Use Pascal’s triangle to expand binomial powers

Ontario Grade 11 Mathematics

Discrete Functions

MCR3U – C1.6 | Using patterns in Pascal's triangle to expand expressions of the form (a + b)ⁿ

You have already worked with polynomials in Grade 10 — you know how to multiply two brackets together and collect like terms. Now imagine doing that for

(x + 1)^6

. Expanding step by step would take a very long time and leave many opportunities for errors. In this lesson you will discover a faster, organised method: using Pascal's triangle to read off the coefficients of a binomial expansion directly. This approach turns a long multiplication into a neat, predictable pattern — and once you understand why it works, you can expand any integer power of a binomial with confidence.

What you will learn

Construct Pascal's triangle and describe how each row is built from the one above it.
Identify which row of Pascal's triangle gives the coefficients for a chosen power of a binomial.
Expand binomial powers using the coefficients, variable degrees, and sign patterns from Pascal's triangle.
Identify and simplify individual terms in a binomial expansion, including cases with negative terms or coefficients inside the binomial.

Prerequisite Bridge: Binomials and Expanding by Hand

A binomial is a polynomial with exactly two terms, such as

(x + 3)

,

(2a - 1)

, or

(x - y)

. The word 'bi' means two. When you raise a binomial to a power, you are multiplying it by itself that many times.

Let's quickly recall what happens for small powers. When the exponent is 1, the expression stays the same:

(a + b)^1 = a + b

. For exponent 2, you use FOIL or the distributive property:

(a + b)^2 = a^2 + 2ab + b^2

. For exponent 3 you multiply

(a+b)^2

by

(a+b)

again to get

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

.

Look at the numerical coefficients (the numbers in front of each term) in those expansions: for power 1 the coefficients are CAD 1, 1; for power 2 they are CAD 1, 2, 1; for power 3 they are CAD 1, 3, 3, 1. These numbers are not random — they form a famous pattern called Pascal's triangle, which you can build once and reuse forever.

A binomial has exactly two terms.
Expanding $(a+b)^n$ means multiplying the binomial by itself $n$ times.
The coefficients of the expanded terms follow a predictable numerical pattern.

Building Pascal's Triangle

Pascal's triangle is a triangular arrangement of numbers. The very top entry is 1, which represents the expansion of

(a+b)^0 = 1

. Each row below it corresponds to the next higher power. To find any entry in a new row, add the two numbers directly above it to the left and right. Any position at the edge of the triangle is always 1.

Here are the first seven rows (rows 0 through 6). Row

n

holds the coefficients for

(a+b)^n

. Row 0:

1

. Row 1: 1 \ \ 1. Row 2: 1 \ \ 2 \ \ 1. Row 3: 1 \ \ 3 \ \ 3 \ \ 1. Row 4: 1 \ \ 4 \ \ 6 \ \ 4 \ \ 1. Row 5: 1 \ \ 5 \ \ 10 \ \ 10 \ \ 5 \ \ 1. Row 6: 1 \ \ 6 \ \ 15 \ \ 20 \ \ 15 \ \ 6 \ \ 1.

To verify the addition rule, look at Row 4: the 6 in the middle is found by adding the two 3s from Row 3. The 4 near the edge is found by adding 3 and 1 from Row 3. The edge entries remain 1 because there is only one number above them on their side. This addition rule is the key mechanical skill — practise building two or three rows from scratch until it feels automatic.

Row 0 (the tip) contains just the number 1, matching $(a+b)^0 = 1$ .
Each interior entry equals the sum of the two entries directly above it.
All edge entries are 1.
Row $n$ of Pascal's triangle gives the coefficients for the expansion of $(a+b)^n$ .

The Structure of a Binomial Expansion

Once you have the coefficients from Pascal's triangle, you still need to know how the powers of

a

and

b

behave across the terms. Two rules always hold. First, the powers of

a

start at

n

in the first term and decrease by 1 each term until they reach 0 in the last term. Second, the powers of

b

start at 0 in the first term and increase by 1 each term until they reach

n

in the last term. In every single term, the exponents of

a

and

b

add up to

n

.

Putting it together: to expand

(a + b)^n

, read the

n

-th row of Pascal's triangle for the coefficients, then attach decreasing powers of

a

and increasing powers of

b

. For

(a+b)^4

using Row 4 coefficients CAD 1, 4, 6, 4, 1, the expansion is

(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

.

When one or both terms inside the binomial have their own coefficients or when a term is negative, you must substitute carefully. Replace

a

and

b

with whatever expressions are actually inside the binomial and then simplify each term. The next section works through this in full detail with two examples.

