Binomial is a polynomial that has 2 terms

(x + y)

^{0}= 1

(x + y)

^{1}= 1x + 1y

(x + y)

^{2}= 1x

^{2}+2xy + 1y

^{2}

(x + y)

^{3}= 1x

^{3}+ 3x

^{2}y + 3xy

^{2}+ 1y

^{3}

(x + y)

^{4}= 1x

^{4}+ 4x

^{3}y + 6x

^{2}y

^{2}+ 4xy

^{3}+ 1y

^{4}

(x + y)

^{5}= 1x

^{5}+ 5x

^{4}y + 10 x

^{3}y

^{2}+ 10 x

^{2}y

^{3}+ 5xy

^{4}+ 1y

^{5}

Notice the pattern that is developing in each expansion of the binomial.

1. The sum of the exponents of each term is “n”.

2. In each expression, there are n + 1 terms.

3. In each expansion, x and y have symmetric roles. The powers of x decreases by 1 in successive terms, whereas the powers of y increase by 1.

4. The coefficient increase and decrease in a symmetric pattern.

The coefficients of a binomial expansion are called binomial coefficients.

A display that contains only the coefficients of the terms in the expansions is called Pascal’s Triangle. The first triangle is Pascal’s Triangle, the second is using combinations to find the numbers.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

_{0}C

_{0}

_{1}C

_{0}

_{1}C

_{1}

_{2}C

_{0}

_{2}C

_{1}

_{2}C

_{2}

_{3}C

_{0}

_{3}C

_{1}

_{3}C

_{2}

_{3}C

_{3}

_{4}C

_{0}

_{4}C

_{1}

_{4}C

_{2}

_{4}C

_{3}

_{4}C

_{4}

_{5}C

_{0}

_{5}C

_{1}

_{5}C

_{2}

_{5}C

_{3}

_{5}C

_{4}

_{5}C

_{5}

Therefore in general, The Binomial Theorem: In the expansion of

(x + y)

^{n}=

_{n}C

_{0 }(x)

^{n}y

^{0}+

_{n}C

_{1 }(x)

^{n-1}y

^{1}+

_{n}C

_{2}(x)

^{n-2}y

^{2}+ ... +

_{n}C

_{n-1}(x)

^{1}y

^{n-1}+

_{n}C

_{n}(x)

^{0}y

^{n}

The coefficient of x

^{n-r}y

^{r}is

The symbol ( ) is often used in place of

_{n}C

_{r}to denote binomial coefficients.

**Example 1: Find the following pairs of binomial coefficients**.

a.

_{7}C

_{0},

_{7}C

_{7}= 1 , 1

b.

_{7}C

_{1},

_{7}C

_{6}= 7 , 7

c.

_{8}C

_{2},

_{8}C

_{6}= 28 , 28 same answer

What can you conclude?

_{n}C

_{r}=

_{n}C

_{n - r}

This shows the symmetric property of binomial coefficients.

(2x - 3y)

^{3}=

_{3}C

_{0}(2x)

^{3}(-3y)

^{0}+

_{3}C

_{1}(2x)

^{2}(-3y)

^{1}+

_{3}C

_{2}(2x)

^{1}(-3y)

^{2}+

_{3}C

_{3}(2x)

^{0}(-3y)

^{3}

= (1)(8x

^{3})(1) + (3)(4x

^{2})(-3y) + (3)(2x)(9y

^{2}) + (1)(1)(-27y

^{3})

= 8x

^{3}- 36x

^{2}y + 54xy

^{2}- 27y

^{3}

Remember to put the negative sign in the parenthesis so when you raise it to a power, the answer is appropriately positive or negative!!

Expand (2x - y)

^{5}= (2x)

^{5}+ 5(2x)

^{4}(-y) + 10(2x)

^{3}(-y)

^{2}+ 10(2x)

^{2}(-y)

^{3}+ 5(2x)(-y)

^{4}+ (-y)

^{5}

= 32x

^{5}- 80x

^{4}y + 80 x

^{3}y

^{2}- 40 x

^{2}y

^{3}+ 10xy

^{4}- y

^{5}

Do you notice a pattern for (x - y)

^{n}?

^{}

_{}