1. Learning objectives

Learning objectives

Use Pascal’s Triangle to expand a binomial

Evaluate a binomial coefficient

Use the Binomial Theorem to expand a binomial

2. Use Pascal’s Triangle to Expand a Binomial

Use Pascal’s Triangle to Expand a Binomial

In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded ${(a+b)}^{2}.$

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

To expand ${(a+b)}^{3},$ we recognize that this is ${(a+b)}^{2}(a+b)$ and multiply.

\begin{array}{c} {(a+b)}^{3} \\ {(a+b)}^{2}(a+b) \\ ({a}^{2}+2ab+{b}^{2})(a+b) \\ {a}^{3}+2{a}^{2}b+a{b}^{2}+{a}^{2}b+2a{b}^{2}+{b}^{3} \\ {a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3} \\ {(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3} \\ \end{array}

To find a method that is less tedious that will work for higher expansions like ${(a+b)}^{7},$ we again look for patterns in some expansions.

Number of terms · First term · Last term ${(a+b)}^{1}=a+b$ · 2 · ${a}^{1}$ · ${b}^{1}$ ${(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$ · 3 · ${a}^{2}$ · ${b}^{2}$ ${(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$ · 4 · ${a}^{3}$ · ${b}^{3}$ ${(a+b)}^{4}={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}$ · 5 · ${a}^{4}$ · ${b}^{4}$ ${(a+b)}^{5}={a}^{5}+5{a}^{4}b+10{a}^{3}{b}^{2}+10{a}^{2}{b}^{3}+5a{b}^{4}+{b}^{5}$ · 6 · ${a}^{5}$ · ${b}^{5}$ ${(a+b)}^{n}$ · $+1$ · ${a}^{n}$ · ${b}^{n}$

Notice the first and last terms show only one variable. Recall that ${a}^{0}=1,$ so we could rewrite the first and last terms to include both variables. For example, we could expand ${(a+b)}^{3}$ to show each term with both variables.

Generally, we don’t show the zero exponents, just as we usually write x rather than 1*x*.

The number of terms is $n+1.$
The first term is ${a}^{n}$ and the last term is ${b}^{n}.$
The exponents on a decrease by one on each term going left to right.
The exponents on b increase by one on each term going left to right.
The sum of the exponents on any term is n.

Let’s look at an example to highlight the last three patterns.

From the patterns we identified, we see the variables in the expansion of ${(a+b)}^{n},$ would be

{(a+b)}^{n}={a}^{n}+\mathrm{\_\_\_}{a}^{n-1}{b}^{1}+\mathrm{\_\_\_}{a}^{n-2}{b}^{2}+...+\mathrm{\_\_\_}{a}^{1}{b}^{n-1}+{b}^{n}.

To find the coefficients of the terms, we write our expansions again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients.

The array to the right is called Pascal’s Triangle. Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

This triangle gives the coefficients of the terms when we expand binomials.

In the next example, we will use this triangle and the patterns we recognized to expand the binomial.

3. Example 12.31

Example 12.31

Use Pascal’s Triangle to expand ${(x+y)}^{6}.$

Solution

We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of x will start at six and decrease to one. The nonzero exponents of y will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, $a=x$ and $b=y.$

\begin{array}{l} {(a+b)}^{n}={a}^{n}+\mathrm{\_\_\_}{a}^{n-1}{b}^{1}+\mathrm{\_\_\_}{a}^{n-2}{b}^{2}+...+\mathrm{\_\_\_}{a}^{1}{b}^{n-1}+{b}^{n} \\ {(x+y)}^{6}={x}^{6}+\mathrm{\_\_\_}{x}^{5}{y}^{1}+\mathrm{\_\_\_}{x}^{4}{y}^{2}+\mathrm{\_\_\_}{x}^{3}{y}^{3}+\mathrm{\_\_\_}{x}^{2}{y}^{4}+\mathrm{\_\_\_}{x}^{1}{y}^{5}+{y}^{6} \\ \end{array}

In the next example we want to expand a binomial with one variable and one constant. We need to identify the a and b to carefully apply the pattern.