(a+b)^n = c_0 a^n + c_1 a^{n-1}b + c_2 a^{n-2}b^2 + \cdots + c_n b^n

In $(a+b)^n$ , powers of $a$ decrease from $n$ to $0$ left to right.
Powers of $b$ increase from $0$ to $n$ left to right.
In every term, the exponents on $a$ and $b$ sum to $n$ .
Substitute the actual expressions for $a$ and $b$ , then simplify each term carefully.

Handling Negative Terms and Inner Coefficients

Many exam problems involve a binomial like

(2x - 3)^4

rather than a simple

(a+b)^4

. The method is identical, but you must treat the entire second term — including its negative sign — as

b

. So here

a = 2x

and

b = -3

. When you raise

b

to an even power, the result is positive; when you raise it to an odd power, the result is negative. Keeping track of signs is the most common source of error.

Similarly, when

a

itself has a coefficient, such as

a = 2x

, you must raise the entire expression

2x

to the required power:

(2x)^3 = 8x^3

, not

2x^3

. Write out each factor in full before multiplying. Taking one small extra step to write

(2x)^3 = 2^3 \ \cdot \ x^3 = 8x^3

prevents a very common mistake.

After writing out every term using the triangle's coefficients and the correct powers, multiply the three factors in each term — the Pascal coefficient, the power of

a

, and the power of

b

— and simplify. Collect any like terms if they exist (they usually do not in a standard expansion).

Treat the full second term, including its sign, as $b$ when substituting.
A negative $b$ raised to an odd power gives a negative term; raised to an even power it gives a positive term.
Raise the entire expression (including its coefficient) to the required power.
Multiply the Pascal coefficient, the power of $a$ , and the power of $b$ together to get the final coefficient of each term.

Spotting a Specific Term Without Full Expansion

Sometimes you only need one particular term from an expansion, not all of them. Because the pattern is completely predictable, you can go straight to that term. The terms in the expansion of

(a+b)^n

are numbered starting from term 1. Term 1 has

a^n b^0

, term 2 has

a^{n-1}b^1

, term 3 has

a^{n-2}b^2

, and so on. In general, term

k

(where

k

starts at 1) uses the

(k-1)

-th coefficient from Row

n

of Pascal's triangle (counting the first entry as the 0th), together with

a^{n-(k-1)}

and

b^{k-1}

.

For example, suppose you want the 4th term of

(x + 2)^6

. The 4th term uses the 3rd entry of Row 6 (counting from 0), which is

20

. The powers are

x^{6-3} = x^3

and

2^3 = 8

. So the 4th term is

20 \ \cdot \ x^3 \ \cdot \ 8 = 160x^3

. This targeted approach is very useful on timed assessments.

Practise identifying which row to use, which entry within that row to pick, and how to compute the powers of each part. These three sub-skills together let you find any term quickly and accurately.

Terms are numbered starting from 1; the power of $b$ in term $k$ is $k - 1$ .
The Pascal coefficient for term $k$ is the entry at position $k-1$ in Row $n$ (0-indexed).
Compute the power of the full expression for both $a$ and $b$ before multiplying.

Pascal's Triangle — Rows 0 to 6 with Corresponding Binomial Power

Row (= Power n)	Pascal's Triangle Entries (Coefficients)	Number of Terms in Expansion
0	1	1
1	1 1	2
2	1 2 1	3
3	1 3 3 1	4
4	1 4 6 4 1	5
5	1 5 10 10 5 1	6
6	1 6 15 20 15 6 1	7

Worked example

Expanding $(2x - 3)^4$

Use Pascal's triangle to fully expand

(2x - 3)^4

and simplify every term.

