maths.free › Precalculus › 11. Sequences, Probability and Counting Theory › The binomial theorem
The binomial theorem
Expanding (a + b)ⁿ with Pascal's triangle and binomial coefficients.
(a + b)ⁿ expands to Σ C(n, k) aⁿ⁻ᵏ bᵏ: the coefficients are the row of Pascal's triangle, which are the same “n choose k” numbers that count selections. So a counting problem and an expansion problem are the same problem in two costumes.
Binomial Theorem
- Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)
Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.
To find the coefficients of the terms, we write our expansion 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 that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.
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.
To find the coefficients of the expansion of the binomial \({(a+b)}^{n}\) , go to the row that has the value n as a second entry.
Example
Try it.
Use Pascal’s Triangle to expand \({(x+y)}^{6}\) .
Solution
| Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 6. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of x decreasing from 6 to 0, and the power of y increasing from 0 to 6. | |
| Binomial expansion of \({(x+y)}^{6}\) . |
Example
Try it.
Use Pascal’s Triangle to expand \({(x+3)}^{5}\) .
Solution
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 5. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of x decreasing from 5 to 0, and the power of 3 increasing from 0 to 5. | |
| Binomial expansion of \({(x+3)}^{5}\) . |
Example
Try it.
Use Pascal’s Triangle to expand \({(3x-2)}^{4}\) .
Solution
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 4. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of (3x) decreasing from 4 to 0, and the power of (-2) increasing from 0 to 4. | |
| Binomial expansion of \({(3x-2)}^{4}\) . |
Use Pascal’s Triangle to expand a binomial.
Try it.
Use Pascal’s Triangle to expand \({(a+b)}^{4}\) .
Try it.
Use Pascal’s Triangle to expand \({(y+3)}^{5}\) .
Try it.
Use Pascal’s Triangle to expand \({(2x-5)}^{3}\) .
Identifying Binomial Coefficients
In Counting Principles, we studied combinations. In the shortcut to finding \({(x+y)}^{n},\) we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation \((\begin{array}{l}n \\ r\end{array})\) instead of \(C(n,r),\) but it can be calculated in the same way. So\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]
The combination \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient. An example of a binomial coefficient is \((\begin{array}{l}5 \\ 2\end{array})=C(5,2)=10.\)
Example
Try it.
Find each binomial coefficient.
- ⓐ \((\begin{array}{l}5 \\ 3\end{array})\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})\)
Solution
Use the formula to calculate each binomial coefficient. You can also use the \(n{C}_{r}\) function on your calculator.
\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]- ⓐ \((\begin{array}{l}5 \\ 3\end{array})=\frac{5!}{3!(5-3)!}=\frac{5⋅4⋅3!}{3!2!}=10\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})=\frac{9!}{2!(9-2)!}=\frac{9⋅8⋅7!}{2!7!}=36\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})=\frac{9!}{7!(9-7)!}=\frac{9⋅8⋅7!}{7!2!}=36\)
Using the Binomial Theorem
When we expand \({(x+y)}^{n}\) by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand \({(x+y)}^{52},\) we might multiply \((x+y)\) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.
\[\begin{array}{l}{(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]First, let’s examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the two exponents is \(n\) for each term.
Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:
\[(\begin{array}{l}n \\ 0\end{array}),(\begin{array}{l}n \\ 1\end{array}),(\begin{array}{l}n \\ 2\end{array}),...,(\begin{array}{l}n \\ n\end{array}).\]These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.
\[\begin{array}{ll}{(x+y)}^{n} & =\sum _{k=0}^{n}(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k} \\ & ={x}^{n}+(\begin{array}{l}n \\ 1\end{array}){x}^{n-1}y+(\begin{array}{l}n \\ 2\end{array}){x}^{n-2}{y}^{2}+...+(\begin{array}{l}n \\ n-1\end{array})x{y}^{n-1}+{y}^{n}\end{array}\]Another way to see the coefficients is to examine the expansion of a binomial in general form, \(x+y,\) to successive powers 1, 2, 3, and 4.
\[\begin{array}{l}{(x+y)}^{1}=x+y \\ {(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]Can you guess the next expansion for the binomial \({(x+y)}^{5}?\)
See , which illustrates the following:
- There are \(n+1\) terms in the expansion of \({(x+y)}^{n}.\)
- The degree (or sum of the exponents) for each term is \(n.\)
- The powers on \(x\) begin with \(n\) and decrease to 0.
