Mastering the Binomial Theorem: Pascal's Triangle, Polynomial Expansion, and Coefficients

Expression	Is it an Integer?	Reason
$n$	Yes	It is defined this way.
$n^{2}$	Yes	Closure Property of Multiplication
$6 n^{2}$	Yes	Closure Property of Multiplication
$3 n$	Yes	Closure Property of Multiplication
$6 n^{2} - 3 n$	Yes	Closure Property of Subtraction
$6 n^{2} - 3 n + \frac{1}{2}$	No	$\frac{1}{2}$ is not an integer

Expression

Is it an Integer?

Reason

n

Yes

It is defined this way.

n^{2}

Yes

Closure Property of Multiplication

6 n^{2}

Yes

Closure Property of Multiplication

3 n

Yes

Closure Property of Multiplication

6 n^{2} - 3 n

Yes

Closure Property of Subtraction

6 n^{2} - 3 n + \frac{1}{2}

\frac{1}{2}

is not an integer

$n th$ term
Number of Term	Coefficient	Variables	Value of Term
$1$	$1$	$f^{9} w^{0}$	$f^{9}$
$2$	$9$	$f^{8} w^{1}$	$9 f^{8} w$
$3$	$36$	$f^{7} w^{2}$	$36 f^{7} w^{2}$
$4$	$84$	$f^{6} w^{3}$	$84 f^{6} w^{3}$
$5$	$126$	$f^{5} w^{4}$	$126 f^{5} w^{4}$
$6$	$126$	$f^{4} w^{5}$	$126 f^{4} w^{5}$
$7$	$84$	$f^{3} w^{6}$	$84 f^{3} w^{6}$
$8$	$36$	$f^{2} w^{7}$	$36 f^{2} w^{7}$
$9$	$9$	$f^{1} w^{8}$	$9 f w^{8}$
$10$	$1$	$f^{0} w^{9}$	$w^{9}$

n th

term

Number of Term

Coefficient

Variables

Value of Term

1

1

f^{9} w^{0}

f^{9}

2

9

f^{8} w^{1}

9 f^{8} w

3

36

f^{7} w^{2}

36 f^{7} w^{2}

4

84

f^{6} w^{3}

84 f^{6} w^{3}

5

126

f^{5} w^{4}

126 f^{5} w^{4}

6

126

f^{4} w^{5}

126 f^{4} w^{5}

7

84

f^{3} w^{6}

84 f^{3} w^{6}

8

36

f^{2} w^{7}

36 f^{2} w^{7}

9

9

f^{1} w^{8}

9 f w^{8}

10

1

f^{0} w^{9}

w^{9}

Number of Term	Coefficient	Variables	Value of Term
$1$	$1$	$(2 a)^{6} (3 x)^{0}$	$64 a^{6}$
$2$	$6$	$(2 a)^{5} (3 x)^{1}$	$576 a^{5} x$
$3$	$15$	$(2 a)^{4} (3 x)^{2}$	$2160 a^{4} x^{2}$
$4$	$20$	$(2 a)^{3} (3 x)^{3}$	$4320 a^{3} x^{3}$
$5$	$15$	$(2 a)^{2} (3 x)^{4}$	$4860 a^{2} x^{4}$
$6$	$6$	$(2 a)^{1} (3 x)^{5}$	$2916 a x^{5}$
$7$	$1$	$(2 a)^{0} (3 x)^{6}$	$729 x^{6}$

Number of Term

Coefficient

Variables

Value of Term

1

1

(2 a)^{6} (3 x)^{0}

64 a^{6}

2

6

(2 a)^{5} (3 x)^{1}

576 a^{5} x

3

15

(2 a)^{4} (3 x)^{2}

2160 a^{4} x^{2}

4

20

(2 a)^{3} (3 x)^{3}

4320 a^{3} x^{3}

5

15

(2 a)^{2} (3 x)^{4}

4860 a^{2} x^{4}

6

6

(2 a)^{1} (3 x)^{5}

2916 a x^{5}

7

1

(2 a)^{0} (3 x)^{6}

729 x^{6}

Write the following binomial expansion.

