This article is targeted to those fresh students in precalculus (or further mathematics) who have come across the following questions \((10x + 3)^8\) and don’t know how to go about it. Read and follow along this easy breakdown of the binomial theorem to understand it and know how to use it. Before going into the binomial theorem itself, there are 4 topics we will briefly touch up on first:

– Polynomials

– Series and sigma notation

– Pascal’s triangle

– Combination

Without further ado, let’s jump right in, as this is the type of topic that is understood better with examples and clear illustrations.

## Polynomials

Polynomials are the algebraic expressions we’ve been using since secondary level, so it’s not a new concept. “Poly” stands for “many,” and “nomial” means “term.” They are mathematical expressions composed of variables, constants, and coefficients, all combined by a mathematical operator such as addition.

\[ 5x + 3y – 2^2 \]

The above expression is a trinomial. Let’s break down the expression:

– \(x\) and \(y\) are the variables

– 5 and 3 are the coefficients.

– \(-2\) is the constant

– 2 is the exponent

Each term is separated by an operator.

Let’s say we have the following:

1. \(x^3 \times y^6\)

2. \(4 + a^3\)

3. \(y^3 + y – 12\)

4. \(10 + 10x – 10y + 1^2\)

Can you guess the names of the polynomials?

1. is a monomial

2. is a binomial

3. is a trinomial as we knew from our previous example

4. is a quadrinomial.

You simply just need to replace the mathematical term for the number of terms with “poly” then add the “nomial” accordingly.

## Series and Sigma Notation

Before you can understand how the binomial theorem formula works, you must understand sigma notation. Before you can understand what sigma notation is, you must know what series are. Let’s begin with series.

### Series

If you’ve reached pre-calculus in your academic studies, then you must know what a sequence is. A sequence is simply a pattern of numbers or even different elements like letters. Let’s observe a quick example:

\[ 3, 9, 12, 15, 18, 21, \dots \]

Just by glancing at the above numbers, we can immediately tell that the numbers are progressing by multiplying each previous term by 3. If you can’t find that from just a look, simply divide a term by its previous one \( (9/3, 12/9, 15/12, \dots) \). This 3 is what we call our common ratio. Because of the presence of a common ratio, this can be identified as a geometric sequence.

Now, a series is a sum of a sequence.

\[ 3 + 9 + 12 + 15 + 18 + 21 + \dots \]

The above is called a geometric series. There is a formula for finding the sum of a geometric and arithmetic sequence, but it is not relevant to our topic, so let’s go right into sigma notation.

\[\sum_{i=1}^{n} f(i)\]

What you see above is called sigma notation. It is a way of expressing a series. The number below stands for which number to begin with, and the number above it is where to end (sometimes it can have the infinity sign, which means there is no end point). The expression near it is like our formula for the series or sequence.

It is hard to understand without working on an example for this, so consider the following:

\[\sum_{i=1}^{6} 3^i\]

The way you read this is: The summation of \(3^i\) is equal to…

So our \(n\), the number on top, is 6.

We will start substituting the \(i\) values from 1 to 6.

For sequences, \(i\) is typically 1 but can also be 0 from time to time.

This will be the sequence:

\[ 3^1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + \dots \]

Simplify and solve:

\[ 3 + 9 + 27 + 81 + 243 + 729 + \dots \]

From here, you can figure out which kind of sequence this is and solve whatever question they ask. They can ask for the \(n\)th term formula, which will be:

\[\text{Common ratio} = 3 \\\text{Geometric sequence general formula:} \ a_n = a r^{n-1} \\a = 3\]

Substitute values:

\[ a_n = 3 \times 3^{n-1} \]

If they ask for the partial sum, you can use the following formula:

If \(n < 1\):

\[S_n = \frac{1 – r^n}{r – 1}\]

If \(n > 1\):

\[S_n = \frac{r^n – 1}{r – 1}\]

With that example, you should now understand what sigma notation is, how to use it, and solving questions with it, so let’s move on.