Identify the row of Pascal's triangle needed
The exponent is $4$ , so use Row 4 of Pascal's triangle. Row 4 has five entries: CAD 1, 4, 6, 4, 1. These will be the coefficients of the five terms in the expansion.
$1 4 6 4 1$
Name the two parts of the binomial
Let $a = 2x$ and $b = -3$ . It is essential to include the negative sign as part of $b$ so that every term's sign is handled automatically when you compute the powers.
$a = 2x, b = -3$
Write the general structure of the expansion
Combine the Pascal coefficients with decreasing powers of $a$ and increasing powers of $b$ . The powers of $a$ go CAD 4, 3, 2, 1, 0 and the powers of $b$ go CAD 0, 1, 2, 3, 4.
$1(2x)^4(-3)^0 + 4(2x)^3(-3)^1 + 6(2x)^2(-3)^2 + 4(2x)^1(-3)^3 + 1(2x)^0(-3)^4$
Simplify each power of a
Raise the full expression $2x$ to each required power. Remember to apply the exponent to both the coefficient $2$ and the variable $x$ : $(2x)^4 = 16x^4$ , $(2x)^3 = 8x^3$ , $(2x)^2 = 4x^2$ , $(2x)^1 = 2x$ , $(2x)^0 = 1$ .
$16x^4, \ 8x^3, \ 4x^2, \ 2x, \ 1$
Simplify each power of b
Raise $-3$ to each required power. An even exponent gives a positive result; an odd exponent gives a negative result: $(-3)^0 = 1$ , $(-3)^1 = -3$ , $(-3)^2 = 9$ , $(-3)^3 = -27$ , $(-3)^4 = 81$ .
$1, \ -3, \ 9, \ -27, \ 81$
Multiply the three factors in each term
For each term, multiply the Pascal coefficient by the power of $a$ and by the power of $b$ . Term 1: $1 \cdot 16x^4 \cdot 1 = 16x^4$ . Term 2: $4 \cdot 8x^3 \cdot (-3) = -96x^3$ . Term 3: $6 \cdot 4x^2 \cdot 9 = 216x^2$ . Term 4: $4 \cdot 2x \cdot (-27) = -216x$ . Term 5: $1 \cdot 1 \cdot 81 = 81$ .
$16x^4 - 96x^3 + 216x^2 - 216x + 81$

Answer:

(2x - 3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81

Check: Substitute

x = 1

: the expansion gives

16 - 96 + 216 - 216 + 81 = 1

. The original expression gives

(2(1) - 3)^4 = (-1)^4 = 1

. The values match, confirming the expansion is correct.

Worked example

Finding the 3rd Term of $\left(x^2 + \frac{1}{2}\right)^5$

Without expanding the full expression, find and simplify the 3rd term of

\left(x^2 + \frac{1}{2}\right)^5

.

Identify the row and the target entry
The exponent is $5$ , so use Row 5 of Pascal's triangle: CAD 1, 5, 10, 10, 5, 1. The 3rd term uses the entry at position $2$ (counting from position 0), which is $10$ .
$1 5 10 10 5 1$
Name the two parts and determine the powers for the 3rd term
Let $a = x^2$ and $b = \frac{1}{2}$ . For the 3rd term (term $k = 3$ ), the power of $b$ is $k - 1 = 2$ and the power of $a$ is $n - (k-1) = 5 - 2 = 3$ .
$a = x^2, b = \frac{1}{2}, a^3 · b^2$
Compute the power of a
Raise $x^2$ to the power of $3$ . When you raise a power to a power, multiply the exponents: $(x^2)^3 = x^{2 \times 3} = x^6$ .
$(x^2)^3 = x^6$
Compute the power of b
Raise $\frac{1}{2}$ to the power of $2$ . Square both the numerator and denominator: $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ .
$(\frac{1}{2})^2 = \frac{1}{4}$
Multiply the Pascal coefficient by the powers
Multiply the three factors together: the Pascal coefficient $10$ , the power of $a$ which is $x^6$ , and the power of $b$ which is $\frac{1}{4}$ .
$10 · x^6 · \frac{1}{4} = \frac{10}{4}x^6 = \frac{5}{2}x^6$

Answer: The 3rd term of

\left(x^2 + \frac{1}{2}\right)^5

is

\frac{5}{2}x^6

.

Check: Re-examine: Row 5 position 2 gives

10

. Powers:

(x^2)^3 = x^6

and

(\frac{1}{2})^2 = \frac{1}{4}

. Product:

10 \times \frac{1}{4} = \frac{10}{4} = \frac{5}{2}

. Term:

\frac{5}{2}x^6

. Confirmed.

Common mistakes and how to avoid them

Using the wrong row of Pascal's triangle — for example, using Row 3 for a power of 4.

Correction: The row number equals the exponent. For

(a+b)^4

, use Row 4, which contains five entries: CAD 1, 4, 6, 4, 1.

Forgetting to include the negative sign as part of

b

, leading to wrong signs in the expanded terms.

Correction: If the binomial is

(a - c)^n

, write

b = -c

and raise that full expression to each power. The sign of each term will then come out automatically.

Applying the exponent only to the variable and not to the coefficient inside the binomial — writing

(3x)^2 = 3x^2

instead of

9x^2

.

Correction: Raise the entire expression to the power.

(3x)^2 = 3^2 \cdot x^2 = 9x^2

. Write it out as a product of separate factors before simplifying.

