a^m b^m = (a b)^m
4n^2 + 2(2n)(1) + 1^2
BaseOne
1^a=1
4n^2 + 2(2n)(1) + 1
Multiply
Multiply
4n^2 + 4n + 1
An even number is always divisible by 2. This means that, if dividing the above expression by 2 results in an integer, the expression represents an even number. Otherwise, it represents an odd number.
4n^2 + 4n + 1/2
WriteSumFrac
Write as a sum of fractions
4n^2/2 + 4n/2 + 1/2
MovePartNumRight
a* b/c=a/c* b
4/2n^2 + 4/2 n+ 1/2
CalcQuot
Calculate quotient
2n^2 + 2n +1/2
Since n is an integer, the expression 2n^2+2n also represents an integer. Adding a non-integer to an integer results in a non-integer. Therefore, the expression 2n^2+2n+ 12 is not an integer. This means that 4n^2+4n+1 is an odd number, proving the statement.
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Discussion
Pascal's Triangle and the Binomial Theorem
Pascal's triangle is a triangular representation of the coefficients in the expansion of a binomial expression. In other words, it contains the coefficients that result when expanding the expression (a+b)^n, with n=0,1,2,...
The rows are numbered from top to bottom starting with 0. The number in each cell equals the sum of the numbers in the two neighboring cells above.
The numbers in the Pascal's Triangle have an interesting application. Consider the following binomial expansions.
Looking at the coefficients of the binomial expansions, it can be noted that every number of a row in the Pascal's Triangle is the same as the coefficients of a binomial expansion.
This is expressed in the Binomial Theorem.
Binomial Theorem
Consider a binomial raised to the power of n, where n is a positive integer. Let a and b be real numbers. Expanding the binomial (a+b)^n results in the following expression.
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Example
The Volume of a Room
Maya, feeling like she challenged Diego well, remembers that her mom asked her to measure her room for new wallpaper. She has to take a phone call, so she asks Diego if he can take the measurements. The room is in the shape of a cube. Diego hands over the measurements to her — he wrote them as a challenge!
Diego wrote that each side of the room has a length of (x-3)^2 . Write the expanded polynomial for the volume of the room.
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Hint
Use the Binomial Theorem to expand the binomial.
Solution
The room has the shape of a room with side length (x-3)^2. To find its volume, it is necessary to use the formula for the volume of a cube.
V = s^3
In this formula, V is the volume of the cube and s its side length. Substituting (x-3)^2 for s gives the formula for the volume of the room.
Instead of multiplying the binomial by itself 6 times, the Binomial Theorem can be used to expand the binomial. This theorem indicates that the coefficients of a binomial to the sixth power coincide with the numbers in the sixth row of Pascal's Triangle.
Each number in the sixth row corresponds to a coefficient of the binomial expansion, considering that the result is written in standard form.
V = 1x^6 + 6x^5(-3) + 15x^4(-3)^2 + 20x^3(-3)^3 + 15x^2(-3)^4 + 6x(-3)^5 + 1(-3)^6
▼
Simplify right-hand side
IdPropMult
Identity Property of Multiplication
V = x^6 + 6x^5(-3) + 15x^4(-3)^2 + 20x^3(-3)^3 + 15x^2(-3)^4 + 6x(-3)^5 + (-3)^6
CalcPow
Calculate power
V = x^6 + 6x^5(-3) + 15x^4(9) + 20x^3(-27) + 15x^2(81) + 6x(-243) + (729)
Multiply
Multiply
V=x^6-18x^5+135x^4-540x^3+1215x^2-1458x+729
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Example
Finding a Certain Coefficient
It's Diego's birthday, and without fail, Maya has gotten him a cool present. Diego is so excited to open it, but there is a problem. The gift has a lock that asks for a combination that he does not have!
Maya, of course, wants to have a bit of fun and tells Diego that the combination is the fifth term in the expansion (f+w)^9. Help Diego open his gift!
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Hint
Consider the Binomial Theorem.
Solution
The Binomial Theorem indicates that the coefficients in the expansion of (f+w)^9 coincide with the numbers in the ninth row of Pascal's Triangle.