(z + 1)^{4}

Instead of multiplying the binomial by itself four times, we can use the Binomial Theorem to write the binomial expansion. This theorem 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 expression to the 4^\text{th} degree, let's look at the fourth row of Pascal's 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. ( z + 1)^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
( z + 1)^4 = 1 z^4( 1)^0 + 4 z^3( 1)^1 + 6 z^2( 1)^2 + 4 z( 1)^3 + 1 z^0( 1)^4

Finally, let's simplify the expression.

Now we have found the binomial expansion. Good job!

Write the following binomial expansion.

(2 x + y)^{6}

Instead of multiplying the binomial by itself six times, we can use the Binomial Theorem to write the binomial expansion. This theorem 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 expression to the 6^\text{th} degree, let's look at the sixth row of Pascal's 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. ( 2x + y)^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
( 2x + y)^6 = 1( 2x)^6 y^0+ 6( 2x)^5 y^1 + 15( 2x)^4 y^2 + 20( 2x)^3 y^3 + 15( 2x)^2 y^4 + 6( 2x)^1 y^5 + 1( 2x)^0 y^6

Finally, let's simplify the expression.

Now we have found the binomial expansion. Good job!

Write the following binomial expansion.

(d - 5)^{5}

Instead of multiplying the binomial by itself five times, we can use the Binomial Theorem to write the binomial expansion. This theorem 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 expression to the 5^\text{th} degree, let's look at the fifth row of Pascal's 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. ( d + ( -5))^5 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
( d + ( -5))^5 = 1( d)^5( -5)^0 + 5( d)^4( -5)^1 + 10( d)^3( -5)^2 + 10( d)^2( -5)^3 + 5( d)^1( -5)^4 + 1( d)^0( -5)^5

Finally, let's simplify the expression.

Now we have found the binomial expansion. Good job!

Write the following binomial expansion.

(- a^{3} + b^{2})^{7}

Instead of multiplying the binomial by itself seven times, we can use the Binomial Theorem to write the binomial expansion. This theorem 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 expression to the 7^\text{th} degree, let's look at the seventh row of Pascal's 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. ( - a ^3 + b^2)^7 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
( - a ^3 + b^2)^7 = 1( - a^3)^7( b^2)^0 + 7( - a^3)^6( b^2)^1 + 21( - a^3)^5( b^2)^2 + 35( - a^3)^4( b^2)^3 + 35( - a^3)^3( b^2)^4 + 21( - a^3)^2( b^2)^5 + 7( - a^3)^1( b^2)^6 + 1( - a^3)^0( b^2)^7

Finally, let's simplify the expression. We should be mindful with the term - a^3. Remember that if we raise a negative number to an even power, the result is a positive number. On the other hand, raising a negative number to an odd power results in a negative number.

Now we have found the binomial expansion. Good job!

Expand the following power of a complex number.

(2 + i)^{7}

Instead of multiplying the complex number by itself seven times, we can use the Binomial Theorem to write the expansion, similar to what is done with binomials. This theorem 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 expression to the 7^\text{th} degree, let's look at the seventh row of Pascal's 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. ( 2 + i)^7 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
( 2 + i)^7 = 1( 2)^7( i)^0 + 7( 2)^6( i)^1 + 21( 2)^5( i)^2 + 35( 2)^4( i)^3 + 35( 2)^3( i)^4 + 21( 2)^2( i)^5 + 7( 2)^1( i)^6 + 1( 2)^0( i)^7

Before simplifying the expression, let's remember some powers of i. i^0 &= 1 & i^4 &= 1 i^1 &= i & i^5 &= i i^2 &= -1 & i^6 &= -1 i^3 &= -i & i^7 &= -i Now, let's simplify the expression.

Now we wrote the power of the complex number. Good job!

Expand the following power of a complex number.

(- 1 + i)^{4}

Instead of multiplying the complex number by itself four times, we can use the Binomial Theorem to write the expansion, similar to what is done with binomials. This theorem 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 expression to the 4th degree, let's look at the fourth row of Pascal's 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. ( -1 + i)^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
( -1 + i)^4 = 1( -1)^4( i)^0 + 4( -1)^3( i)^1 + 6( -1)^2( i)^2 + 4( -1)^1( i)^3 + 1( -1)^0( i)^4

Before simplifying the expression, let's remember some powers of i. i^0 &= 0 & i^1 &= i i^2 &= -1 & i^3 &= -i i^4 &= 1 Now, let's simplify the expression.