- The powers on \(y\) begin with 0 and increase to \(n.\)
- The coefficients are symmetric.
- Introduce \({x}^{5},\) and then for each successive term reduce the exponent on \(x\) by 1 until \({x}^{0}=1\) is reached.
- Introduce \({y}^{0}=1,\) and then increase the exponent on \(y\) by 1 until \({y}^{5}\) is reached. \[{x}^{5},\ {x}^{4}y,\ {x}^{3}{y}^{2},\ {x}^{2}{y}^{3},\ x{y}^{4},\ {y}^{5}\]
Condensed — the full section is in OpenStax College Algebra 2e.
Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as \({(x+2y)}^{16}\) can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.
Note the pattern of coefficients in the expansion of \({(x+y)}^{5}.\)
\[{(x+y)}^{5}={x}^{5}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}y+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{y}^{3}+(\begin{array}{l}5 \\ 4\end{array})x{y}^{4}+{y}^{5}\]The second term is \((\begin{array}{l}5 \\ 1\end{array}){x}^{4}y.\) The third term is \((\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}.\) We can generalize this result.
\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]Example
Try it.
Find the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.
Solution
Because we are looking for the tenth term, \(r+1=10,\) we will use \(r=9\) in our calculations.
\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]\[(\begin{array}{l}16 \\ 9\end{array}){x}^{16-9}{(2y)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}\]Key Equations
| Binomial Theorem | \({(x+y)}^{n}=\sum _{k-0}^{n}(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k}\) |
| \((r+1)th\) term of a binomial expansion | \((\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\) |
Key Concepts
- \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient and is equal to \(C(n,r).\) See .
- The Binomial Theorem allows us to expand binomials without multiplying. See .
- We can find a given term of a binomial expansion without fully expanding the binomial. See .
Binomial Theorem
- Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)
Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.
To find the coefficients of the terms, we write our expansion 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 that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.
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.
To find the coefficients of the expansion of the binomial \({(a+b)}^{n}\) , go to the row that has the value n as a second entry.
Example
Try it.
Use Pascal’s Triangle to expand \({(x+y)}^{6}\) .
Solution
| Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 6. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of x decreasing from 6 to 0, and the power of y increasing from 0 to 6. | |
| Binomial expansion of \({(x+y)}^{6}\) . |
Example
Try it.
Use Pascal’s Triangle to expand \({(x+3)}^{5}\) .
Solution
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 5. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of x decreasing from 5 to 0, and the power of 3 increasing from 0 to 5. | |
| Binomial expansion of \({(x+3)}^{5}\) . |
Example
Try it.
Use Pascal’s Triangle to expand \({(3x-2)}^{4}\) .
Solution
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 4. | |
| Write the expansion with the coefficients. | |
| Fill in the variable with the power of (3x) decreasing from 4 to 0, and the power of (-2) increasing from 0 to 4. | |
| Binomial expansion of \({(3x-2)}^{4}\) . |
Use Pascal’s Triangle to expand a binomial.
Try it.
Use Pascal’s Triangle to expand \({(a+b)}^{4}\) .
Try it.
Use Pascal’s Triangle to expand \({(y+3)}^{5}\) .
Try it.
Use Pascal’s Triangle to expand \({(2x-5)}^{3}\) .
Identifying Binomial Coefficients
In Counting Principles, we studied combinations. In the shortcut to finding \({(x+y)}^{n},\) we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation \((\begin{array}{l}n \\ r\end{array})\) instead of \(C(n,r),\) but it can be calculated in the same way. So\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]
The combination \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient. An example of a binomial coefficient is \((\begin{array}{l}5 \\ 2\end{array})=C(5,2)=10.\)
Example
Try it.