4. Example 12.32

Example 12.32

Use Pascal’s Triangle to expand ${(x+3)}^{5}.$

Solution

We identify the a and b of the pattern.

In our pattern, $a=x$ and $b=3.$

We know the variables for this expansion will follow the pattern we identified. The sum of the exponents in each term will be five.

\begin{array}{c} {(a+b)}^{n}={a}^{n}+\mathrm{\_\_\_}{a}^{n-1}{b}^{1}+\mathrm{\_\_\_}{a}^{n-2}{b}^{2}+...+\mathrm{\_\_\_}{a}^{1}{b}^{n-1}+{b}^{n} \\ {(x+3)}^{5}={x}^{5}+\mathrm{\_\_\_}{x}^{4}\cdot{3}^{1}+\mathrm{\_\_\_}{x}^{3}\cdot{3}^{2}+\mathrm{\_\_\_}{x}^{2}\cdot{3}^{3}+\mathrm{\_\_\_}{x}^{1}\cdot{3}^{4}+{3}^{5} \\ \end{array}

In the next example, the binomial is a difference and the first term has a constant times the variable. Once we identify the a and b of the pattern, we must once again carefully apply the pattern.

5. Example 12.33

Example 12.33

Use Pascal’s Triangle to expand ${(3x-2)}^{4}.$

Solution

We identify the a and b of the pattern.

In our pattern, $a=3x$ and $b=-2.$

\begin{array}{l} {(a+b)}^{n}={a}^{n}+\mathrm{\_\_\_}{a}^{n-1}{b}^{1}+\mathrm{\_\_\_}{a}^{n-2}{b}^{2}+...+\mathrm{\_\_\_}{a}^{1}{b}^{n-1}+{b}^{n} \\ {(3x-2)}^{4}=1\cdot{(3x)}^{4}+4{(3x)}^{3}{(-2)}^{1}+6{(3x)}^{2}{(-2)}^{2}+4{(3x)}^{1}{(-2)}^{3}+1\cdot{(-2)}^{4} \\ {(3x-2)}^{4}=81{x}^{4}+4(27{x}^{3})(-2)+6(9{x}^{2})(4)+4(3x)(-8)+1\cdot{}16 \\ {(3x-2)}^{4}=81{x}^{4}-216{x}^{3}+216{x}^{2}-96x+16 \\ \end{array}

6. Evaluate a Binomial Coefficient

Evaluate a Binomial Coefficient

While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more factorial notation. This notation is not only used to expand binomials, but also in the study and use of probability.

To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation $(\begin{array}{l} n \\ r \\ \end{array})$ which is called a binomial coefficient. We read $(\begin{array}{l} n \\ r \\ \end{array})$ as “n choose r” or “n taken r at a time”.

A binomial coefficient $(\begin{array}{l} n \\ r \\ \end{array}),$ where r and n are integers with $0\le{}r\le{}n,$ is defined as

(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!}

We read $(\begin{array}{l} n \\ r \\ \end{array})$ as “n choose r” or “n taken r at a time”.

7. Example 12.34

Example 12.34

Evaluate: ⓐ $(\begin{array}{l} 5 \\ 1 \\ \end{array})$ ⓑ $(\begin{array}{l} 7 \\ 7 \\ \end{array})$ ⓒ $(\begin{array}{l} 4 \\ 0 \\ \end{array})$ ⓓ $(\begin{array}{l} 8 \\ 5 \\ \end{array}).$

Solution

ⓐ We will use the definition of a binomial coefficient, $(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!}.$

$\,(\begin{array}{l} 5 \\ 1 \\ \end{array})$

Use the definition, $(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!},$ where $n=5,r=1.$ · $\,\frac{5!}{1!(5-1)!}$ Simplify. · $\,\frac{5!}{1!(4)!}$ Rewrite $5!\,\mathrm{as}\,5\cdot{}4!$ · $\,\frac{5\cdot{}4!}{1!\cdot{}4!}$ Simplify, by removing common factors. · $\,\frac{5\cdot{}4!}{1\cdot{}4!}$ Simplify. · $\,5$

