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## 9.4 Binomial Theorem

### Learning Objectives

1. Evaluate expressions involving factorials.
2. Calculate binomial coefficients.
3. Expand powers of binomials using the binomial theorem.

## Factorials and the Binomial Coefficient

We begin by defining the factorialThe product of all natural numbers less than or equal to a given natural number, denoted n!. of a natural number n, denoted n!, as the product of all natural numbers less than or equal to n.

$n!=n(n−1)(n−2)⋯3⋅2⋅1$

For example,

$7!=7⋅6⋅5⋅4⋅3⋅2⋅1=5,040 Seven factorial5!=5⋅4⋅3⋅2⋅1=120 Five factorial3!=3⋅2⋅1=6 Three factorial1!=1=1 One factorial$

We define zero factorialThe factorial of zero is defined to be equal to 1; $0!=1.$ to be equal to 1,

$0!=1 Zero factorial$

The factorial of a negative number is not defined.

Note: On most modern calculators you will find a factorial function. Some calculators do not provide a button dedicated to it. However, it usually can be found in the menu system if one is provided.

The factorial can also be expressed using the following recurrence relation,

$n!=n(n−1)!$

For example, the factorial of 8 can be expressed as the product of 8 and 7!:

$8!=8⋅7!=8⋅7⋅6⋅5⋅4⋅3⋅2⋅1=40,320$

When working with ratios involving factorials, it is often the case that many of the factors cancel.

### Example 1

Evaluate: $12!6!$.

Solution:

$12!6!=12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1=12⋅11⋅10⋅9⋅8⋅7⋅6!6!=12⋅11⋅10⋅9⋅8⋅7=665,280$

The binomial coefficientAn integer that is calculated using the formula: $(nk)=n!k!(n−k)!.$, denoted $nCk=(nk)$, is read “n choose k” and is given by the following formula:

$nCk=(nk)=n!k!(n−k)!$

This formula is very important in a branch of mathematics called combinatorics. It gives the number of ways k elements can be chosen from a set of n elements where order does not matter. In this section, we are concerned with the ability to calculate this quantity.

### Example 2

Calculate: $(73)$.

Solution:

Use the formula for the binomial coefficent,

$(nk)=n!k!(n−k)!$

where $n=7$ and $k=3.$ After substituting, look for factors to cancel.

$(73)=7!3!(7−3)!=7!3! 4!=7⋅6⋅5⋅4!3! 4!=2106=35$

Note: Check the menu system of your calculator for a function that calculates this quantity. Look for the notation $nCk$ in the probability subsection.

Try this! Calculate: $(85)$.

Consider the following binomial raised to the 3rd power in its expanded form:

$(x+y)3=x3+3x2y+3xy2+y3$

Compare it to the following calculations,

$(30)=3!0!(3−0)!=3!1⋅3!=1(31)=3!1!(3−1)!=3⋅2!1⋅2!=3(32)=3!2!(3−2)!=3⋅2!2! 1!=3(33)=3!3!(3−3)!=3!3! 0!=1$

Notice that there appears to be a connection between these calculations and the coefficients of the expanded binomial. This observation is generalized in the next section.

## Binomial Theorem

Consider expanding $(x+2)5$:

$(x+2)5=(x+2)(x+2)(x+2)(x+2)(x+2)$

One quickly realizes that this is a very tedious calculation involving multiple applications of the distributive property. The binomial theoremDescribes the algebraic expansion of binomials raised to powers: $(x+y)n= Σk=0n (nk) xn−kyk.$ provides a method of expanding binomials raised to powers without directly multiplying each factor:

$(x+y)n=(n0)xny0+(n1)xn−1y1+(n2)xn−2y2+…+(nn−1)x1yn−1+(nn)x0yn$

More compactly we can write,

$(x+y)n= Σk=0n (nk) xn−kyk Binomial theorem$

### Example 3

Expand using the binomial theorem: $(x+2)5.$

Solution:

Use the binomial theorem where $n=5$ and $y=2.$

$(x+2)5=(50)x520+(51)x421+(52)x322+(53)x223+(54)x124+(55)x025$

Sometimes it is helpful to identify the pattern that results from applying the binomial theorem. Notice that powers of the variable x start at 5 and decrease to zero. The powers of the constant term start at 0 and increase to 5. The binomial coefficents can be calculated off to the side and are left to the reader as an exercise.

