# 7.5: Binomial Theorem

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We end this chapter with one more application of combinations. Combinations are used in determining the coefficients of a binomial expansion such as $$(x + y)^n$$. Expanding a binomial expression by multiplying it out is a very tedious task, and is not practiced. Instead, a formula known as the Binomial Theorem is utilized to determine such an expansion. Before we introduce the Binomial Theorem, however, consider the following expansions.

$\begin{array}{l} (x+y)^{2}=x^{2}+2 x y+y^{2} \\ (x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \\ (x+y)^{4}=x^{4}+4 x^{2} y+6 x^{2} y^{2}+4 x y^{3}+y^{4} \\ (x+y)^{5}=x^{5}+5 x^{4} y+10 x^{3} y^{2}+10 x^{2} y^{3}+5 x y^{4}+y^{5} \\ (x+y)^{6}=x 6+6 x^{5} y+15 x^{4} y^{2}+20 x^{3} y^{3}+15 x^{2} y^{4}+6 x y^{5}+y^{6} \end{array} \nonumber$

We make the following observations.

1. There are $$n + 1$$ terms in the expansion $$(x + y)^n$$
2. The sum of the powers of $$x$$ and $$y$$ is $$n$$.
3. The powers of $$x$$ begin with $$n$$ and decrease by one with each successive term.
The powers of $$y$$ begin with 0 and increase by one with each successive term.

Suppose we want to expand $$(x + y)^3$$. We first write the expansion without the coefficients. We temporarily substitute a blank in place of the coefficients.

$(x+y)^{3}=\square x^{3}+\square x^{2} y+\square x y^{2}+\square y^{3} \label{I}$

Our next job is to replace each of the blanks in equation (\ref{I}) with the corresponding coefficients that belong to this expansion. Clearly,

$(x+y)^{3}=(x+y)(x+y)(x+y) \nonumber$

If we multiply the right side and do not collect terms, we get the following.

$xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy \nonumber$

Each product in the above expansion is the result of multiplying three variables by picking one from each of the factors $$(x+y)(x+y)(x+y)$$. For example, the product $$xxy$$ is gotten by choosing $$x$$ from the first factor, $$x$$ from the second factor, and $$y$$ from the third factor. There are three such products that simplify to $$x^2y$$, namely $$xxy$$, $$xyx$$, and $$yxx$$. These products take place when we choose an $$x$$ from two of the factors and choose a $$y$$ from the other factor. Clearly this can be done in 3C2, or 3 ways. Therefore, the coefficient of the term $$x^2y$$ is 3. The coefficients of the other terms are obtained in a similar manner.

We now replace the blanks with the coefficients in equation (\ref{I}), and we get

$(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \nonumber$

## Example $$\PageIndex{1}$$

Find the coefficient of the term $$x^2y^5$$ in the expansion $$(x+y)^7$$.

Solution

The expansion $$(x + y)^7 = (x + y) (x + y) (x + y) (x + y) (x + y) (x + y) (x + y)$$

In multiplying the right side, each product is gotten by picking an $$x$$ or $$y$$ from each of the seven factors $$(x + y) (x + y) (x + y) (x + y) (x + y) (x + y) (x + y)$$.

The term $$x^2y^5$$ is obtained by choosing an $$x$$ from two of the factors and a $$y$$ from the other five factors. This can be done in 7C2, or 21 ways.

Therefore, the coefficient of the term $$x^2y^5$$ is 21.

## Example $$\PageIndex{2}$$

Expand $$(x + y)^7$$

Solution

We first write the expansion without the coefficients.

$(x+y)^{7}=\square x^{7}+\square x^{6} y+ \square x^{5} y^{2}+ \square x^{4} y^{3}+ \square x^{3} y^{4}+\square x^{2} y^{5}+\square x y^{6}+\square y^{7} \nonumber$

Now we determine the coefficient of each term as we did in Example $$\PageIndex{1}$$.

The coefficient of the term $$x^7$$ is 7C7 or 7CO which equals 1.

The coefficient of the term $$x^6y$$ is 7C6 or 7C1 which equals 7.

The coefficient of the term $$x^5y^2$$ is 7C5 or 7C2 which equals 21.

The coefficient of the term $$x^4y^3$$ is 7C4 or 7C3 which equals 35,

and so on.

Substituting, we get: $$(x+y)^{7}=x^{7}+7 x^{6} y+21 x^{5} y^{2}+35 x^{4} y^{3}+35 x^{3} y^{4}+21 x^{2} y^{5}+7 x y^{6}+y^{7}$$

We generalize the result.

## Binomial Theorem

$(x+y)^{n}=_{n} C_{0} x^{n}+_{n} C_{1} x^{n-1} y+_{n} C_{2} x^{n-2} y^{2}+\cdots \cdot+_{n} C_{n-1} x y^{n-1+}_{n} C_{n} y^{n} \nonumber$

## Example $$\PageIndex{3}$$

Expand $$(3a-2b)^4$$

Solution

If we let $$x = 3a$$ and $$y = - 2b$$, and apply the Binomial Theorem, we get

\begin{aligned}
(3 a-2 b)^{4} &=4 \operatorname{Co}(3 a)^{4}+4 C l(3 a)^{3}(-2 b)+4 C 2(3 a)^{2}(-2 b)^{2}+4 C 3(3 a)(-2 b)^{3}+4 C 4(-2 b)^{4} \\
&=1\left(81 a^{4}\right)+4\left(27 a^{3}\right)(-2 b)+6\left(9 a^{2}\right)\left(4 b^{2}\right)+4(3 a)\left(-8 b^{3}\right)+1\left(16 b^{3}\right) \\
&=81 a^{4}-216 a^{3} b+216 a^{2} b^{2}-96 a b^{3}+16 b^{4}
\end{aligned}

## Example $$\PageIndex{4}$$

Find the fifth term of the expansion $$(3a-2b)^7$$.

Solution

The Binomial theorem tells us that in the r-th term of an expansion, the exponent of the $$y$$ term is always one less than $$r$$, and, the coefficient of the term is $$_{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}$$.

$$n = 7$$ and $$r - 1 = 5 - 1= 4$$, so the coefficient is $$7 \mathrm{C} 4=35$$

Thus, the fifth term is $$(7 \mathrm{C} 4)(3 \mathrm{a})^{3}(-2 \mathrm{b})^{4}=35\left(27 \mathrm{a}^{3}\right)\left(16 \mathrm{b}^{4}\right)=15120 \mathrm{a}^{3} \mathrm{b}^{4}$$

This page titled 7.5: Binomial Theorem is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Rupinder Sekhon and Roberta Bloom via source content that was edited to the style and standards of the LibreTexts platform.