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Lesson 8.6 Exercises, pages 743–749
A
3. Expand using Pascal’s triangle.
a) (x + 1)5
The exponent is 5, so use the terms in row 6 of Pascal’s triangle as
coefficients: 1, 5, 10, 10, 5, 1
(x ⴙ 1)5 ⴝ 1(x)5 ⴙ 5(x)4(1) ⴙ 10(x)3(1)2 ⴙ 10(x)2(1)3 ⴙ 5(x)1(1)4 ⴙ 1(1)5
ⴝ x5 ⴙ 5x4 ⴙ 10x3 ⴙ 10x2 ⴙ 5x ⴙ 1
b) (x - 1)6
The exponent is 6, so use the terms in row 7 of Pascal’s triangle as
coefficients: 1, 6, 15, 20, 15, 6, 1
(x ⴚ 1)6 ⴝ 1(x)6 ⴙ 6(x)5(ⴚ1) ⴙ 15(x)4(ⴚ1)2 ⴙ 20(x)3(ⴚ1)3
ⴙ 15(x)2(ⴚ1)4 ⴙ 6(x)(ⴚ1)5 ⴙ 1(ⴚ1)6
ⴝ x6 ⴚ 6x5 ⴙ 15x4 ⴚ 20x3 ⴙ 15x2 ⴚ 6x ⴙ 1
c) (x + y)4
The exponent is 4, so use the terms in row 5 of Pascal’s triangle
as coefficients: 1, 4, 6, 4, 1
(x ⴙ y)4 ⴝ 1(x)4 ⴙ 4(x)3(y) ⴙ 6(x)2(y)2 ⴙ 4(x)(y)3 ⴙ 1(y)4
ⴝ x4 ⴙ 4x3y ⴙ 6x2y2 ⴙ 4xy3 ⴙ y4
d) (x - y)8
The exponent is 8, so use the terms in row 9 of Pascal’s triangle
as coefficients: 1, 8, 28, 56, 70, 56, 28, 8, 1
(x ⴚ y)8 ⴝ 1(x)8 ⴙ 8(x)7(ⴚy) ⴙ 28(x)6(ⴚy)2 ⴙ 56(x)5(ⴚy)3 ⴙ 70(x)4(ⴚy)4
ⴙ 56(x)3(ⴚy)5 ⴙ 28(x)2(ⴚy)6 ⴙ 8(x)(ⴚy)7 ⴙ 1(ⴚy)8
ⴝ x8 ⴚ 8x7y ⴙ 28x6y2 ⴚ 56x5y3 ⴙ 70x4y4 ⴚ 56x3y5
ⴙ 28x2y6 ⴚ 8xy7 ⴙ y8
4. Determine each missing number in the expansion of (x + y)7.
x7 + nx6y + 21x5y2 + 35x ny3 + nx3y4 + 21x ny n + 7xy6 + y n
The exponent is 7, so the coefficients of the terms in the expansion
are the terms in row 8 of Pascal’s triangle: 1, 7, 21, 35, 35, 21, 7, 1
The exponents in each term must add to 7.
The exponents of the powers of x start at 7 and decrease by 1 each time.
The exponents of the powers of y start at 0 and increase by 1 each time.
So, the missing numbers are: 7, 4, 35, 2, 5, 7
5. Determine the indicated term in each expansion.
a) the last term in (x + 1)9
The last term in the expansion of (x ⴙ y)n is yn.
So, the last term in the expansion of (x ⴙ 1)9 is 19, or 1.
b) the 1st term in (x - 1)12
The first term in the expansion of (x ⴙ y)n is xn.
So, the first term in the expansion of (x ⴚ 1)12 is x12.
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B
6. a) Multiply 4 factors of (x - 5).
(x ⴚ 5)4 ⴝ
ⴝ
ⴝ
ⴝ
(x ⴚ 5)(x ⴚ 5)(x ⴚ 5)(x ⴚ 5)
(x2 ⴚ 10x ⴙ 25)(x2 ⴚ 10x ⴙ 25)
x4 ⴚ 10x3 ⴙ 25x2 ⴚ 10x3 ⴙ 100x2 ⴚ 250x ⴙ 25x2 ⴚ 250x ⴙ 625
x4 ⴚ 20x3 ⴙ 150x2 ⴚ 500x ⴙ 625
b) Use the binomial theorem to expand (x - 5)4.