$\,(\begin{array}{l} 5 \\ 1 \\ \end{array})=5$

ⓑ

$\,(\begin{array}{l} 7 \\ 7 \\ \end{array})$

Use the definition, $(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!},$ where $n=7,r=7.$ · $\,\frac{7!}{7!(7-7)!}$ Simplify. · $\,\frac{7!}{7!(0)!}$ Simplify. Remember $0!=1.$ · $\,1$

$\,(\begin{array}{l} 7 \\ 7 \\ \end{array})=1$

ⓒ

$\,(\begin{array}{l} 4 \\ 0 \\ \end{array})$

Use the definition, $(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!},$ where $n=4,r=0.$ · $\,\frac{4!}{0!(4-0)!}$ Simplify. · $\,\frac{4!}{0!(4)!}$ Simplify. · $\,1$

$\,(\begin{array}{l} 4 \\ 0 \\ \end{array})=1$

ⓓ

$\,(\begin{array}{l} 8 \\ 5 \\ \end{array})$

Use the definition, $(\begin{array}{l} n \\ r \\ \end{array})=\frac{n!}{r!(n-r)!},$ where $n=8,r=5.$ · $\,\frac{8!}{5!(8-5)!}$ Simplify. · $\,\frac{8!}{5!(3)!}$ Rewrite $8!\,\mathrm{as}\,8\cdot{}7\cdot{}6\cdot{}5!$ and remove common factors. · $\,\frac{8\cdot{}7\cdot{}6\cdot{}5!}{5!\cdot{}3\cdot{}2\cdot{}1}$ Simplify. · $\,56$

$\,(\begin{array}{l} 8 \\ 5 \\ \end{array})=56$

In the previous example, parts (a), (b), (c) demonstrate some special properties of binomial coefficients.

(\begin{array}{l} n \\ 1 \\ \end{array})=n\,(\begin{array}{l} n \\ n \\ \end{array})=1\,(\begin{array}{l} n \\ 0 \\ \end{array})=1

8. Use the Binomial Theorem to Expand a Binomial

Use the Binomial Theorem to Expand a Binomial

We are now ready to use the alternate method of expanding binomials. The Binomial Theorem uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term.

For any real numbers a and b, and positive integer n,

{(a+b)}^{n}=(\begin{array}{l} n \\ 0 \\ \end{array}){a}^{n}+(\begin{array}{l} n \\ 1 \\ \end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l} n \\ 2 \\ \end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l} n \\ r \\ \end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l} n \\ n \\ \end{array}){b}^{n}

9. Example 12.35

Example 12.35

Use the Binomial Theorem to expand ${(p+q)}^{4}.$

Solution

We identify the a and b of the pattern.

In our pattern, $a=p$ and $b=q.$

We use the Binomial Theorem.

{(a+b)}^{n}=(\begin{array}{l} n \\ 0 \\ \end{array}){a}^{n}+(\begin{array}{l} n \\ 1 \\ \end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l} n \\ 2 \\ \end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l} n \\ r \\ \end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l} n \\ n \\ \end{array}){b}^{n}

Substitute in the values $a=p,$ $b=q$ and $n=4.$

{(p+q)}^{4}=(\begin{array}{l} 4 \\ 0 \\ \end{array}){p}^{4}+(\begin{array}{l} 4 \\ 1 \\ \end{array}){p}^{4-1}{q}^{1}+(\begin{array}{l} 4 \\ 2 \\ \end{array}){p}^{4-2}{q}^{2}+(\begin{array}{l} 4 \\ 3 \\ \end{array}){p}^{4-3}{q}^{3}+(\begin{array}{l} 4 \\ 4 \\ \end{array}){q}^{4}

Simplify the exponents.