## Pascal’s Triangle

Pascal’s triangle is a pattern of numbers in the shape of a triangle, named after the 17th-century French mathematician Blaise Pascal. The Pascal’s triangle is an extremely fundamental tool in the binomial theorem. Each number (except the edges, which should all be 1) is an addition of the number directly above it.

Observe the triangle above. You will notice that we start our first row with just 1. Then go on and add extra numbers in a triangular pattern. The second row can only be 1 and 1. Row three is when things start to mix up. Leave the edges (the first and last number) as 1, then add the numbers in the middle. For row three, you’ll be left with the numbers 1, 2, 1. Keep going like that. Row 4 will be 1 at the edges, then add 1 and 2 and 2 and 1, which will leave you with the pattern 1, 3, 3, 1 for row 4. It keeps going on like that.

The Pascal triangle is interesting because it has many patterns you can pick at. For example, the third diagonals are all triangle numbers \( (1, 3, 6, 10, \dots) \).

Additionally, the triangle is symmetrical because the numbers on the left side are identical to those on the right. It’s like a mirror image.

Another thing to note is, Pascal’s triangle is actually eleven being raised from the exponent 0 to infinity (in order).

– First row: \(11^0 = 1\) (which is why row 1 is usually referred to as row 0)

– Second row: \(11^2 = 121\). Hence 1, 2, 1

– Third row: \(11^3 = 1331\). Hence 1, 3, 3, 1

– Fourth row: \(11^4 = 14641\). Hence 1, 4, 6, 4, 1

From the fifth row, things start to get just a bit different.

– Fifth row: \(11^5 = 161051\) while our 5th row is 1, 5, 10, 10, 5, 1

– Sixth row: \(11^6 = 1771561\) while our 6th row is 1, 6, 15, 20, 15, 6, 1

Do you notice what’s happening? The numbers are overlapping because, from the fifth row in Pascal’s triangle, we have 2-digit numbers, while the \(11^5\) value is considered as one number.

– Seventh row: \(11^7 = 19487171\). Hence 1, 7, 21, 35, 35, 21, 7, 1

That’s all from this section. Just remember the numbers in Pascal’s triangle because we’ll get to see how it plays out in our theorem.

## Combinations

Ideally, this section should be titled “combinations and permutations” since they are under the same category, but because the one useful to the binomial theorem is combinations, it’ll be our main focus.

It’s simple;

Permutations is a mathematical calculation of the number of ways a particular set can be arranged, where the order of the arrangement matters.

Combination is the calculation of the number of ways a particular set can be arranged, where the order of the arrangement does not matter.

Imagine a kid has 6 toys. From toy A to toy F, he is only allowed to carry 3 with him on a road trip. Permutations allow us to find the number of orders of toys this kid can choose from if the order mattered to us.

The formula:

\[\text{Permutations} = \frac{n!}{(n – k)!}\]

Where \(n\) is the number of toys, \(k\) is the number of toys to select from, and \( ! \) is called factorial.

Factorial means multiplying the number with all the other numbers less than it

\( n! \) is equal to \( n \times (n-1) \times (n-2) \times (n-3) \times (n-4) \times \ldots \times 1 \).

Let’s substitute our values into the formula:

\[\frac{6!}{(6-3)!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3!}\]

\[\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}\]

Because the \(3 \times 2 \times 1\) in the numerator and denominator will cancel, we can rewrite the fraction as:

\[6 \times 5 \times 4 = 120\]

There are 120 ways the kid can choose toy A to toy F from.

In reality, the order a kid will choose 3 out of 6 toys does not matter, right? So using combinations is more suitable for this situation. The formula of combination can be represented as either of the following:

\[C(n,r) = \frac{n!}{k!(n-r)!} = \frac{n!}{(n-r)! \times r!}\]

Let’s substitute our given information into the formula:

Again, \(n\) is 6 and \(k\) is 3.

\[\frac{6!}{(6-3)! \times 3!}\]

From our last solution, we know we can cancel out \(3!\):

\[\frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20\]

Our possible order for 3 toys out of 6 is 20.