Find each binomial coefficient.
- ⓐ \((\begin{array}{l}5 \\ 3\end{array})\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})\)
Solution
Use the formula to calculate each binomial coefficient. You can also use the \(n{C}_{r}\) function on your calculator.
\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]- ⓐ \((\begin{array}{l}5 \\ 3\end{array})=\frac{5!}{3!(5-3)!}=\frac{5⋅4⋅3!}{3!2!}=10\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})=\frac{9!}{2!(9-2)!}=\frac{9⋅8⋅7!}{2!7!}=36\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})=\frac{9!}{7!(9-7)!}=\frac{9⋅8⋅7!}{7!2!}=36\)
Using the Binomial Theorem
When we expand \({(x+y)}^{n}\) by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand \({(x+y)}^{52},\) we might multiply \((x+y)\) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.
\[\begin{array}{l}{(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]First, let’s examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the two exponents is \(n\) for each term.
Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:
\[(\begin{array}{l}n \\ 0\end{array}),(\begin{array}{l}n \\ 1\end{array}),(\begin{array}{l}n \\ 2\end{array}),...,(\begin{array}{l}n \\ n\end{array}).\]These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.
\[\begin{array}{ll}{(x+y)}^{n} & =\sum _{k=0}^{n}(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k} \\ & ={x}^{n}+(\begin{array}{l}n \\ 1\end{array}){x}^{n-1}y+(\begin{array}{l}n \\ 2\end{array}){x}^{n-2}{y}^{2}+...+(\begin{array}{l}n \\ n-1\end{array})x{y}^{n-1}+{y}^{n}\end{array}\]Another way to see the coefficients is to examine the expansion of a binomial in general form, \(x+y,\) to successive powers 1, 2, 3, and 4.
\[\begin{array}{l}{(x+y)}^{1}=x+y \\ {(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]Can you guess the next expansion for the binomial \({(x+y)}^{5}?\)
See , which illustrates the following:
- There are \(n+1\) terms in the expansion of \({(x+y)}^{n}.\)
- The degree (or sum of the exponents) for each term is \(n.\)
- The powers on \(x\) begin with \(n\) and decrease to 0.
- The powers on \(y\) begin with 0 and increase to \(n.\)
- The coefficients are symmetric.
- Introduce \({x}^{5},\) and then for each successive term reduce the exponent on \(x\) by 1 until \({x}^{0}=1\) is reached.
- Introduce \({y}^{0}=1,\) and then increase the exponent on \(y\) by 1 until \({y}^{5}\) is reached. \[{x}^{5},\ {x}^{4}y,\ {x}^{3}{y}^{2},\ {x}^{2}{y}^{3},\ x{y}^{4},\ {y}^{5}\]
Condensed — the full section is in OpenStax Precalculus 2e.
Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as \({(x+2y)}^{16}\) can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.
Note the pattern of coefficients in the expansion of \({(x+y)}^{5}.\)
\[{(x+y)}^{5}={x}^{5}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}y+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{y}^{3}+(\begin{array}{l}5 \\ 4\end{array})x{y}^{4}+{y}^{5}\]The second term is \((\begin{array}{l}5 \\ 1\end{array}){x}^{4}y.\) The third term is \((\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}.\) We can generalize this result.
\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]Example
Try it.
Find the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.
Solution
Because we are looking for the tenth term, \(r+1=10,\) we will use \(r=9\) in our calculations.
\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]\[(\begin{array}{l}16 \\ 9\end{array}){x}^{16-9}{(2y)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}\]Key Equations
| Binomial Theorem | \({(x+y)}^{n}=\sum _{k-0}^{n}(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k}\) |
| \((r+1)th\) term of a binomial expansion | \((\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\) |
Key Concepts
- \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient and is equal to \(C(n,r).\) See .
- The Binomial Theorem allows us to expand binomials without multiplying. See .