{(p+q)}^{4}=(\begin{array}{l} 4 \\ 0 \\ \end{array}){p}^{4}+(\begin{array}{l} 4 \\ 1 \\ \end{array}){p}^{3}{q}^{}+(\begin{array}{l} 4 \\ 2 \\ \end{array}){p}^{2}{q}^{2}+(\begin{array}{l} 4 \\ 3 \\ \end{array}){p}^{}{q}^{3}+(\begin{array}{l} 4 \\ 4 \\ \end{array}){q}^{4}

Evaluate the coefficients. Remember, $(\begin{array}{l} n \\ 1 \\ \end{array})=n,$ $(\begin{array}{l} n \\ n \\ \end{array})=1,$ $(\begin{array}{l} n \\ 0 \\ \end{array})=1.$

\begin{array}{l} {(p+q)}^{4}=1{p}^{4}+4{p}^{3}{q}^{1}+\frac{4!}{2!(2)!}{p}^{2}{q}^{2}+\frac{4!}{3!(4-3)!}{p}^{1}{q}^{3}+1{q}^{4} \\ {(p+q)}^{4}={p}^{4}+4{p}^{3}{q}^{}+6{p}^{2}{q}^{2}+4{p}^{}{q}^{3}+{q}^{4} \\ \end{array}

Notice that when we expanded ${(p+q)}^{4}$ in the last example, using the Binomial Theorem, we got the same coefficients we would get from using Pascal’s Triangle.

The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern.

10. Example 12.36

Example 12.36

Use the Binomial Theorem to expand ${(x-2)}^{5}.$

Solution

We identify the a and b of the pattern.

In our pattern, $a=x$ and $b=-2.$

We use the Binomial Theorem.

{(a+b)}^{n}=(\begin{array}{l} n \\ 0 \\ \end{array}){a}^{n}+(\begin{array}{l} n \\ 1 \\ \end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l} n \\ 2 \\ \end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l} n \\ r \\ \end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l} n \\ n \\ \end{array}){b}^{n}

Substitute in the values $a=x,$ $b=-2,$ and $n=5.$

{(x-2)}^{5}=(\begin{array}{l} 5 \\ 0 \\ \end{array}){x}^{5}+(\begin{array}{l} 5 \\ 1 \\ \end{array}){x}^{5-1}{(-2)}^{1}+(\begin{array}{l} 5 \\ 2 \\ \end{array}){x}^{5-2}{(-2)}^{2}+(\begin{array}{l} 5 \\ 3 \\ \end{array}){x}^{5-3}{(-2)}^{3}+(\begin{array}{l} 5 \\ 4 \\ \end{array}){x}^{5-4}{(-2)}^{4}+(\begin{array}{l} 5 \\ 5 \\ \end{array}){(-2)}^{5}

Simplify the exponents and evaluate the coefficients. Remember, $(\begin{array}{l} n \\ 1 \\ \end{array})=n,$ $(\begin{array}{l} n \\ n \\ \end{array})=1,$ $(\begin{array}{l} n \\ 0 \\ \end{array})=1.$

\begin{array}{l} {(x-2)}^{5}=(\begin{array}{l} 5 \\ 0 \\ \end{array}){x}^{5}+(\begin{array}{l} 5 \\ 1 \\ \end{array}){x}^{4}{(-2)}^{}+(\begin{array}{l} 5 \\ 2 \\ \end{array}){x}^{3}{(-2)}^{2}+(\begin{array}{l} 5 \\ 3 \\ \end{array}){x}^{2}{(-2)}^{3}+(\begin{array}{l} 5 \\ 4 \\ \end{array})x{(-2)}^{4}+(\begin{array}{l} 5 \\ 5 \\ \end{array}){(-2)}^{5} \\ {(x-2)}^{5}=1{x}^{5}+5(-2){x}^{4}+\frac{5!}{2!\cdot{}3!}{(-2)}^{2}{x}^{3}+\frac{5!}{3!\cdot{}2!}{(-2)}^{3}{x}^{2}+\frac{5!}{4!\cdot{}1!}{(-2)}^{4}x+1{(-2)}^{5} \\ {(x-2)}^{5}={x}^{5}+5(-2){x}^{4}+10\cdot{}4\cdot{x}^{3}+10(-8){x}^{2}+5\cdot{}16\cdot{}x+1(-32) \\ {(x-2)}^{5}={x}^{5}-10{x}^{4}+40{x}^{3}-80{x}^{2}+80x-32 \\ \end{array}

Things can get messy when both terms have a coefficient and a variable.