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 4, y ⴝ ⴚ5
(x ⴚ 5)4 ⴝ 4C0(x)4 ⴙ 4C1(x)4 ⴚ 1(ⴚ5) ⴙ 4C2(x)4 ⴚ 2(ⴚ5)2
ⴙ 4C3(x)4 ⴚ 3(ⴚ5)3 ⴙ 4C4(ⴚ5)4
ⴝ 1(x)4 ⴙ 4(x)3(ⴚ5) ⴙ 6(x)2(25) ⴙ 4(x)1(ⴚ125) ⴙ 1(625)
ⴝ x4 ⴚ 20x3 ⴙ 150x2 ⴚ 500x ⴙ 625
c) Compare the two methods. What conclusions can you make?
I find it easier to use the binomial theorem; it saves time and it is less
cumbersome than multiplying 4 factors.
7. Expand using the binomial theorem.
a) (x + 2)6
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2x n ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 6, y ⴝ 2
(x ⴙ 2)6 ⴝ 6C0(x)6 ⴙ 6C1(x)6 ⴚ 1(2) ⴙ 6C2(x)6 ⴚ 2(2)2 ⴙ 6C3(x)6 ⴚ 3(2)3
ⴙ 6C4(x)6 ⴚ 4(2)4 ⴙ 6C5(x)6 ⴚ 5(2)5 ⴙ 6C6(2)6
ⴝ 1(x)6 ⴙ 6(x)5(2) ⴙ 15(x)4(4) ⴙ 20(x)3(8) ⴙ 15(x)2(16)
ⴙ 6(x)1(32) ⴙ 1(64)
ⴝ x6 ⴙ 12x5 ⴙ 60x4 ⴙ 160x3 ⴙ 240x2 ⴙ 192x ⴙ 64
b) (x2 - 3)5
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 5, x ⴝ x2, y ⴝ ⴚ3
(x2 ⴚ 3)5 ⴝ 5C0(x2)5 ⴙ 5C1(x2)4(ⴚ3) ⴙ 5C2(x2)3(ⴚ3)2 ⴙ 5C3(x2)2(ⴚ3)3
ⴙ 5C4(x2)1(ⴚ3)4 ⴙ 5C5(ⴚ3)5
ⴝ 1(x10) ⴙ 5(x2)4(ⴚ3) ⴙ 10(x2)3(9) ⴙ 10(x2)2(ⴚ27)
ⴙ 5(x2)1(81) ⴙ 1(ⴚ243)
ⴝ x10 ⴚ 15x8 ⴙ 90x6 ⴚ 270x4 ⴙ 405x2 ⴚ 243
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c) (3x - 2)4
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 4, x ⴝ 3x, y ⴝ ⴚ2
(3x ⴚ 2)4 ⴝ 4C0(3x)4 ⴙ 4C1(3x)3(ⴚ2) ⴙ 4C2(3x)2(ⴚ2)2 ⴙ 4C3(3x)(ⴚ2)3
ⴙ 4C4(ⴚ2)4
ⴝ 1(81x4) ⴙ 4(27x3)(ⴚ2) ⴙ 6(9x2)(4) ⴙ 4(3x)(ⴚ8) ⴙ 1(16)
ⴝ 81x4 ⴚ 216x3 ⴙ 216x2 ⴚ 96x ⴙ 16
d) (-2 + 2x)4
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 4, x ⴝ ⴚ2, y ⴝ 2x
(ⴚ2 ⴙ 2x)4 ⴝ 4C0(ⴚ2)4 ⴙ 4C1(ⴚ2)3(2x) ⴙ 4C2(ⴚ2)2(2x)2
ⴙ 4C3(ⴚ2)(2x)3 ⴙ 4C4(2x)4
ⴝ 1(16) ⴙ 4(ⴚ8)(2x) ⴙ 6(4)(4x2) ⴙ 4(ⴚ2)(8x3) ⴙ 1(16x4)
ⴝ 16 ⴚ 64x ⴙ 96x2 ⴚ 64x3 ⴙ 16x4
e) (-4 + 3x4)5
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 5, x ⴝ ⴚ4, y ⴝ 3x4
(ⴚ4 ⴙ 3x4)5 ⴝ 5C0(ⴚ4)5 ⴙ 5C1(ⴚ4)4(3x4) ⴙ 5C2(ⴚ4)3(3x4)2 ⴙ 5C3(ⴚ4)2(3x4)3
ⴙ 5C4(ⴚ4)(3x4)4 ⴙ 5C5(3x4)5
ⴝ 1(ⴚ1024) ⴙ 5(256)(3x4) ⴙ 10(ⴚ64)(9x8) ⴙ 10(16)(27x12)
ⴙ 5(ⴚ4)(81x16) ⴙ 1(243x20)
ⴝ ⴚ1024 ⴙ 3840x4 ⴚ 5760x8 ⴙ 4320x12 ⴚ 1620x16 ⴙ 243x20
8. a) Write the terms in row 7 of Pascal’s triangle.