Now we wrote the power of the complex number.

Consider the following binomial expansion.

(2 x + 4)^{5}

What is the coefficient of the fourth term?

To find the coefficient of the 4^(th) term of the binomial expansion, 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 expression to the 5 th degree, let's look at the fifth row of Pascal's Triangle.

Note that the first and last number in each row is 1. Any other number found in the row is the sum of the two numbers diagonally above it. Now, consider the given binomial. ( 2x+ 4 )^5 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
( 2x+ 4)^5= 1( 2x)^5( 4)^0+ 5( 2x)^4( 4)^1+ 10( 2x)^3( 4)^2+ 10( 2x)^2( 4)^3+ 5( 2x)^1( 4)^4+ 1( 2x)^0( 4)^5

Looking at the equation above, we can see that the 4^(th) term is the one where 2x is raised to the power of two. Let's simplify this term to find the coefficient!

The 4^(th) term of this expansion is 2560x^2. Therefore, the coefficient of the fourth term is 2560.

Consider the following binomial expansion.

(3 y - 7)^{4}

Write the third term of the expansion.

To find the coefficient of the third term of the binomial expansion, 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 expression to the 4 th degree, let's look at the fourth row of Pascal's 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. ( 3y+( - 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
( 3y+( - 7) )^4 = 1( 3y)^4( - 7)^0+ 4( 3y)^3( - 7)^1 + 6( 3y)^2( - 7)^2 + 4( 3y)^1( - 7)^3+ 1( 3y)^0( - 7)^4

Looking at the expression above, we can see that the third term is the one where 3y is raised to the power of two. Let's simplify it.

The third term of this expansion is 2646y^2. We did it!

Consider the following figure of a cube.

The length of each side is

(c - 1)^{3} .

Write an expanded expression for the volume of the cube.

The volume of a cube is equal to length of a side cubed. We are told that the side's length is (c-1)^3. Therefore, to find the volume of the cube, first we need to find the cube of (c-1)^3.

To expand the expression above, we can use the Binomial Theorem, which relates the coefficients of a binomial expansion with the numbers 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 Let's rewrite the given expression as the sum of two terms raised to 9. V=( c + ( -1))^9 In our case, a= c, b= -1, and n= 9, so the coefficients P_0, P_1,..., P_n are the ones in the ninth row of Pascal's Triangle.

Using the information above, we can expand the expression for the volume of the cube as follows.

(a + b)^9 = P_0a^9b^0+P_1a^8b^1 + ⋯ + P_8a^1b^8 + P_9 a^0b^9
V= 1( c)^9( -1)^0 + 9( c)^8( -1)^1 + 36( c)^7( -1)^2 + 84( c)^6( -1)^3 + 126( c)^5( -1)^4 + 126( c)^4( -1)^5 + 84( c)^3( -1)^6 + 36( c)^2( -1)^7 + 9( c)^1( -1)^8 + 1( c)^0( -1)^9

Finally, let's simplify this expression a bit.

Now we found the expression for the volume of the cube!

The Binomial Theorem and Other Polynomial Identities

Catch-Up and Review

Recognizing Patterns

Difference of Squares

Answer

Hint

Solution

Difference of Cubes

Answer

Hint

Solution

$2 n$ and $2 n - 1$

$2 n + 1$ and $2 n$

Square of the Sum of Two Consecutive Integers

Answer

Hint

Solution

Pascal's Triangle and the Binomial Theorem

Binomial Theorem

The Volume of a Room

Hint

Solution

Finding a Certain Coefficient

Hint

Solution

Using the Binomial Theorem to Solve a Problem

Hint

Solution

The Binomial Theorem and Other Polynomial Identities

Recommended exercises

	9 Theory slides
	15 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

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Hint

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Solution

2n and 2n−1

2n+1 and 2n

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Solution

Binomial Theorem

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