11. Example 12.37

Example 12.37

Use the Binomial Theorem to expand ${(2x-3y)}^{4}.$

Solution

We identify the a and b of the pattern.

In our pattern, $a=2x$ and $b=-3y.$

We use the Binomial Theorem.

{(a+b)}^{n}=(\begin{array}{l} n \\ 0 \\ \end{array}){a}^{n}+(\begin{array}{l} n \\ 1 \\ \end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l} n \\ 2 \\ \end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l} n \\ r \\ \end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l} n \\ n \\ \end{array}){b}^{n}

Substitute in the values $a=2x,$ $b=-3y$ and $n=4.$

{(2x-3y)}^{4}=(\begin{array}{l} 4 \\ 0 \\ \end{array}){(2x)}^{4}+(\begin{array}{l} 4 \\ 1 \\ \end{array}){(2x)}^{4-1}{(-3y)}^{1}+(\begin{array}{l} 4 \\ 2 \\ \end{array}){(2x)}^{4-2}{(-3y)}^{2}+(\begin{array}{l} 4 \\ 3 \\ \end{array}){(2x)}^{4-3}{(-3y)}^{3}+(\begin{array}{l} 4 \\ 4 \\ \end{array}){(-3y)}^{4}

Simplify the exponents.

{(2x-3y)}^{4}=(\begin{array}{l} 4 \\ 0 \\ \end{array}){(2x)}^{4}+(\begin{array}{l} 4 \\ 1 \\ \end{array}){(2x)}^{3}{(-3y)}^{1}+(\begin{array}{l} 4 \\ 2 \\ \end{array}){(2x)}^{2}{(-3y)}^{2}+(\begin{array}{l} 4 \\ 3 \\ \end{array}){(2x)}^{1}{(-3y)}^{3}+(\begin{array}{l} 4 \\ 4 \\ \end{array}){(-3y)}^{4}

Evaluate the coefficients. Remember, $(\begin{array}{l} n \\ 1 \\ \end{array})=n,\,(\begin{array}{l} n \\ n \\ \end{array})=1,\,(\begin{array}{l} n \\ 0 \\ \end{array})=1.$

${(2x-3y)}^{4}=1{(2x)}^{4}+4{(2x)}^{3}{(-3y)}^{1}+\frac{4!}{2!(2)!}{(2x)}^{2}{(-3y)}^{2}+\frac{4!}{3!(4-3)!}{(2x)}^{1}{(-3y)}^{3}+1{(-3y)}^{4}$

${(2x-3y)}^{4}=16{x}^{4}+4\cdot{}8{x}^{3}(-3y)+6(4{x}^{2})(9{y}^{2})+4(2x)(-27{y}^{3})+81{y}^{4}$

${(2x-3y)}^{4}=16{x}^{4}-96{x}^{3}y+216{x}^{2}{y}^{2}-216x{y}^{3}+81{y}^{4}$

The real beauty of the Binomial Theorem is that it gives a formula for any particular term of the expansion without having to compute the whole sum. Let’s look for a pattern in the Binomial Theorem.

Notice, that in each case the exponent on the b is one less than the number of the term. The $(r+1)st$ term is the term where the exponent of b is r. So we can use the format of the $(r+1)st$ term to find the value of a specific term.

The ${(r+1)}^{st}$ term in the expansion of ${(a+b)}^{n}$ is

(\begin{array}{l} n \\ r \\ \end{array}){a}^{n-r}{b}^{r}

12. Example 12.38

Example 12.38

Find the fourth term of ${(x+y)}^{7}.$

Solution

13. Example 12.39

Example 12.39

Find the coefficient of the ${x}^{6}$ term of ${(x+3)}^{9}.$

Solution

14. Section 12.4 Exercises

Section 12.4 Exercises

Use Pascal’s Triangle to Expand a Binomial

In the following exercises, expand each binomial using Pascal’s Triangle.

