# 25.2: Binomial Expansion

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Using the binomial theorem, we can also expand more general powers of sums or differences.

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

Expand the expression.

1. $$(x^2+2y^3)^5$$
2. $$(2xy^2-\frac{4}{y^2})^3$$
3. $$(\sqrt{2}+1)^6$$
4. $$(i-3)^4$$

Solution

1. We use the binomial theorem with $$a=x^2$$ and $$b=2y^3$$:

\begin{aligned} (x^2+2y^3)^5 &= (x^2)^5+\dbinom{5}{1} (x^2)^{4}(2y^3)+\dbinom{5}{2} (x^2)^{3}(2y^3)^2+\dbinom{5}{3} (x^2)^{2}(2y^3)^{3}+\dbinom{5}{4} (x^2)(2y^3)^{4}+(2y^3)^5 \\ &= x^{10}+5x^8\cdot 2y^3+10x^6\cdot 4y^6+10x^4\cdot 2^3y^9+5x^2\cdot 2^4y^{12}+2^5y^{15}\\ &= x^{10}+10x^8y^3+40x^6y^6+80x^4y^9+80x^2y^{12}+32 y^{15}\end{aligned} \nonumber

1. For part (b), it is $$a=2xy^2$$ and $$b=-\dfrac{4}{y^2}$$.

\begin{aligned} \left(2xy^2-\dfrac{4}{y^2}\right)^3 & = (2xy^2)^3+\dbinom{3}{1}(2xy^2)^2\left(-\dfrac{4}{y^2}\right)+\dbinom{3}{2}(2xy^2)\left(-\dfrac{4}{y^2}\right)^2+\left(-\dfrac{4}{y^2}\right)^3 \\ &= 2^3x^3y^6+3\cdot 2^2x^2y^4\left(-\dfrac{4}{y^2}\right)+3(2xy^2)(-1)^2\dfrac{4^2}{y^4}+(-1)^3\dfrac{4^3}{y^6} \\ &=8x^3y^6-48x^2y^2+96x\cdot \dfrac{1}{y^2}-64\cdot \dfrac{1}{y^6}\end{aligned} \nonumber

1. Similarly, for part (c), we now have $$a=\sqrt{2}$$ and $$b=1$$:

\begin{aligned} (\sqrt{2}+1)^6 &= (\sqrt{2})^6+\dbinom{6}{1} (\sqrt{2})^{5}\cdot 1+\dbinom{6}{2} (\sqrt{2})^{4}\cdot 1^2 +\dbinom{6}{3} (\sqrt{2})^{3}\cdot 1^{3}+\dbinom{6}{4} (\sqrt{2})^2\cdot 1^{4}+\dbinom{6}{5}(\sqrt{2})\cdot 1^5+1^6 \\ &= \sqrt{64}+6\cdot \sqrt{32}+15\cdot \sqrt{16}+20\cdot \sqrt{8}+15\cdot \sqrt{4} +6\cdot \sqrt{2}+1\\ &= 8+6\cdot \sqrt{16\cdot 2}+15\cdot 4+20\cdot \sqrt{4\cdot 2}+15\cdot 2 +6\cdot \sqrt{2}+1\\ &= 8+24\sqrt{2}+60+40\sqrt{2}+30+6\sqrt{2}+1\\ &= 99+70\sqrt{2}\end{aligned} \nonumber

Note that the last expression cannot be simplified any further (due to the order of operations).

1. Finally, we have $$a=i$$ and $$b=-3$$, and we use the fact that $$i^2=-1$$, and therefore, $$i^3=-i$$ and $$i^4=+1$$:

\begin{aligned} (i-3)^4 &= i^4+\dbinom{4}{1}\cdot i^3\cdot (-3)+\dbinom{4}{2}\cdot i^2\cdot (-3)^2 +\dbinom{4}{3}\cdot i\cdot (-3)^{3}+(-3)^4 \\ &= 1+4\cdot (-i)\cdot (-3)+6 \cdot (-1)\cdot 9 +4 \cdot i\cdot (-27)+81 \\ &= 1+12i-54-108i+81\\ &= 28-96i\end{aligned} \nonumber

In some instances it is not necessary to write the full binomial expansion, but it is enough to find a particular term, say the $$k$$th term of the expansion.

## Observation: $$k$$th term of expansion

Recall, for example, the binomial expansion of $$(a+b)^6$$:

$\begin{array}{ccccccc}\\ \dbinom{6}{0} a^6b^0 + & \dbinom{6}{1} a^{5}b^1 + & \dbinom{6}{2} a^{4}b^2 + & \dbinom{6}{3} a^{3}b^{3} + & \dbinom{6}{4} a^{2}b^{4} + & \dbinom{6}{5} a^{1}b^{5} + & \dbinom{6}{6}a^0b^6\\ \text{first} & \text{second} &\text{third} &\text{fourth} &\text{fifth} &\text{sixth} &\text{seventh} \\ \text{term} & \text{term} &\text{term} &\text{term} &\text{term} &\text{term} &\text{term} \end{array}\nonumber$

Note that the exponents of the $$a$$’s and $$b$$’s for each term always add up to $$6$$, and that the exponents of $$a$$ decrease from $$6$$ to $$0$$, and the exponents of $$b$$ increase from $$0$$ to $$6$$. Furthermore observe that in the above expansion the fifth term is $$\dbinom{6}{4} a^{2}b^{4}$$.