- We can find a given term of a binomial expansion without fully expanding the binomial. See .
Worked example: expand (a + b)^5
Step by step
- \left(a + b\right)^{5}
Start from the expression.
- a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5}
Distribute (multiply out) the products.
Reveal the answer
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Use Pascal’s Triangle to expand \({(x+y)}^{6}\) .
Reveal the answer
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 6. Write the expansion with the coefficients. Fill in the variable with the power of x decreasing from 6 to 0, and the power of y increasing from 0 to 6. Binomial expansion of \({(x+y)}^{6}\) . -
Use Pascal’s Triangle to expand \({(x+3)}^{5}\) .
Reveal the answer
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 5.Write the expansion with the coefficients. Fill in the variable with the power of x decreasing from 5 to 0, and the power of 3 increasing from 0 to 5. Binomial expansion of \({(x+3)}^{5}\) . -
Use Pascal’s Triangle to expand \({(3x-2)}^{4}\) .
Reveal the answer
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 4.Write the expansion with the coefficients. Fill in the variable with the power of (3x) decreasing from 4 to 0, and the power of (-2) increasing from 0 to 4. Binomial expansion of \({(3x-2)}^{4}\) . -
Use Pascal’s Triangle to expand \({(a+b)}^{4}\) .
-
Use Pascal’s Triangle to expand \({(y+3)}^{5}\) .
-
Use Pascal’s Triangle to expand \({(2x-5)}^{3}\) .
-
Find each binomial coefficient.
- ⓐ \((\begin{array}{l}5 \\ 3\end{array})\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})\)
Reveal the answer
Use the formula to calculate each binomial coefficient. You can also use the \(n{C}_{r}\) function on your calculator.
\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]- ⓐ \((\begin{array}{l}5 \\ 3\end{array})=\frac{5!}{3!(5-3)!}=\frac{5⋅4⋅3!}{3!2!}=10\)
- ⓑ \((\begin{array}{l}9 \\ 2\end{array})=\frac{9!}{2!(9-2)!}=\frac{9⋅8⋅7!}{2!7!}=36\)
- ⓒ \((\begin{array}{l}9 \\ 7\end{array})=\frac{9!}{7!(9-7)!}=\frac{9⋅8⋅7!}{7!2!}=36\)
-
- ⓐ \((\begin{array}{l}7 \\ 3\end{array})\)
- ⓑ \((\begin{array}{l}11 \\ 4\end{array})\)
Reveal the answer
- ⓐ35
- ⓑ330
-
Write in expanded form.
- ⓐ \({(x+y)}^{5}\)
- ⓑ \({(3x-y)}^{4}\)
Reveal the answer
- ⓐSubstitute \(n=5\) into the formula. Evaluate the \(k=0\) through \(k=5\) terms. Simplify. \[\begin{array}{ll}{(x+y)}^{5} & =(\begin{array}{l}5 \\ 0\end{array}){x}^{5}{y}^{0}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}{y}^{1}+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{y}^{3}+(\begin{array}{l}5 \\ 4\end{array}){x}^{1}{y}^{4}+(\begin{array}{l}5 \\ 5\end{array}){x}^{0}{y}^{5} \\ {(x+y)}^{5} & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\end{array}\]
- ⓑSubstitute \(n=4\) into the formula. Evaluate the \(k=0\) through \(k=4\) terms. Notice that \(3x\) is in the place that was occupied by \(x\) and that \(-y\) is in the place that was occupied by \(y.\) So we substitute them. Simplify. \[\begin{array}{ll}{(3x-y)}^{4} & =(\begin{array}{l}4 \\ 0\end{array}){(3x)}^{4}{(-y)}^{0}+(\begin{array}{l}4 \\ 1\end{array}){(3x)}^{3}{(-y)}^{1}+(\begin{array}{l}4 \\ 2\end{array}){(3x)}^{2}{(-y)}^{2}+(\begin{array}{l}4 \\ 3\end{array}){(3x)}^{1}{(-y)}^{3}+(\begin{array}{l}4 \\ 4\end{array}){(3x)}^{0}{(-y)}^{4} \\ {(3x-y)}^{4} & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\end{array}\]
-
Write in expanded form.
