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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.**

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.See solution.

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.

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.
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.
Since $n$ is an integer number and $21 $ is not, the number $n+21 $ 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.

$Numbersnandn+1 Squares of the Numbersn_{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

$2n+1$

$22n+1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$22n +21 $

MovePartNumRight

$ca⋅b =ca ⋅b$

$22 n+21 $

QuotOne

$aa =1$

$1n+21 $

IdPropMult

Identity Property of Multiplication

$n+21 $

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!See solution.

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 $2n−1.$

Any even number can be written as $2n,$ where $n$ is an integer. Consider the cube of this number.
*before* or *after* $2n.$ These cases can be expressed by subtracting $1$ from or adding $1$ to $2n,$ respectively.
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 $2n.$
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.
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 $212n_{2}−6n+1 .$

### $2n+1$ 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.
Just like in the previous case, the expression that represents the difference will be divided by $2$ to determine if the result is even.
Similar to the previous case, $6n_{2}+3n$ is an integer. Because of the $21 ,$ the sum $6n_{2}+3n+21 $ is not an integer. Therefore, the difference of the cubes is odd. Now that the two cases are considered, the statement has been proved.

$(2n)_{3}=8n_{3} $

Now, when considering two consecutive integers, one is even and the other is odd. The odd number can be $Number Beforen:Number Aftern: 2n−12n+1 $

Recall the formula for the cube of a binomial.
$(a±b)_{3}=a_{3}±3a_{2}b+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 $b.$
$(a±b)_{3}=a_{3}±3a_{2}b+3ab_{2}±b_{3}$

SubstituteII

$a=2n$, $b=1$

$(2n±1)_{3}=(2n)_{3}±3(2n)_{2}(1)+3(2n)(1)_{2}±(1)_{3}$

$(2n±1)_{3}=8n_{3}±12n_{2}+6n±1$

$8n_{3}−(8n_{3}−12n_{2}+6n−1)$

$12n_{2}−6n+1$

$212n_{2}−6n+1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$212n_{2} −26n +21 $

MovePartNumRight

$ca⋅b =ca ⋅b$

$212 n_{2}−26 n+21 $

CalcQuot

Calculate quotient

$6n_{2}−3n+21 $

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+21 $ | No | $21 $ 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.

$(8n_{3}+12n_{2}+6n+1)−8n_{3}$

$12n_{2}+6n+1$

$212n_{2}+6n+1 $

Simplify

WriteSumFrac

Write as a sum of fractions

$212n_{2} +26n +21 $

MovePartNumRight

$ca⋅b =ca ⋅b$

$212 n_{2}+26 n+21 $

CalcQuot

Calculate quotient

$6n_{2}+3n+21 $

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!See solution.

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.

Let $n$ and $n+1$ be two consecutive integer numbers. To prove the given statement, the numbers first must be added.
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.
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+21 $ is not an integer. This means that $4n_{2}+4n+1$ is an odd number, proving the statement.

$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

$a_{m}b_{m}=(ab)_{m}$

$4n_{2}+2(2n)(1)+1_{2}$

BaseOne

$1_{a}=1$

$4n_{2}+2(2n)(1)+1$

Multiply

Multiply

$4n_{2}+4n+1$

$24n_{2}+4n+1 $

WriteSumFrac

Write as a sum of fractions

$24n_{2} +24n +21 $

MovePartNumRight

$ca⋅b =ca ⋅b$

$24 n_{2}+24 n+21 $

CalcQuot

Calculate quotient

$2n_{2}+2n+21 $

The 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.

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.

In this expansion, the numbers $P_{0},$ $P_{1},$ $…,$ $P_{n}$ are the numbers in the $n$th row of Pascal's Triangle.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.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"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:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.22222em;\">V<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["x^6-18x^5+135x^4-540x^3+1215x^2-1458x+729"]}}

Use the Binomial Theorem to expand the binomial.

