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  • 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

The next expansion would be

( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5 .

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle , shown in [link] .

Pascal's Triangle

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3 rd row, flank the ends of the rows with 1’s, and add 1 + 1 to find the middle number, 2. In the n th row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents

The binomial theorem

The Binomial Theorem    is a formula that can be used to expand any binomial.

( x + y ) n = k = 0 n ( n k ) x n k y k = x n + ( n 1 ) x n 1 y + ( n 2 ) x n 2 y 2 + ... + ( n n 1 ) x y n 1 + y n

Given a binomial, write it in expanded form.

  1. Determine the value of n according to the exponent.
  2. Evaluate the k = 0 through k = n using the Binomial Theorem formula.
  3. Simplify.

Expanding a binomial

Write in expanded form.

  1. ( x + y ) 5
  2. ( 3 x y ) 4
  1. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify.
    ( x + y ) 5 = ( 5 0 ) x 5 y 0 + ( 5 1 ) x 4 y 1 + ( 5 2 ) x 3 y 2 + ( 5 3 ) x 2 y 3 + ( 5 4 ) x 1 y 4 + ( 5 5 ) x 0 y 5 ( x + y ) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5
  2. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3 x 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.
    ( 3 x y ) 4 = ( 4 0 ) ( 3 x ) 4 ( y ) 0 + ( 4 1 ) ( 3 x ) 3 ( y ) 1 + ( 4 2 ) ( 3 x ) 2 ( y ) 2 + ( 4 3 ) ( 3 x ) 1 ( y ) 3 + ( 4 4 ) ( 3 x ) 0 ( y ) 4 ( 3 x y ) 4 = 81 x 4 108 x 3 y + 54 x 2 y 2 12 x y 3 + y 4
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Write in expanded form.

  1. ( x y ) 5
  2. ( 2 x + 5 y ) 3
  1. x 5 5 x 4 y + 10 x 3 y 2 10 x 2 y 3 + 5 x y 4 y 5
  2. 8 x 3 + 60 x 2 y + 150 x y 2 + 125 y 3
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Using the binomial theorem to find a single term

Expanding a binomial with a high exponent such as ( x + 2 y ) 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 + ( 5 1 ) x 4 y + ( 5 2 ) x 3 y 2 + ( 5 3 ) x 2 y 3 + ( 5 4 ) x y 4 + y 5

The second term is ( 5 1 ) x 4 y . The third term is ( 5 2 ) x 3 y 2 . We can generalize this result.

( n r ) x n r y r

The (r+1)th term of a binomial expansion

The ( r + 1 ) th term of the binomial expansion    of ( x + y ) n is:

( n r ) x n r y r

Given a binomial, write a specific term without fully expanding.

  1. Determine the value of n according to the exponent.
  2. Determine ( r + 1 ) .
  3. Determine r .
  4. Replace r in the formula for the ( r + 1 ) th term of the binomial expansion.

Writing a given term of a binomial expansion

Find the tenth term of ( x + 2 y ) 16 without fully expanding the binomial.

Because we are looking for the tenth term, r + 1 = 10 , we will use r = 9 in our calculations.

( n r ) x n r y r
( 16 9 ) x 16 9 ( 2 y ) 9 = 5 , 857 , 280 x 7 y 9
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Find the sixth term of ( 3 x y ) 9 without fully expanding the binomial.

10 , 206 x 4 y 5

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Access these online resources for additional instruction and practice with binomial expansion.

Practice Key Terms 3

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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