13.5 The Binomial Theorem There are several theorems and strategies that allow us to expand binomials raised to powers such as (x.
Download ReportTranscript 13.5 The Binomial Theorem There are several theorems and strategies that allow us to expand binomials raised to powers such as (x.
13.5 The Binomial Theorem
There are several theorems and strategies that allow us to expand binomials raised to powers such as (
x
+
y
) 4 or (2
x
– 5
y
) 7 .
One of these is Pascal’s Triangle your turn: again: 1 1 1 6 1 5 1 1 1 3 4 15 10 2 6 20 1 3 10 1 4 15 1 5 1 6 1 1 • • each row starts & ends with a 1 to get each term, you add the two numbers diagonally above that spot *we can continue like this forever!
Another is the idea of factorials
n
!
*remember
n C r
n
!
( )! !
1) (
n
it can also be written as
n
(
n
!
Ex 1) Find the lie and fix it!
A) 10!
362,880 10 B) 10 210 C) 524,160 No 4368 There is a connection between the numbers in Pascal’s Triangle and Take for instance Row 4 1 4 6 4 1
n
notice that 4 This helps us both expand binomials as well as find a particular term of an expansion without expanding the whole thing.
How to expand a binomial: (
a
+
b
)
n
(1) Coefficients: use Pascal’s Triangle or
n r
(2) Powers of each variable: the powers on the first term descend from
n
…. 0 the powers on the second term ascend from 0 ….
n
Ex 2) Using Pascal’s Triangle, expand (
x
+
y
) 4 1
x
4
y
0 + 4
x
3
y
1 + 6
x
2
y
2 + 4
x
1
y
3 + 1
x
0
y
4
x
4 + 4
x
3
y
+ 6
x
2
y
2 + 4
xy
3 +
y
4
Ex 3) Expand 2
x
y
2 5 5
y
2 0 4
y
2 1 3
y
2 2 2
y
2 3 1
y
2 4 0
y
2 5 32
x
5 80 2 80 4 40 6 10
xy
8
y
10 Ex 4) Find the coefficient of the indicated term & identify the missing exponent.
x
?
y
9 ; (
x
+
y
) 11 *each set of powers adds to the total of 11 ? + 9 = 11
x
2
y
9 ? = 2 total!
this is simply exponent!exponent!
11!
2!9!
55 55 2
x y
9
Ex 5) Find the term involving the specified variable.
b
6 in (
a
–
b
) 14 14!
8!6!
6 3003 8
a b
6 Ex 6) Find the indicated term of the expansion.
a) the third term of (
a
4!
2!2!
2 6 + 5
b
) 4 *the third term would have
a
2 25
b
2 150 2
a b
2 (count down) b) the fourth term of (2
a
– 6
b
) 11 11!
8!3!
6
b
3 *fourth term
a
11
a
10
a
9
a
8
a
8 216
b
3 9123840 8
a b
3