# Algebra Rules Review Sheet Page 19

REVIEW OF ALGEBRA
19
69. 2x
2
+ 3x + 4 is irreducible because its discriminant is negative: b
2
− 4ac = 9 − 4(2)(4) = −23 < 0.
2
+ 9x + 4 is not irreducible because b
2
2
− 4ac = 9
− 4(2)(4) = 49 > 0.
71. 3x
2
+ x − 6 is not irreducible because its discriminant is nonnegative: b
2
− 4ac = 1 − 4(3)(−6) = 73 > 0.
2
+ 3x + 6 is irreducible because b
2
2
− 4ac = 3
− 4(1)(6) = −15 < 0.
73. Using the Binomial Theorem with k = 6 we have
6 · 5
6 · 5 · 4
6 · 5 · 4 · 3
6
6
5
4
2
3
3
2
4
5
6
(a + b)
= a
+ 6a
b +
a
b
+
a
b
+
a
b
+ 6ab
+ b
1 · 2
1 · 2 · 3
1 · 2 · 3 · 4
6
5
4
2
3
3
2
4
5
6
= a
+ 6a
b + 15a
b
+ 20a
b
+ 15a
b
+ 6ab
+ b
74. Using the Binomial Theorem with k = 7 we have
7 · 6
7 · 6 · 5
7 · 6 · 5 · 4
7 · 6 · 5 · 4 · 3
7
7
6
5
2
4
3
3
4
2
5
6
7
(a + b)
= a
+ 7a
b +
a
b
+
a
b
+
a
b
+
a
b
+ 7ab
+ b
1 · 2
1 · 2 · 3
1 · 2 · 3 · 4
1 · 2 · 3 · 4 · 5
7
6
5
2
4
3
3
4
2
5
6
7
= a
+ 7a
b + 21a
b
+ 35a
b
+ 35a
b
+ 21a
b
+ 7ab
+ b
75. Using the Binomial Theorem with a = x
2
, b = −1, k = 4 we have
4 · 3
2
4
2
4
2
4
2
3
2
2
2
2
3
4
(x
− 1)
= [x
+ (−1)]
= (x
)
+ 4(x
)
(−1) +
(x
)
(−1)
+ 4(x
)(−1)
+ (−1)
1 · 2
8
6
4
2
= x
− 4x
+ 6x
− 4x
+ 1
76. Using the Binomial Theorem with a = 3, b = x
2
, k = 5 we have
5 · 4
5 · 4 · 3
2
5
5
4
2
1
3
2
2
2
2
3
2
4
2
5
(3 + x
)
= 3
+ 5(3)
(x
)
+
(3)
(x
)
+
(3)
(x
)
+ 5(3)(x
)
+ (x
)
1 · 2
1 · 2 · 3
2
4
6
8
10
= 243 + 405x
+ 270x
+ 90x
+ 15x
+ x
77. Using Equation 10,
32
2 =
32 · 2 =
64 = 8.
3
3
−2
−2
−1
−1
−1
1
78.
3
3
=
=
=
=
= −
3
54
27
3
3
3
54
27
4
4
4
4
4
32x
32
x
32
79. Using Equation 10,
4
4
4
4
=
=
x
=
16 |x| = 2 |x|.
4
4
2
2
2
80.
2
3
3
4
2
xy
x
y =
(xy)(x
y) =
x
y
= x
|y|
81. Using Equation 10,
2
3/2
2
1/2
2
4
3
4
3
16a
b
=
16
a
b
= 4a
b
= 4a
b b
= 4a
b
b.
5
6
96a
6
96a
82.
5
5
5
=
=
32a
= 2a
5
3a
3a
83. Using Laws 3 and 1 of Exponents respectively, 3
10
8
10
2
8
10
2 · 8
10 + 16
26
.
× 9
= 3
× (3
)
= 3
× 3
= 3
= 3
84. Using Laws 3 and 1, 2
16
10
6
16
2
10
4
6
16
20
24
60
.
× 4
× 16
= 2
× (2
)
× (2
)
= 2
× 2
× 2
= 2
9
4
9
4
4
9 + 4
x
(2x)
x
(2
)x
16x
85. Using Laws 4, 1, and 2 of Exponents respectively,
9 + 4 − 3
10
.
=
=
= 16x
= 16x
3
3
3
x
x
x
n
2n + 1
n + 2n + 1
3n + 1
a
× a
a
a
86. Using Laws 1 and 2,
3n + 1−(n − 2)
2n + 3
.
=
=
= a
= a
a
n − 2
a
n − 2
a
n − 2
−3
4
2
a
b
a
87. Using Law 2 of Exponents,
−3 − (−5)
4 − 5
2
−1
.
= a
b
= a
b
=
−5
5
a
b
b
−1
−1
2
x
+ y
1
1
y + x
(y + x)
88.
= (x + y)
+
= (x + y)
=
−1
(x + y)
x
y
xy
xy
1
1
89. By deﬁnitions 3 and 4 for exponents respectively, 3
−1/2
.
=
=
1/2
3
3
90. 96
1/5
5
5
5
5
5
=
96 =
32 · 3 =
32
3 = 2
3
2
91. Using deﬁnition 4 for exponents, 125
2/3
3
2
=
125
= 5
= 25.
1
1
1
1
92. 64
−4/3
=
=
=
=
4
4/3
4
64
4
256
3
64