$(x+2)5=(50)x520+(51)x421+(52)x322+(53)x223+(54)x124+(55)x025=1x5×1+5x4×2+10x3×4+10x2×8+5x1×16+1×1×32=x5+10x4+40x3+80x2+80x+32$

Answer: $x5+10x4+40x3+80x2+80x+32$

The binomial may have negative terms, in which case we will obtain an alternating series.

### Example 4

Expand using the binomial theorem: $(u−2v)4.$

Solution:

Use the binomial theorem where $n=4$, $x=u$, and $y=−2v$ and then simplify each term.

$(u−2v)4=(40)u4(−2v)0+(41)u3(−2v)1+(42)u2(−2v)2+(43)u1(−2v)3+(44)u0(−2v)4=1×u4×1+4u3(−2v)+6u2(4v2)+4u(−8v3)+1×1(16v4)=u4−8u3v+24u2v2−32uv3+16v4$

Answer: $u4−8u3v+24u2v2−32uv3+16v4$

Try this! Expand using the binomial theorem: $(a2−3)4.$

Answer: $a8−12a6+54a4−108a2+81$

Next we study the coefficients of the expansions of $(x+y)n$ starting with $n=0$:

$(x+y)0=1(x+y)1=x+y(x+y)2=x2+2xy+y2(x+y)3=x3+3x2y+3xy2+y3(x+y)4=x4+4x3y+6x2y2+4xy3+y4$

Write the coefficients in a triangular array and note that each number below is the sum of the two numbers above it, always leaving a 1 on either end.

This is Pascal’s triangleA triangular array of numbers that correspond to the binomial coefficients.; it provides a quick method for calculating the binomial coefficients. Use this in conjunction with the binomial theorem to streamline the process of expanding binomials raised to powers. For example, to expand $(x−1)6$ we would need two more rows of Pascal’s triangle,

The binomial coefficients that we need are in blue. Use these numbers and the binomial theorem to quickly expand $(x−1)6$ as follows:

$(x−1)6=1x6(−1)0+6x5(−1)1+15x4(−1)2+20x3(−1)3+15x2(−1)4+6x(−1)5+1x0(−1)6=x6−6x5+15x4−20x3+15x2−6x+1$

### Example 5

Expand using the binomial theorem and Pascal’s triangle: $(2x−5)4.$

Solution:

From Pascal’s triangle we can see that when $n=4$ the binomial coefficients are 1, 4, 6, 4, and 1. Use these numbers and the binomial theorem as follows:

$(2x−5)4=1(2x)4(−5)0+4(2x)3(−5)1+6(2x)2(−5)2+4(2x)1(−5)3+1(2x)0(−5)4=16x4⋅1+4⋅8x3(−5)+6⋅4x2⋅25+4⋅2x(−125)+1⋅625=16x4−160x3+600x2−1,000x+625$

Answer: $16x4−160x3+600x2−1,000x+625$

### Key Takeaways

• To calculate the factorial of a natural number, multiply that number by all natural numbers less than it: $5!=5⋅4⋅3⋅2⋅1=120.$ Remember that we have defined $0!=1.$
• The binomial coefficients are the integers calculated using the formula: $(nk)=n!k!(n−k)!$.
• The binomial theorem provides a method for expanding binomials raised to powers without directly multiplying each factor: $(x+y)n= Σk=0n (nk) xn−kyk$.
• Use Pascal’s triangle to quickly determine the binomial coefficients.

### Part A: Factorials and the Binomial Coefficient

Evaluate.