The fifth term in the expansion of the binomial raised to the ninth power matches the fifth number in the ninth row of Pascal's Triangle.
The terms of the polynomial are counted starting from the term f^9w^0, this being the first one. The variables for each consecutive term are obtained by subtracting one from the exponent of f^9 and adding one to the exponent of w. The coefficients are the same as the numbers in the ninth row of Pascal's Triangle.
| nth term | |||
|---|---|---|---|
| Number of Term | Coefficient | Variables | Value of Term |
| 1 | 1 | f^9w^0 | f^9 |
| 2 | 9 | f^8w^1 | 9f^8w |
| 3 | 36 | f^7w^2 | 36f^7w^2 |
| 4 | 84 | f^6w^3 | 84f^6w^3 |
| 5 | 126 | f^5w^4 | 126f^5w^4 |
| 6 | 126 | f^4w^5 | 126f^4w^5 |
| 7 | 84 | f^3w^6 | 84f^3w^6 |
| 8 | 36 | f^2w^7 | 36f^2w^7 |
| 9 | 9 | f^1w^8 | 9fw^8 |
| 10 | 1 | f^0w^9 | w^9 |
Therefore, the fifth term is 126f^5w^4.
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Closure
Using the Binomial Theorem to Solve a Problem
To finish this lesson, a more complex exercise will be solved using the Binomial Theorem. Consider the following expression. (2a+3x)^6 It is known that the coefficient in the x^3-term in the expansion of the above expression is the same as the coefficient in the x^5-term. Find the possible values of a. Write the exact answer in its simplest form, with no radicals in denominators.
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Hint
Use the Binomial Theorem to find the coefficients of every term.
Solution
The Binomial Theorem indicates that the coefficients in the expansion of (2a+3x)^6 are the numbers in the sixth row of Pascal's Triangle.
Now the terms for x^5 and x^3 have to be identified. To identify these terms, every term of the binomial expansion is determined using the Binomial Theorem.
| Number of Term | Coefficient | Variables | Value of Term |
|---|---|---|---|
| 1 | 1 | (2a)^6(3x)^0 | 64a^6 |
| 2 | 6 | (2a)^5(3x)^1 | 576a^5x |
| 3 | 15 | (2a)^4(3x)^2 | 2160a^4x^2 |
| 4 | 20 | (2a)^3(3x)^3 | 4320a^3x^3 |
| 5 | 15 | (2a)^2(3x)^4 | 4860a^2x^4 |
| 6 | 6 | (2a)^1(3x)^5 | 2916ax^5 |
| 7 | 1 | (2a)^0(3x)^6 | 729x^6 |
It is given that the coefficients of the x^5- and the x^3-terms are the same. Therefore, 2916a must be equal to 4320a^3. This can be written as an equation which can be solved for a.
2916a=4320a^3
▼
Solve for a
DivEqn
.LHS /108a.=.RHS /108a.
2916a/108a=4320a^3/108a
WriteProdFrac
Write as a product of fractions
2916/108* a/a=4320/108* a^3/a
CalcQuot
Calculate quotient
27* a/a=40* a^3/a
QuotOne
a/a=1
27(1)=40* a^3/a
IdPropMult
Identity Property of Multiplication
27=40* a^3/a
a^m/a=a^(m-1)
27=40a^2
DivEqn
.LHS /40.=.RHS /40.
27/40=a^2
RearrangeEqn
Rearrange equation
a^2 = 27/40
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(a^2) = sqrt(27/40)
sqrt(a^2)=± a
a = ±sqrt(27/40)
▼
Simplify right-hand side
SplitIntoFactors
Split into factors
a = ±sqrt(9(3)/4(10))
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
a = ± sqrt(9(3))/sqrt(4(10))
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
a = ± 3sqrt(3)/2sqrt(10)
ExpandFrac
a/b=a * sqrt(10)/b * sqrt(10)
a = ± 3sqrt(3)*sqrt(10)/2sqrt(10)*sqrt(10)
ProdSqrt
sqrt(a)*sqrt(b)=sqrt(a* b)
a = ± 3sqrt(30)/2sqrt(100)
CalcRoot
Calculate root
a = ± 3sqrt(30)/2(10)
Multiply
Multiply
a = ± 3sqrt(30)/20
Therefore, the possible values for a are 3sqrt(30)20 and - 3sqrt(30)20.
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The Binomial Theorem and Other Polynomial Identities
Exercise 2.1
Consider the following expression. (2x + 4a)^5 The coefficient of the x^4-term in the expansion of the expression is the same as the coefficient of the x^2-term. Find the possible values of a that are different from zero. Write the exact answer in its simplest form, with no radicals in denominators.
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The Binomial Theorem indicates that the coefficients in the expansion of (2x + 4a)^5 are associated with the numbers in the fifth row of Pascal's Triangle.
Now the x^4- and x^2-terms have to be identified. To do so, every term of the binomial expansion is determined using the Binomial Theorem.
| Number of term | Coefficient | Variables | Value of term |
|---|---|---|---|
| 1 | 1 | (2x)^5(4a)^0 | 32x^5 |
| 2 | 5 | (2x)^4(4a)^1 | 320x^4a |
| 3 | 10 | (2x)^3(4a)^2 | 1280x^3a^2 |
| 4 | 10 | (2x)^2(4a)^3 | 2560x^2a^3 |
| 5 | 5 | (2x)^1(4a)^4 | 2560x a^4 |
| 6 | 1 | (2x)^0(4a)^5 | 1024a^5 |
We are told that the coefficients of the x^4- and the x^2-terms are the same, so 320a must be equal to 2560a^3. This can be written as an equation which can be solved for a. Since a is different from zero, we can divide both sides of the equation by a.
We have found the possible values for a, but we are told that the denominator should not include any radicals. Let's rewrite the value of a to eliminate the radicals!
Therefore, the possible values for a are sqrt(2)4 and - sqrt(2)4.
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Solution
Consider the following expression. (3a + 2x)^6 The coefficient of the x^5-term in the expansion of the expression is the same as the coefficient of the x^3-term. Find the possible values of a that are different from zero. Write the exact answer in its simplest form, with no radicals in denominators.
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The Binomial Theorem indicates that the coefficients in the expansion of (3a + 2x)^6 are associated with the numbers in the sixth row of Pascal's Triangle.
We have to identify the terms for x^5 and x^3. To identify these terms, first we will determine every term of the binomial expansion the Binomial Theorem.
| Number of Term | Coefficient | Variables | Value of Term |
|---|---|---|---|
| 1 | 1 | (3a)^6(2x)^0 | 729a^6 |
| 2 | 6 | (3a)^5(2x)^1 | 2916a^5x |
| 3 | 15 | (3a)^4(2x)^2 | 4860a^4x^2 |
| 4 | 20 | (3a)^3(2x)^3 | 4320a^3x^3 |
| 5 | 15 | (3a)^2(2x)^4 | 2160a^2x^4 |
| 6 | 6 | (3a)^1(2x)^5 | 576ax^5 |
| 7 | 1 | (3a)^0(2x)^6 | 64x^6 |
We are told that the coefficients of the x^5-term and the x^3-term are the same. Therefore, 576a must be equal to 4320a^3. We can write this as an equation that can be solved for a. Since a is different from zero, we can divide both sides of the equation by a.
We have found the possible values for a, but we are told that the denominator should not include any radicals. Let's rewrite the value of a to eliminate the radicals!
Therefore, the possible values for a are sqrt(30)15 and - sqrt(30)15.
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Solution
Dylan expanded (4x - 7)^4 as shown below.
Which of the following best describes Dylan's mistake?
{"type":"choice","form":{"alts":["He did not multiply the terms correctly.","He wrote the wrong Pascal's Triangle numbers.","He ignored the negative sign."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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To check for Dylan's mistake, let's first do the expansion. To expand the given binomial, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle. cc (a+b)^n= & P_0a^nb^0+P_1a^(n-1)b^1 & + & ... & + & P_(n-1)a^1b^(n-1)+P_na^0b^n In the above formula, P_0, P_1, ..., P_n are the numbers in the n^(th) row of Pascal's triangle. Since we are raising our binomial to the 4^\text{th} degree, we will look at the fourth row of the triangle.