## Combining it all – the Binomial Theorem

Now we will combine everything we went over into the binomial theorem – A formula made for expanding binomials multiplied into themselves several times. For the sake of demonstration, let’s use the simple example of \((a+b)^2\).

Here is our formula:

\[(a+b)^2 = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} a^{n-k} b^k\]

To use the formula, we simply substitute the values, but before that, to show that this formula actually makes sense, we will solve this by using the traditional method we’ve been using up to the secondary level – foiling. Or you can say expanding.

### Expand or foil:

\[(a+b)^2 = (a+b)(a+b)\]

You are simply going to multiply every term in the first bracket with the second bracket. Like this:

\[a \times a = a^2 \quad \text{plus} \quad a \times b = ab \quad \text{plus} \quad b \times a = ab \quad \text{plus} \quad b \times b = b^2\]

Collect the like terms and you get:

\[a^2 + 2ab + b^2\]

Try \((a+b)^3\):

\[(a+b)(a+b)(a+b)\]

We already did \((a+b)(a+b)\), so we can just substitute it to become:

\[(a^2 + 2ab + b^2)(a+b)\]

Again, multiply all three terms in the first bracket with the second one:

\[a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3\]

Collect the like terms:

\[a^3 + 3a^2b + 3ab^2 + b^3\]

### The Coefficients

Look at the two expressions:

\[a^2 + 2ab + b^2\]

\[a^3 + 3a^2b + 3ab^2 + b^3\]

First, look at their coefficients:

For expression one with exponent 2, the coefficients are \(1, 2, 1\).

For expression two with the exponent 3, the coefficients are \(1, 3, 3, 1\).

Do you notice what’s happening here?

For each binomial, the exponent that it’s being raised to determines which row of Pascal’s triangle numbers will be used. So if you get the expression \((a+b)\) raised to the power of 4, the coefficients will be \(1, 4, 6, 4, 1\).

### The Power

In expression number 1 with exponent 2, the exponents for term one are \(2\) and \(0\). We consider \(b\) to have the power \(0\) because anything raised to the power of \(0\) is just one and is canceled out. The second term has \(1\) and \(1\), then the last term is \(2\) and \(0\).

There’s a pattern here also:

The first term, \(a\), is decreasing from \(2\) (the exponent of the entire expression), while the second term, \(b\), is increasing from \(0\) up till the exponent of the entire expression.

Let’s see if this pattern continues for the second expression with an exponent of \(3\).

Our first terms have the power \(3\) and \(0\).

The third term is \(2\) and \(1\).

The fourth term is \(1\) and \(2\).

The fifth and last term is \(0\) and \(3\).

The same pattern is recurring:

The first term in our binomial, \(a\), is decreasing from the number \(3\) to \(0\) – our exponent, while the second term, \(b\), is increasing from the number \(0\) up till our original exponent.

### The Binomial Theorem

The formula, as shown above, is:

\[\sum_{k=0}^{n} \frac{n!}{k!(n-k)!} a^{n-k} b^k\]

Now that we have expanded two binomials, we can see how the formula will work.

The summation is telling us to start from \(0\) because the exponents of the terms either end or start with zero.

The combination formula allows us to get the numbers from Pascal’s triangle for our coefficients.

Let’s try expanding \((a+b)\) raised to the power of 4.

Before you use the formula, expand it either by the traditional way or by using the pattern we’ve seen recurring. So, since our exponent is 4, our coefficients should be from the row in Pascal’s triangle that starts with the number \(4\) (after number \(1\)). That’s \(1, 4, 6, 4, 1\).

We will use these numbers in order:

For the power of \(a\), since it is the first term, the power will start from 4 and decrease all the way to 0.

For the power of \(b\), since it is the second term, the power will start from \(0\) and increase until the number 4.