1, 6, 15, 20, 15, 6, 1
b) Use your answer to part a to write the first 3 terms in each
expansion.
i) (x - 3)6
The exponent is 6, so the terms in row 7 of Pascal’s triangle are the
coefficients of the terms in the expansion of the binomial.
So, (x ⴙ y)6 ⴝ 1x6 ⴙ 6x5y ⴙ 15x4y2 ⴙ . . .
Substitute: y ⴝ ⴚ3
(x ⴚ 3)6 ⴝ 1x6 ⴙ 6x5(ⴚ3) ⴙ 15x4(ⴚ3)2 ⴙ . . .
ⴝ x6 ⴚ 18x5 ⴙ 135x4 ⴙ . . .
So, the first 3 terms in the expansion are: x6 ⴚ 18x5 ⴙ 135x4
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ii) (a + 4b)6
(x ⴙ y)6 ⴝ 1x6 ⴙ 6x5y ⴙ 15x4y2 ⴙ . . .
Substitute: x ⴝ a, y ⴝ 4b
(a ⴙ 4b)6 ⴝ 1a6 ⴙ 6a5(4b) ⴙ 15a4(4b)2 ⴙ . . .
ⴝ a6 ⴙ 24a5b ⴙ 240a4b2 ⴙ . . .
So, the first 3 terms in the expansion are: a6 ⴙ 24a5b ⴙ 240a4b2
iii) (-2a + 1)6
(x ⴙ y)6 ⴝ 1x6 ⴙ 6x5y ⴙ 15x4y2 ⴙ . . .
Substitute: x ⴝ ⴚ2a, y ⴝ 1
(ⴚ2a ⴙ 1)6 ⴝ 1(ⴚ2a)6 ⴙ 6(ⴚ2a)5(1) ⴙ 15( ⴚ2a)4(1)2 ⴙ . . .
ⴝ 64a6 ⴚ 192a5 ⴙ 240a4 ⴙ . . .
So, the first 3 terms in the expansion are: 64a6 ⴚ 192a5 ⴙ 240a4
iv) (2x + 5y2)6
(x ⴙ y)6 ⴝ 1x6 ⴙ 6x5y ⴙ 15x4y2 ⴙ . . .
Substitute: x ⴝ 2x, y ⴝ 5y2
(2x ⴙ 5y2)6 ⴝ 1(2x)6 ⴙ 6(2x)5(5y2) ⴙ 15(2x)4(5y2)2 ⴙ . . .
ⴝ 64x6 ⴙ 960x5y2 ⴙ 6000x4y4 ⴙ . . .
So, the first 3 terms in the expansion are: 64x6 ⴙ 960x5y2 ⴙ 6000x4y4
9. Determine the coefficient of each term.
a) x5 in (x + 1)8
x5 is the 4th term in (x ⴙ 1)8.
The coefficient of the 4th term is:
8C4ⴚ 1 or 8C3
8C3 ⴝ 56
So, the coefficient of x5 is 56.
c) x2y in (x + y)3
The coefficients of the terms
will be the terms in row 4 of
Pascal’s triangle: 1, 3, 3, 1
x2y is the 2nd term in the
expansion of (x ⴙ y)3.
So, the coefficient of x2y is 3.
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b) x9 in (x + y)9
x9 is the first term in the expansion
of (x ⴙ y)9.
So, the coefficient of x9 is 1.
d) x2y3 in (x + y)5
The coefficients of the terms
will be the terms in row 6 of
Pascal’s triangle: 1, 5, 10, 10, 5, 1
x2y3 is the 4th term in the
expansion of (x ⴙ y)5.
So, the coefficient of x2y3 is 10.
8.6 The Binomial Theorem—Solutions
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10. Explain why the coefficients of the 3rd term and the 3rd-last term in
the expansion of (x + y)n, n ≥ 2, are the same.
Since each of x and y has coefficient 1, the coefficients of the terms in the
expansion of (x ⴙ y)n correspond to the terms in row (n ⴙ 1) of Pascal’s
triangle.
In any row of Pascal’s triangle, the 3rd term and 3rd-last term are the same.
11. Determine the indicated term in each expansion.
a) the last term in (3x + 2)5
The last term in the expansion
of (x ⴙ y)n is yn.
Substitute: y ⴝ 2, n ⴝ 5
25 ⴝ 32
So, the last term is 32.
c) the 2nd term in (3x - 3)4
The second term in the expansion
of (x ⴙ y)n is nxn ⴚ 1y.