Note that the first and last number in each row is one. Any other number found in the row is the sum of the two numbers diagonally above it. Now, consider the given binomial. ( 4x+( -7) )^4 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and the coefficients from Pascal's Triangle.
| (a+b)^n=P_0a^nb^0+P_1a^(n-1)b^1+... +P_(n-1)a^1b^(n-1)+P_na^0b^n |
|---|
| ( 4x+( -7) )^4 = 1( 4x)^4( -7)^0 + 4( 4x)^3( -7)^1 + 6( 4x)^2( -7)^2 + 4( 4x)^1( -7)^3 + 1( 4x)^0( -7)^4 |
Finally, let's simplify the expression.
Now we completed the binomial expansion. Comparing our result with Dylan's, we can see that every coefficient is different, except for the last one.
| Dylan's Result | x^4 - 28 x^3 + 294x^2 - 1372x + 2401 |
|---|---|
| Our Result | 256x^4 - 1792x^3 + 4704x^2 - 5488x + 2401 |
We can see that most of the coefficients of our expansion are greater than the coefficients of Dylan's expansion. Let's divide each pair of coefficients to see if we can find anything.
| Term 1 | Term 2 | Term 3 | Term 4 | Term 5 | |
|---|---|---|---|---|---|
| Our Expansion | 256x^4 | -1792x^3 | 4704x^2 | - 5488x | 2401 |
| Dylan Expansion | x^4 | -28x^3 | 294x^2 | - 1372x | 2401 |
| Division | 256 | 64 | 16 | 4 | 1 |
| Rewrite | 4^4 | 4^3 | 4^2 | 4^1 | 4^0 |
We can see that each of Dylan's terms can be corrected by multiplying it by a certain power of four. It seems like Dylan ignored the fact that the x-term in (4x - 7)^4 is multiplied by 4. Therefore, Dylan's mistake is that he did not multiply the terms correctly.
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Solution
Dylan expanded (- x + 3)^6 as shown below.
Which of the following best describes Dylan's mistake?
{"type":"choice","form":{"alts":["He did not multiply the terms correctly.","He wrote the wrong Pascal's Triangle numbers.","He ignored the negative sign."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
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To check for Dylan's mistake, let's first do the expansion. To expand the given binomial, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle. cc (a+b)^n= & P_0a^nb^0+P_1a^(n-1)b^1 & + & ... & + & P_(n-1)a^1b^(n-1)+P_na^0b^n In the above formula, P_0, P_1, ..., P_n are the numbers in the n^(th) row of Pascal's triangle. Since we are raising the binomial to the 6^\text{th} power, we will look at the sixth row of the triangle.
Note that the first and last number in each row is one. Any other number found in the row is the sum of the two numbers diagonally above it. Now, consider the given binomial. ( - x+ 3 )^6 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and the coefficients from Pascal's Triangle.
| (a+b)^n=P_0a^nb^0+P_1a^(n-1)b^1+... +P_(n-1)a^1b^(n-1)+P_na^0b^n |
|---|
| ( - x+ 3 )^6 = 1( - x)^6( 3)^0 + 6( - x)^5( 3)^1 + 15( - x)^4( 3)^2 + 20( - x)^3( 3)^3 + 15( - x)^2( 3)^4 + 6( - x)^1( 3)^5 + 1( - x)^0( 3)^6 |
Finally, let's simplify the expression.
Now we completed the binomial expansion. Comparing our result with Dylan's, we can see that Dylan did not consider the negative signs.
| Dylan's Result | x^6 + 18x^5 + 135x^4 + 540x^3 + 1215x^2 + 1458x + 729 |
|---|---|
| Our Result | x^6 - 18x^5 + 135x^4 - 540x^3 + 1215x^2 - 1458x + 729 |
Therefore, Dylan's mistake is that he ignored the negative sign.
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The Binomial Theorem and Other Polynomial Identities
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≤
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
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