25.2: Binomial Expansion

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 $b=2y^3$:

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

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

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

3. 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).

4. 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:

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

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

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

2. 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

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

3. 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

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

2. 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

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

3. 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 $-119$.