So our final expression will be:

\[a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\]

Finally, let’s use the formula. We’ll be using \((a+b)\) raised to the power of 4 to see if our solution was correct.

\[\sum_{k=0}^{4} \frac{n!}{k!(4-k)!} a^{4-k} b^k\]

Now let’s simplify them:

\[\text{Note: } 0! = 1\]

\[\frac{4!}{0!(4-0)!} \times a^{4-0} \times b^0 = \frac{4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \times a^4 \times 1 = 1 \times a^4 = a^4\]

\[\frac{4!}{1!(4-1)!} \times a^{4-1} \times b^1 = \frac{4 \times 3 \times 2 \times 1}{3!} \times a^3 \times b = 4 \times a^3b = 4a^3b\]

\[\frac{4!}{2!(4-2)!} \times a^{4-2} \times b^2 = \frac{4 \times 3 \times 2 \times 1}{2! \times 2!} \times a^2 \times b^2 = 6 \times a^2b^2 = 6a^2b^2\]

\[\frac{4!}{3!(4-3)!} \times a^{4-3} \times b^3 = \frac{4 \times 3!}{3! \times 1!} \times a^1 \times b^3 = 4 \times ab^3 = 4ab^3\]

\[\frac{4!}{4!(4-4)!} \times a^{4-4} \times b^4 = 1 \times b^4 = b^4\]

Now to add them all together

\[ a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \]

It matches our answer above so we have shown that our calculations are correct, and hence, shown that the formula works.

Try expanding \((3 – 5y)^7\)

Let’s determine our values:

\( n \) is 7,

\( a \) is 3,

\( b \) is -5y.

Now let’s solve it:

\[\frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1 \cdot (3)^{7-0} = (3)^7 = 2187 \cdot (-5y)^0 = 1\]

\[\frac{7!}{1!(7-1)!} = \frac{7 \times 6!}{6!} = 7 \cdot (3)^{7-1} = (3)^6 = 729 \cdot (-5y)^1 = -5y\]

\[\frac{7!}{2!(7-2)!} = \frac{7 \times 6 \times 5!}{5! \cdot 2!} = \frac{42}{2} = 21 \cdot (3)^{7-2} = (3)^5 = 243 \cdot (-5y)^2 = -25y^2\]

\[\frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \cdot (3)^{7-3} = (3)^4 = 81 \cdot (-5y)^3 = -135y^3\]

\[\frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5 \times 4!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \cdot (3)^{7-4} = (3)^3 = 27 \cdot (-5y)^4 = 675y^4\]

\[\frac{7!}{5!(7-5)!} = \frac{7 \times 6 \times 5!}{2! \cdot 5!} = \frac{42}{2} = 21 \cdot (3)^{7-5} = (3)^2 = 9 \cdot (-5y)^5 = -2625y^5\]

\[\frac{7!}{6!(7-6)!} = \frac{7 \times 6!}{1! \cdot 6!} = 7 \cdot (3)^{7-6} = (3)^1 = 3 \cdot (-5y)^6 = 7875y^6\]

\[\frac{7!}{7!(7-7)!} = \frac{7!}{0! \cdot 7!} = 1 \cdot (3)^{7-7} = (3)^0 = 1 \cdot (-5y)^7 = -78125y^7\]

Add the terms together:

\[2187 – 25,515y + 127,575y^2 – 383,175y^3 + 637,875y^4 – 496,125y^5 + 165,375y^6 – 78,125y^7\]

## Conclusion

Now that you’ve seen the binomial theorem, how and why it works, you will be able to solve any binomial expansion that comes your way. Due to this being a lengthy calculation, it requires extreme care and focus. One silly miscalculation and you will become completely lost.

### Recap:

Binomials are algebraic expressions with two terms. In mathematics, you’ll often find them being multiplied to themselves multiple times. For the ease of calculation and to save time, we use the binomial theorem. This theorem is the product of all the patterns and principles we’ve explored. By thoroughly understanding combinations and the structure of Pascal’s triangle, smart mathematicians derived a powerful formula that allows us to expand binomials raised to any power. The pattern consists of:

The coefficients of the specific expression are based on its exponent. The exponent is the number you’ll look for in a row of Pascal’s triangle.

The first term’s power is decreasing from the entire expression’s exponent down to the number 0.

The second term’s power is increasing from the number 0 to the entire expression’s exponent.

With these patterns, we were able to get the following formula:

\[(a + b)^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} a^{n-k} b^k\]