In general, we define the $$k$$th term by the following formula:

$\text{The kth term in the expansion of (a+b)^n is: }\boxed{\dbinom{n}{k-1}a^{n-k+1}b^{k-1} }$

Note in particular, that the $$k$$th term has a power of $$b$$ given by $$b^{k-1}$$ (and not $$b^k$$), it has a binomial coefficient $$\dbinom{n}{k-1}$$, and the exponents of $$a$$ and $$b$$ add up to $$n$$.

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

Determine:

1. the $$4$$th term in the binomial expansion of $$(p+3q)^{5}$$
2. the $$8$$th term in the binomial expansion of $$(x^3y-2x^2)^{10}$$
3. the $$12$$th term in the binomial expansion of $$\left(-\dfrac{5a}{b^7}-b\right)^{15}$$

Solution

1. We have $$a=p$$ and $$b=3q$$, and $$n=5$$ and $$k=4$$. Thus, the binomial coefficient of the $$4$$th term is $$\dbinom{5}{3}$$, the $$b$$-term is $$(3q)^3$$, and the $$a$$-term is $$p^2$$. The $$4$$th term is therefore given by

$\dbinom{5}{3}\cdot p^2\cdot (3q)^3=10\cdot p^2\cdot 3^3q^3=270p^2q^3 \nonumber$

1. In this case, $$a=x^3y$$ and $$b=-2x^2$$, and furthermore, $$n=10$$ and $$k=8$$. The binomial coefficient of the $$8$$th term is $$\dbinom{10}{7}$$, the $$b$$-term is $$(-2x^2)^7$$, and the $$a$$-term is $$(x^3y)^3$$. Therefore, the $$8$$th term is

$\dbinom{10}{7} \cdot (x^3y)^3\cdot (-2x^2)^7 = 120\cdot x^{9}y^3\cdot (-128)x^{14} = -15360\cdot x^{23}y^3 \nonumber$

1. Similarly, we obtain the $$12$$th term of $$\left(-\dfrac{5a}{b^7}-b\right)^{15}$$ as

\begin{aligned} \dbinom{15}{11}\cdot \left(-\dfrac{5a}{b^7}\right)^{4}\cdot (-b)^{11}&= 1365\cdot \dfrac{5^4a^4}{b^{28}} \cdot (-b^{11})\\&=1365 \cdot \dfrac{625\cdot a^4\cdot (-b^{11})}{b^{28}}\\ &=-853125 \cdot \dfrac{a^4}{b^{17}}\end{aligned} \nonumber

Here is a variation of the above problem.

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

Determine:

1. the $$x^4y^{12}$$-term in the binomial expansion of $$(5x^2+2y^3)^{6}$$
2. the $$x^{15}$$-term in the binomial expansion of $$(x^3-x)^{7}$$
3. the real part of the complex number $$(3+2i)^4$$

Solution

1. In this case we have $$a=5x^2$$ and $$b=2y^3$$. The term $$x^4y^{12}$$ can be rewritten as $$x^4y^{12}=(x^2)^2\cdot (y^3)^4$$, so that the full term $$\dbinom{n}{k-1}a^{n-k+1}b^{k-1}$$ (including the coefficients) is

$\dbinom{6}{4}\cdot (5x^2)^2\cdot(2y^3)^4=15\cdot 25x^4\cdot 16y^{12}=6000 \cdot x^4 y^{12} \nonumber$

1. The various powers of $$x$$ in $$(x^3-x)^{7}$$ (in the order in which they appear in the binomial expansion) are:

$(x^3)^7=x^{21}, \quad (x^3)^6\cdot x^1=x^{19}, \quad (x^3)^5\cdot x^2=x^{17}, \quad (x^3)^4\cdot x^3=x^{15}, \quad \dots \nonumber$

The last term is precisely the $$x^{15}$$-term, that is we take the fourth term, $$k=4$$. We obtain a total term (including the coefficients) of

$\dbinom{7}{3}\cdot (x^3)^4 \cdot (-x)^3=35\cdot x^{12}\cdot (-x)^3= -35\cdot x^{15} \nonumber$

1. Recall that $$i^n$$ is real when $$n$$ is even, and imaginary when $$n$$ is odd:

\begin{aligned} i^{1} &=i \\ i^{2} &=-1 \\ i^{3} &=-i \\ i^{4} &=1 \\ i^{5} &=i \\ i^{6} &=-1 \\ & \vdots \end{aligned} \nonumber

The real part of $$(3+2i)^4$$ is therefore given by the first, third, and fifth term of the binomial expansion:

\begin{aligned} \text{real part}&=\dbinom{4}{0}\cdot 3^4 \cdot (2i)^0+ \dbinom{4}{2}\cdot 3^2 \cdot(2i)^2+\dbinom{4}{4} \cdot 3^0\cdot (2i)^4\\ &= 1\cdot 81\cdot 1+6\cdot 9 \cdot 4 i^2+ 1\cdot 1\cdot 16 i^4\\&=81+216\cdot (-1)+16\cdot 1\\ &= 81-216+16\\& =-119\end{aligned} \nonumber

The real part of $$(3+2i)^4$$ is $$-119$$.

This page titled 25.2: Binomial Expansion is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.