- ⓐ \({(x-y)}^{5}\)
- ⓑ \({(2x+5y)}^{3}\)
Reveal the answer
- ⓐ \({x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}\)
- ⓑ \(8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}\)
-
Find the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.
Reveal the answer
Because we are looking for the tenth term, \(r+1=10,\) we will use \(r=9\) in our calculations.
\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]\[(\begin{array}{l}16 \\ 9\end{array}){x}^{16-9}{(2y)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}\] -
Find the sixth term of \({(3x-y)}^{9}\) without fully expanding the binomial.
Reveal the answer
\(-10,206{x}^{4}{y}^{5}\)
-
What is a binomial coefficient, and how it is calculated?
Reveal the answer
A binomial coefficient is an alternative way of denoting the combination \(C(n,r).\) It is defined as \((\begin{array}{l}n \\ r\end{array})=\ C(n,r)\ =\frac{n!}{r!(n-r)!}.\)
-
What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?
-
What is the Binomial Theorem and what is its use?
Reveal the answer
The Binomial Theorem is defined as \({(x+y)}^{n}=\sum _{k=0}^{n}(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k}\) and can be used to expand any binomial.
-
When is it an advantage to use the Binomial Theorem? Explain.
-
\((\begin{array}{l}6 \\ 2\end{array})\)
Reveal the answer
15
-
\((\begin{array}{l}5 \\ 3\end{array})\)
-
\((\begin{array}{l}7 \\ 4\end{array})\)
Reveal the answer
35
-
\((\begin{array}{l}9 \\ 7\end{array})\)
-
\((\begin{array}{l}10 \\ 9\end{array})\)
Reveal the answer
10
-
\((\begin{array}{l}25 \\ 11\end{array})\)
-
\((\begin{array}{l}17 \\ 6\end{array})\)
Reveal the answer
12,376
-
\((\begin{array}{l}200 \\ 199\end{array})\)
-
\({(4a-b)}^{3}\)
Reveal the answer
\(64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}\)
-
\({(5a+2)}^{3}\)
-
\({(3a+2b)}^{3}\)
Reveal the answer
\(27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}\)
-
\({(2x+3y)}^{4}\)
-
\({(4x+2y)}^{5}\)
Reveal the answer
\(1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}\)
-
\({(3x-2y)}^{4}\)
-
\({(4x-3y)}^{5}\)
Reveal the answer
\(1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}\)
-
\({(\frac{1}{x}+3y)}^{5}\)
-
\({({x}^{-1}+2{y}^{-1})}^{4}\)
Reveal the answer
\(\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}\)
-
\({(\sqrt{x}-\sqrt{y})}^{5}\)
-
\({(a+b)}^{17}\)
Reveal the answer
\({a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}\)
-
\({(x-1)}^{18}\)
-
\({(a-2b)}^{15}\)
Reveal the answer
\({a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}\)
-
\({(x-2y)}^{8}\)
-
\({(3a+b)}^{20}\)
Reveal the answer
\(3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}\)
-
\({(2a+4b)}^{7}\)
Symbols used here
Add a_k for k = 1 up to n.
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.
Inequalities that allow equality; < and > exclude it.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
Number of k-element subsets of n things: n!/(k!(n−k)!).
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
i² = −1.
The usual name for an angle.
The exponent b must be raised to for x; ln uses base e.
A quantity with magnitude and direction; a column of numbers.
How to: The binomial theorem
- Apply the Binomial Theorem.
- Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)
- There are
- The degree (or sum of the exponents) for each term is
- The powers on
- The powers on
- The coefficients are symmetric.
- Introduce
Questions people ask
What is a function, really?
A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.
Why do we need complex numbers?
Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.
Try your own
Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
More in Precalculus
Complex numbersPolynomial functionsRational functionsSequences and seriesConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates