ALGEBRA II HONORS/GIFTED
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SECTION 5-7 : THE
BINOMIAL THEOREM
Work out the problem that corresponds to your group
number.
9) What do you notice
about the signs before
each term?
1) (x + y)0
=1
2) (x – y)0
=1
3) (x + y)1
=x+y
4) (x – y)1
=x–y
5) (x + y)2
= x2 + 2xy + y2
6) (x – y)2
= x2 – 2xy + y2
7) (x + y)3
= x3 + 3x2y + 3xy2 + y3
8) (x – y)3
= x3 – 3x2y + 3xy2 – y3
10) And, those
exponents. What about
them?
11) Hey, what’s going
on with the coefficients?
1
1
1
1
1
1
1
1
1
8
This is Pascal’s
Triangle.
3
5
7
6
15
1
4
1
10
20
35
56
1
3
10
21
28
2
4
6
1
5
15
35
70
1
6
21
56
1
7
28
1
8
1
The triangle’s row
are numbered
beginning with row
0.
Expand.
a) How many terms are in
the expansion?
9a) (x + y)4
+ (1)
(x)4
(y)0
ANSWER : 5
b) How are the signs of
the terms arranged?
ANSWER : all +
+ (4) (x)3 (y)1
c) Lay down the coefficients
from Pascal’s Triangle.
+ (6) (x)2 (y)2
d) Place the terms of
the binomial down
noting their exponents.
+ (4) (x)1 (y)3
+ (1) (x)0 (y)4
e) Simplify.
= x4 + 4x3y + 6x2y2 + 4xy3 + y4
f) (x – y)5
ANSWER : x5 - 5x4y + 10x3y2 - 10x2y3 + 5xy4 - y5
g) (x – y)6
ANSWER : x6 – 6x5y + 15x4y2 – 20x3y3 + 15x2y4 – 6xy5 + y6
Find the specified term.
10a) Third term of (x + 2y)6
b) Fourth term of (2x2 – y)5
ANSWER : 60x4y2
ANSWER : -40x4y3
11)a) Expand using Pascal’s Triangle : (x – 4)5
ANSWER : x5 – 20x4 + 160x3 – 640x2 + 1280x - 1024
b) Now, find : 5C0, 5C1, 5C2, 5C3, 5C4, and 5C5.
ANSWERS : 1, 5, 10, 10, 5, 1
c) How do these answers compare with the
coefficients found in Pascal’s Triangle?
ANSWER : They are the same.
12) Expand.
a) (3x – 2)4
SOLUTION :
+4C0 (3x)4(2)0
-4C1(3x)3(2)1
+4C2(3x)2(2)2
-4C3(3x)1(2)3
+4C4(3x)0(2)4
= 81x4 – 216x3 + 216x2 – 96x + 16
b) (4x + y2)3
ANSWER : 64x3 + 48x2y2 + 12xy4 + y6
c) (a + a-1)6
ANSWER : a6 + 6a4 + 15a2 + 20 + 15a-2 + 6a-4 + a-6
d) (1 – i)5
ANSWER : -4 + 4i
13) Find the specified term.
Must add to 9.
a) (x + y)9 ; y4
SOLUTION :
Where does
the “9” come
from?
9C4
(x)5 (y)4
=126x5y4
These numbers are
always the same.
b) (2x – 3y)7 ; y4
ANSWER : 7C4 (2x)3 (-3y)4 = 22680x3y4
c) (a – 2b)8 ; a3
ANSWER : 8C5 (a)3 (-2b)5 = -1792a3b5
d) (a + b)n ; br
ANSWER : nCr (a)n-r (b)r
14) Find the specified term.
a) (m + n)20 ; 12th term
SOLUTION :
20C11
(m)9 (n)11
= 167960m9n11
The choose number is always 1
less than the term number. The
choose number is also known
as the “magic number”.
b) (r3 – 2s)9 ; 5th term
ANSWER : 9C4 (r3)5 (-2s)4 = 2016r15s4
c) (2x – y)12 ; middle term
ANSWER : 12C6 (2x)6 (-y)6 = 59136x6y6
d) (a + 2b3)10 ; middle term
ANSWER : 10C5 (a)5 (2b)5 = 8064a5b5