Counting the first entry of a Pascal's triangle row as entry 1 instead of entry 0, causing the wrong coefficient to be selected for a specific term.

Correction: The first entry in every row is counted as position 0. For term

k

(starting at 1), the Pascal coefficient sits at position

k - 1

in the row.

Thinking the expansion of

(a - b)^n

must always alternate signs, and forcing alternation even when an error is made elsewhere.

Correction: The signs alternate only when

b

itself is positive. Rely on correctly computing each power of the actual

b

expression — do not assume the pattern without computing it.

Lesson summary

Pascal's triangle is built by placing 1s on both edges and making each interior entry the sum of the two entries above it. Row $n$ gives the coefficients for the expansion of $(a+b)^n$ .
In the expansion of $(a+b)^n$ , there are $n+1$ terms. The power of $a$ decreases from $n$ to $0$ and the power of $b$ increases from $0$ to $n$ across the terms.
Substitute the actual expressions for $a$ and $b$ — including any coefficients or negative signs — and raise them to the required powers before multiplying.
To find a specific term $k$ , use position $k-1$ in Row $n$ of Pascal's triangle, together with $a^{n-(k-1)}$ and $b^{k-1}$ .
Always verify your expansion by substituting a simple number such as $x = 1$ into both the original expression and the expansion and confirming that the results match.

Check your understanding

Question 1

Which row of Pascal's triangle provides the coefficients needed to expand

(x + 5)^3

?

Row 2: CAD 1, 2, 1
Row 3: CAD 1, 3, 3, 1
Row 4: CAD 1, 4, 6, 4, 1
Row 5: CAD 1, 5, 10, 10, 5, 1

Show answer and explanation

Row 3: CAD 1, 3, 3, 1

The exponent is 3, so you need Row 3. Its entries CAD 1, 3, 3, 1 are the four coefficients of the expansion of any binomial raised to the power of 3.

Question 2

What is the correct expansion of

(x - 2)^3

?

$x^3 + 6x^2 + 12x + 8$
$x^3 - 2x^2 - 4x - 8$
$x^3 - 6x^2 + 12x - 8$
$x^3 - 3x^2 + 3x - 8$

Show answer and explanation

x^3 - 6x^2 + 12x - 8

Using Row 3 coefficients CAD 1, 3, 3, 1 with

a = x

and

b = -2

:

x^3 + 3x^2(-2) + 3x(-2)^2 + (-2)^3 = x^3 - 6x^2 + 12x - 8

. Signs alternate because

b

is negative.

Question 3

What entry from Pascal's triangle is the coefficient of the 4th term in the expansion of

(a + b)^6

?

15
20
6
10

Show answer and explanation

20

Row 6 is CAD 1, 6, 15, 20, 15, 6, 1. The 4th term uses the entry at position 3 (0-indexed), which is

20

.

Question 4

In the expansion of

(3x + 1)^4

, what is the simplified value of the first term?

$12x^4$
$3x^4$
$81x^4$
$27x^4$

Show answer and explanation

81x^4

The first term uses the Pascal coefficient

1

, then

(3x)^4 = 3^4 x^4 = 81x^4

, and

(1)^0 = 1

. Multiplying:

1 \times 81x^4 \times 1 = 81x^4

. Remember to raise the full expression

3x

, not just

x

, to the power.

Key terms

Binomial: A polynomial expression with exactly two terms, such as $(x + 3)$ or $(2a - b)$ .
Pascal's Triangle: A triangular arrangement of numbers where each interior entry is the sum of the two entries directly above it, and every edge entry is 1. Row n gives the coefficients for the expansion of $(a+b)^n$ .
Coefficient: The numerical factor multiplied by the variable part of a term. In $6x^2$ , the coefficient is 6.
Expansion: The result of multiplying out a power or product of polynomials and writing it as a sum of simplified terms.
Exponent: The small raised number that tells you how many times a base is multiplied by itself. In $x^4$ , the exponent is 4.
Term: A single part of a polynomial separated from others by addition or subtraction. In $x^3 - 6x^2 + 12x - 8$ , each of the four parts is a term.
Row n of Pascal's Triangle: The row corresponding to the power n. Row 0 is the single 1 at the top; Row 1 is the next row CAD 1, 1; and so on. The row number equals the exponent of the binomial power.
Position (0-indexed): A way of numbering entries in a row starting from 0 rather than 1. The first entry in any row is at position 0, the second at position 1, and so on.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation C1.6. It is a study resource, not an official curriculum publication.

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