1. $6!$

2. $4!$

3. $10!$

4. $9!$

5. $6!3!$

6. $8!4!$

7. $13!9!$

8. $15!10!$

9. $12!3! 7!$

10. $10!2! 5!$

11. $n!(n−2)!$

12. $(n+1)!(n−1)!$

1. $4! +3!$
2. $(4+3)!$
1. $4!−3!$
2. $(4−3)!$

Rewrite using factorial notation.

1. $1×2×3×4×5×6×7$

2. $1×2×3×4×5$

3. $15×14×13$

4. $10×9×8×7$

5. 13

6. $8×7$

7. $n(n−1)(n−2)$

8. $1×2×3×⋯×n×(n+1)$

Calculate the indicated binomial coefficient.

1. $(64)$
2. $(84)$
3. $(72)$
4. $(95)$
5. $(90)$
6. $(1312)$
7. $(n0)$
8. $(nn)$
9. $(n1)$
10. $(nn−1)$
11. $10C8$

12. $5C1$

13. $12C12$

14. $10C5$

15. $nCn−2$

16. $nCn−3$

### Part B: Binomial Theorem

Expand using the binomial theorem.

1. $(4x−3)3$

2. $(2x−5)3$

3. $(x2+y)3$

4. $(x+1y)3$

5. $(x+3)4$

6. $(x+5)4$

7. $(x−4)4$

8. $(x−2)4$

9. $(x+2y)4$

10. $(x3−y)4$

11. $(x+1)5$

12. $(x−3)5$

13. $(x−2)6$

14. $(x+1)6$

15. $(x−1)7$

16. $(x+1)7$

17. $(5x−1)4$

18. $(3x−2)4$

19. $(4u+v)4$

20. $(3u−v)4$

21. $(u−5v)5$

22. $(2u+3v)5$

23. $(a−b2)5$

24. $(a2+b2)4$

25. $(a2+b4)6$

26. $(a5+b2)5$

27. $(x+2)3$

28. $(x−2)4$

29. $(x−y)4$, $x,y≥0$

30. $(x+2y)5$, $x,y≥0$

31. $(x+y)7$

32. $(x+y)8$

33. $(x+y)9$

34. $(x−y)7$

35. $(x−y)8$

36. $(x−y)9$

### Part C: Discussion Board

1. Determine the factorials of the integers 5, 10, 15, 20, and 25. Which grows faster, the common exponential function $an=10n$ or the factorial function $an=n!$? Explain.

2. Research and discuss the history of the binomial theorem.

1. 720

2. 3,628,800

3. 120

4. 17,160

5. 15,840

6. $n2−n$

1. 30
2. 5,040
7. $7!$

8. $15!12!$

9. $13!12!$

10. $n!(n−3)!$

11. 15

12. 21

13. 1

14. 1

15. $n$

16. 45

17. 1

18. $n2−n2$

1. $64x3−144x2+108x−27$

2. $x38+3x2y4+3xy22+y3$

3. $x4+12x3+54x2+108x+81$

4. $x4−16x3+96x2−256x+256$

5. $x4+8x3y+24x2y2+32xy3+16y4$

6. $x5+5x4+10x3+10x2+5x+1$

7. $x6−12x5+60x4−160x3+240x2−192x+64$

8. $x7−7x6+21x5−35x4+35x3−21x2+7x−1$

9. $625x4−500x3+150x2−20x+1$

10. $256u4+256u3v+96u2v2+16uv3+v4$

11. $u5−25u4v+250u3v2−1,250u2v3+3,125uv4−3,125v5$
12. $a5−5a4b2+10a3b4−10a2b6+5ab8−b10$

13. $a12+6a10b4+15a8b8+20a6b12+15a4b16+6a2b20+b24$
14. $x3+32x2+6x+22$

15. $x2−4xxy+6xy−4yxy+y2$

16. $x7+7x6y+21x5y2+35x4y3+35x3y4+21x2y5+7xy6+y7$
17. $x9+9x8y+36x7y2+84x6y3+126x5y4+126x4y5+84x3y6+36x2y7+9xy8+y9$
18. $x8−8x7y+28x6y2−56x5y3+70x4y4−56x3y5+28x2y6−8xy7+y8$