Evaluate a Binomial Coefficient

In the following exercises, evaluate.

Use the Binomial Theorem to Expand a Binomial

In the following exercises, expand each binomial.

In the following exercises, find the indicated term in the expansion of the binomial.

In the following exercises, find the coefficient of the indicated term in the expansion of the binomial.

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

15. Practice 1

Use Pascal’s Triangle to expand ${(x+y)}^{5}.$

Use Pascal’s Triangle to expand ${(p+q)}^{7}.$

Use Pascal’s Triangle to expand ${(x+2)}^{4}.$

16. Practice 2

Use Pascal’s Triangle to expand ${(x+1)}^{6}.$

Use Pascal’s Triangle to expand ${(2x-3)}^{4}.$

Use Pascal’s Triangle to expand ${(2x-1)}^{6}.$

17. Practice 3

Evaluate each binomial coefficient:

ⓐ $(\begin{array}{l} 6 \\ 1 \\ \end{array})$ ⓑ $(\begin{array}{l} 8 \\ 8 \\ \end{array})$ ⓒ $(\begin{array}{l} 5 \\ 0 \\ \end{array})$ ⓓ $(\begin{array}{l} 7 \\ 3 \\ \end{array}).$

Evaluate each binomial coefficient:

ⓐ $(\begin{array}{l} 2 \\ 1 \\ \end{array})$ ⓑ $(\begin{array}{l} 11 \\ 11 \\ \end{array})$ ⓒ $(\begin{array}{l} 9 \\ 0 \\ \end{array})$ ⓓ $(\begin{array}{l} 6 \\ 5 \\ \end{array}).$

Use the Binomial Theorem to expand ${(x+y)}^{5}.$

18. Practice 4

Use the Binomial Theorem to expand ${(m+n)}^{6}.$

Use the Binomial Theorem to expand ${(x-3)}^{5}.$

Use the Binomial Theorem to expand ${(y-1)}^{6}.$

19. Practice 5

Use the Binomial Theorem to expand ${(3x-2y)}^{5}.$

Use the Binomial Theorem to expand ${(4x-3y)}^{4}.$

Find the third term of ${(x+y)}^{6}.$

20. Practice 6

Find the fifth term of ${(a+b)}^{8}.$

Find the coefficient of the ${x}^{5}$ term of ${(x+4)}^{8}.$

${(a+b)}^{8}$

21. Practice 7

${(p+q)}^{9}$

${(a-b)}^{6}$

${(x+5)}^{3}$

22. Practice 8

${(y+1)}^{7}$

${(z-2)}^{6}$

${(3x-1)}^{5}$

23. Practice 9

${(3x-5)}^{3}$

${(3x+5y)}^{3}$

ⓐ $(\begin{array}{l} 7 \\ 1 \\ \end{array})$ ⓑ $(\begin{array}{l} 4 \\ 4 \\ \end{array})$ ⓒ $(\begin{array}{l} 3 \\ 0 \\ \end{array})$ ⓓ $(\begin{array}{l} 10 \\ 8 \\ \end{array})$

24. Practice 10

ⓐ $(\begin{array}{l} 4 \\ 1 \\ \end{array})$ ⓑ $(\begin{array}{l} 5 \\ 5 \\ \end{array})$ ⓒ $(\begin{array}{l} 8 \\ 0 \\ \end{array})$ ⓓ $(\begin{array}{l} 11 \\ 9 \\ \end{array})$

${(m+n)}^{5}$

${(s+t)}^{7}$

25. Practice 11

${(y-3)}^{4}$

${(q-4)}^{3}$

${(5x-2y)}^{4}$

26. Practice 12

${(3x+4y)}^{5}$

Fifth term of ${(a+b)}^{9}$

Seventh term of ${(x-y)}^{11}$

27. Practice 13

${p}^{5}{q}^{4}$ term of ${(3p+q)}^{9}$

In your own words, explain the pattern of exponents for each variable in the expansion of.

In your own words, explain how to find a specific term in the expansion of a binomial without expanding the whole thing. Use an example to help explain.