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

$V=x_{6}−18x_{5}+135x_{4}−540x_{3}+1215x_{2}−1458x+729$

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!{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["f","w"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["126f^5w^4"]}}

Consider the Binomial Theorem.

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 $f_{9}w_{0},$ 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 |

$1$ | $1$ | $f_{9}w_{0}$ | $f_{9}$ |

$2$ | $9$ | $f_{8}w_{1}$ | $9f_{8}w$ |

$3$ | $36$ | $f_{7}w_{2}$ | $36f_{7}w_{2}$ |

$4$ | $84$ | $f_{6}w_{3}$ | $84f_{6}w_{3}$ |

$5$ | $126$ | $f_{5}w_{4}$ | $126f_{5}w_{4}$ |

$6$ | $126$ | $f_{4}w_{5}$ | $126f_{4}w_{5}$ |

$7$ | $84$ | $f_{3}w_{6}$ | $84f_{3}w_{6}$ |

$8$ | $36$ | $f_{2}w_{7}$ | $36f_{2}w_{7}$ |

$9$ | $9$ | $f_{1}w_{8}$ | $9fw_{8}$ |

$10$ | $1$ | $f_{0}w_{9}$ | $w_{9}$ |

Therefore, the fifth term is $126f_{5}w_{4}.$

To finish this lesson, a more complex exercise will be solved using the Binomial Theorem. Consider the following expression.
### Hint

### Solution

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 $a.$
Therefore, the possible values for $a$ are $20330 $ and $-20330 .$

$(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. {"type":"text","form":{"type":"roots","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]},"hideNoSolution":false,"variable":"a"},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"texts":["\\dfrac{3\\sqrt{30}}{20}","-\\dfrac{3\\sqrt{30}}{20}"]}}

Use the Binomial Theorem to find the coefficients of every term.

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 |
---|---|---|---|

$1$ | $1$ | $(2a)_{6}(3x)_{0}$ | $64a_{6}$ |

$2$ | $6$ | $(2a)_{5}(3x)_{1}$ | $576a_{5}x$ |

$3$ | $15$ | $(2a)_{4}(3x)_{2}$ | $2160a_{4}x_{2}$ |

$4$ | $20$ | $(2a)_{3}(3x)_{3}$ | $4320a_{3}x_{3}$ |

$5$ | $15$ | $(2a)_{2}(3x)_{4}$ | $4860a_{2}x_{4}$ |

$6$ | $6$ | $(2a)_{1}(3x)_{5}$ | $2916ax_{5}$ |

$7$ | $1$ | $(2a)_{0}(3x)_{6}$ | $729x_{6}$ |

$2916a=4320a_{3}$

Solve for $a$

DivEqn

$LHS/108a=RHS/108a$

$108a2916a =108a4320a_{3} $

WriteProdFrac

Write as a product of fractions

$1082916 ⋅aa =1084320 ⋅aa_{3} $

CalcQuot

Calculate quotient

$27⋅aa =40⋅aa_{3} $

QuotOne

$aa =1$

$27(1)=40⋅aa_{3} $

IdPropMult

Identity Property of Multiplication

$27=40⋅aa_{3} $

$aa_{m} =a_{m−1}$

$27=40a_{2}$

DivEqn

$LHS/40=RHS/40$

$4027 =a_{2}$

RearrangeEqn

Rearrange equation

$a_{2}=4027 $

SqrtEqn

$LHS =RHS $

$a_{2} =4027 $

$a_{2} =±a$

$a=±4027 $

Simplify right-hand side

SplitIntoFactors

Split into factors

$a=±4(10)9(3) $

SqrtQuot

$ba =b a $

$a=±4(10) 9(3) $

SqrtProd

$a⋅b =a ⋅b $

$a=±210 33 $

ExpandFrac

$ba =b⋅10 a⋅10 $

$a=±210 ⋅10 33 ⋅10 $

ProdSqrt

$a ⋅b =a⋅b $

$a=±2100 330 $

CalcRoot

Calculate root

$a=±2(10)330 $

Multiply

Multiply

$a=±20330 $