Substitute: x ⴝ 3x, y ⴝ ⴚ3,
nⴝ4
4(3x)3(ⴚ3) ⴝ ⴚ324x3
So, the 2nd term is ⴚ324x3.
b) the 1st term in (-2x + 5)7
The first term in the expansion
of (x ⴙ y)n is xn.
Substitute: x ⴝ ⴚ2x, n ⴝ 7
(ⴚ2x)7 ⴝ ⴚ128x7
So, the 1st term is ⴚ128x7.
d) the 6th term in (4x + 1)8
The kth term is: nCk ⴚ 1xn ⴚ (k ⴚ 1)yk ⴚ 1
Substitute: n ⴝ 8, k ⴝ 6, x ⴝ 4x,
yⴝ1
C (4x)3(1)5 ⴝ 56(64x3)(1)
8 5
ⴝ 3584x3
So, the 6th term is 3584x3.
12. When will the coefficients of the terms in the expansion of (ax + b)n
be the same as the terms in row (n + 1) of Pascal’s triangle?
The coefficients of the terms in the expansion of (x ⴙ y)n correspond
to the terms in row (n ⴙ 1) of Pascal’s triangle. So, both a and b must
equal 1.
13. Expand and simplify (x + 1)8 + (x - 1)8. What strategy did you use?
The coefficients of the terms in the expansion of (x ⴙ 1)8 are the terms in
row 9 of Pascal’s triangle: 1, 8, 28, 56, 70, 56, 28, 8, 1
The coefficients of the terms in the expansion of (x ⴚ 1)8 depend on
whether the term number is odd or even. When the term numbers are odd,
the coefficients are the terms in row 9 of Pascal’s triangle. When the term
numbers are even, the coefficients are the opposites of the terms in row 9.
So, the coefficients of the terms in the expansion of (x ⴚ 1)8 are:
1, ⴚ8, 28, ⴚ56, 70, ⴚ56, 28, ⴚ8, 1
When the terms in the two expansions are added, every second term is
eliminated as the sum of the terms is 0.
(x ⴙ 1)8 ⴙ (x ⴚ 1)8 ⴝ 2x8 ⴙ 56x6 ⴙ 140x4 ⴙ 56x2 ⴙ 2
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14. a) Show that the expansion of (-2x + 1)6 is the same as the
expansion of (2x - 1)6.
Use row 7 of Pascal’s triangle: 1, 6, 15, 20, 15, 6, 1
(ⴚ2x ⴙ 1)6 ⴝ 1(ⴚ2x)6 ⴙ 6(ⴚ2x)5(1) ⴙ 15( ⴚ2x)4(1)2 ⴙ 20( ⴚ2x)3(1)3
ⴙ 15(ⴚ2x)2(1)4 ⴙ 6(ⴚ2x)1(1)5 ⴙ (1)6
ⴝ 64x6 ⴙ 6(ⴚ32x5) ⴙ 15(16x4) ⴙ 20(ⴚ8x3) ⴙ 15(4x2) ⴙ 6(ⴚ2x) ⴙ 1
ⴝ 64x6 ⴚ 192x5 ⴙ 240x4 ⴚ 160x3 ⴙ 60x2 ⴚ 12x ⴙ 1
(2x ⴚ 1)6 ⴝ 1(2x)6 ⴙ 6(2x)5(ⴚ1) ⴙ 15(2x)4(ⴚ1)2 ⴙ 20(2x)3(ⴚ1)3
ⴙ 15(2x)2(ⴚ1)4 ⴙ 6(2x)1(ⴚ1)5 ⴙ (ⴚ1)6
ⴝ 64x6 ⴙ 6(32x5)(ⴚ1) ⴙ 15(16x4) ⴙ 20(8x3)(ⴚ1) ⴙ 15(4x2)
ⴙ 6(2x)(ⴚ1) ⴙ 1
ⴝ 64x6 ⴚ 192x5 ⴙ 240x4 ⴚ 160x3 ⴙ 60x2 ⴚ 12x ⴙ 1
b) Will (-ax + b)n always have the same expansion as (ax - b)n?
Explain.
No, when n is odd, the expansions will not be the same. For example,
look at the first terms in the expansions of (ⴚ2x ⴙ 1)3 and (2x ⴚ 1)3.
For (ⴚ2x ⴙ 1)3: the 1st term in the expansion is: 1(ⴚ2x)3 ⴝ ⴚ8x3
For (2x ⴚ 1)3: the 1st term in the expansion is: 1(2x)3 ⴝ 8x3
Since the 1st terms are different, the expansions are not the same.
15. Which binomial power when expanded results in
16x4 - 32x3 + 24x2 - 8x + 1?
What strategy did you use to find out?
The first term is 16x 4. The exponent of x is 4, so the binomial has the form
(ax ⴙ b)4.
√
The coefficient a must be 4 16: a ⴝ 2, or a ⴝ ⴚ2
The last term is 1, so b is either 1 or ⴚ1 because (ⴚ1)4 ⴝ 14, or 1.
The terms alternate in sign, so the coefficients a and b must have
different signs.
The binomial power is either (ⴚ2x ⴙ 1)4 or (2x ⴚ 1)4.
16. Expand using the binomial theorem.
a) (0.2x - 1.2y)5
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
Substitute: n ⴝ 5, x ⴝ 0.2x, y ⴝ ⴚ1.2y
(0.2x ⴚ 1.2y)5 ⴝ 5C0(0.2x)5 ⴙ 5C1(0.2x)4(ⴚ1.2y) ⴙ 5C2(0.2x)3(ⴚ1.2y)2
ⴙ 5C3(0.2x)2(ⴚ1.2y)3 ⴙ 5C4(0.2x)(ⴚ1.2y)4 ⴙ 5C5(ⴚ1.2y)5
ⴝ 1(0.000 32x5) ⴙ 5(0.0016x4)(ⴚ1.2y) ⴙ 10(0.008x3)(1.44y2)
ⴙ 10(0.04x2)(ⴚ1.728y3) ⴙ 5(0.2x)(2.0736y4) ⴙ 1(ⴚ2.488 32y5)
ⴝ 0.000 32x5 ⴚ 0.0096x4y ⴙ 0.1152x3y2 ⴚ 0.6912x2y3
ⴙ 2.0736xy4 ⴚ 2.488 32y5
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b) a8a + 6bb
3
1
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4
(x ⴙ y)n ⴝ nC0xn ⴙ nC1xn ⴚ 1y ⴙ nC2xn ⴚ 2y2 ⴙ . . . ⴙ nCnyn
3
8
1
6
Substitute: n ⴝ 4, x ⴝ a, y ⴝ b
4
4
3
2
1
3
3
3
1
3
1
a a ⴙ bb ⴝ 4C0 a ab ⴙ 4C1 a ab a bb ⴙ 4C2 a ab a bb
8
6
8
8
6
8
6
3
ⴙ 4C3 a ab a bb ⴙ 4C4 a bb
3
8
1
6
1
6
2
4
ⴝ 1a
81 4
27
1
9
1
a b ⴙ 4 a a3b a bb ⴙ 6 a a2b a b2b
4096
512
6
64
36
ⴙ 4 a ab a
3
8
1 3
1
b b ⴙ 1a
b4b
216
1296
81 4
27 3
9 2 2
3
1 4
a ⴙ
abⴙ
ab ⴙ
ab3 ⴙ
b
4096
768
384
432
1296
81 4
9 3
3 2 2
1
1 4
ⴝ
a ⴙ
abⴙ
ab ⴙ
ab3 ⴙ
b
4096
256
128
144
1296
ⴝ
C
17. Determine the 3rd term in the expansion of (x2 + 2x + 1)6.
(x2 ⴙ 2x ⴙ 1) ⴝ (x ⴙ 1)2
So, (x2 ⴙ 2x ⴙ 1)6 ⴝ ((x ⴙ 1)2)6
ⴝ (x ⴙ 1)12
The coefficients in the expansion of (x ⴙ 1)12 are the terms in row 13 of
Pascal’s triangle.
The 3rd term in row 13 is 66.
So, the 3rd term in the expansion of (x2 ⴙ 2x ⴙ 1)6 is: 66(x)10(1)2, or 66x10
18. a) Show that nC0 + nC1 + nC2 + . . . + nCn - 1 + nCn = 2n
Express 2n as the binomial (1 ⴙ 1)n and expand:
(1 ⴙ 1)n ⴝ nC0(1)n ⴙ nC1(1)n ⴚ 1(1) ⴙ nC2(1)n ⴚ 2(1)2 ⴙ . . .
ⴙ nCn ⴚ 1(1)(1)n ⴚ 1 ⴙ nCn(1)n
So, 2 ⴝ nC0 ⴙ nC1 ⴙ nC2 ⴙ . . . ⴙ nCn ⴚ 1 ⴙ nCn
n
b) What does the relationship in part a indicate about the sum of
the terms in any row of Pascal’s triangle?
nC0 , nC1, nC2, . . . nCn ⴚ 1, nCn are the terms in row (n ⴙ 1) of the triangle.
From part a, the sum of these terms is 2n.
So, the sum of the terms in any row of Pascal’s triangle is a power of 2.
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