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Precalculus
Appendix A
Review of Fundamental Concepts of Algebra
A.1 Real Numbers and Their Properties
Interval and Inequality Notation
Exercises
A. Determine which numbers in the set are (a) natural numbers, (b) whole numbers, (c) integers, (d)
rational numbers, and (e) irrational numbers. Some numbers will be in more than one category.
Solution:
7
2
1. − 9, − , 5, , 2 , 0, 1, − 4, 2, − 11
(a) 1, 2, 5 (b) 0, 1, 2, 5 (c) -11, -9, -4, 0, 1, 2, 5
2
3
(d) -11, -9, -4, − 72 , 0,
2.
2
3
, 1, 2, 5
(e)
2
7
5
5 , − 7, − , 0, 3.12, , − 3,12,5
3
4
1 6
3. − π , − , , 0.5, 2 , − 7.5, − 1, 8, − 22
3 3
B. Sketch the subset on the real number line. If the problem is in inequality notation, also write it in
interval notation. If the problem is in interval notation, also write it in inequality notation.
1. x ≤ 5
2. x < 0
5. [4, ∞ )
Solution: (− ∞,0)
6. (−∞,2)
3. x ≥ −2
7. − 2 < x < 2
4. x > 3
8. 0 ≤ x < 5
Solution: [0,5)
Precalculus
Appendix A
Review of Fundamental Concepts of Algebra
A.2 Exponents and Radicals
3
−2
⎛4⎞ ⎛3⎞
Example 1: Simplify ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟
⎝ y⎠ ⎝ y⎠
−2
3
⎛ 4 ⎞⎛ 3 ⎞ ⎛ 64 ⎞⎛ y 2
Solution: ⎜⎜ 3 ⎟⎟⎜⎜ − 2 ⎟⎟ = ⎜⎜ 3 ⎟⎟⎜⎜
⎝ y ⎠⎝ y ⎠ ⎝ y ⎠⎝ 9
Example 2: Simplify
Solution:
6 x 7 ( x + 3)
⎞ 64
⎟⎟ =
⎠ 9y
2
8 x 3 ( x + 3)
5
Watch out!
(a + b )2 ≠ a 2 + b 2
3x 4
4( x + 3)
3
A. Simplify each expression.
1. (−5 z ) 3
5.
7x2
x3
2. 5 x 4 ( x 2 )
6.
12( x + y ) 3
9( x + y )
3. 6 y 2 (2 y 0 ) 2
7.
r4
r6
4.
3x 5
x3
⎛ x −3 y 4
8. ⎜⎜
⎝ 5
⎞
⎟⎟
⎠
−3
Radicals and Their Properties
Simplifying Radicals
An expression involving radicals is in simplest form when the following conditions are satisfied.
1. All possible factors have been removed from the radical.
2. All fractions have radical-free denominators.
3. The index of the radical is reduced.
Example: Simplify each radical expression exactly.
a)
48 x 5 y 2
Solution:
16 3 x 5 y 2 = 4 3 ⋅ x 2 x ⋅ y
= 4 x 2 y 3x
b) 4 27 − 75
Solution:
4 9 3 − 25 3 = 4 ⋅ 3 3 − 5 3
= 12 3 − 5 3 = 7 3
B. Simplify each radical expression exactly.
1.
72x 3
4. 2 50 + 12 8
2.
54xy 4
5. 10 32 − 6 18
3.
75 x 2 y −4
6. − 2 9 y + 10 y
Rationalizing Denominators and Numerators
To rationalize a denominator or numerator of the form a − b m or a + b m , multiply both the
numerator and denominator by a conjugate ( a − b m and a + b m are conjugates of each other).
Example: Rationalize:
3
a)
6
Solution:
3
6
⋅
6
6
=
3
b)
3 6
6
=
6
2
Solution:
2− 6
3
⋅
2+ 6
2− 6 2+ 6
=
2 3 + 18
4+2 6 −2 6 −6
=
2 3 +3 2
−2
C. Completely simplify. Rationalize the denominator unless otherwise specified.
1.
2.
3.
1
4.
3
5
10
2
5− 3
Rational Exponents
Example: a
2
3
= 3 a2
3
5+ 6
5.
8
2
6.
5+ 3
(rationalize the numerator)
3
D. Fill in the missing form of the expression. Do not simplify.
1.
Solution:
Example: Perform the operations and simplify.
−1
5
5 2 ⋅ 5x 2
(5x )
5
1
Solution:
3
2
52 ⋅ x 2
3
3
52 x 2
=
x1
x
=
1
5
5
E. Perform the operations and simplify.
4
3
1.
(2 x 2 ) 2
2.
1
22 x4
y3
3
1
Solution: y 9 = y 3 = 3 y
F. Reduce the index of each radical.
1.
4
32
2.
6
x3
1
( xy ) 3
Example: Reduce the index of the radical.
9
2
x3 y3
3
64 = 64
1
3
Precalculus
Appendix A
Review of Fundamental Concepts of Algebra
A.3 Polynomials and Factoring
Polynomials with one, two, and three terms are called monomials, binomials, and trinomials,
respectively. In standard form, a polynomial is written with descending powers of x.
Steps to Factor a Polynomial
A.3 – Part B – Graphing Calculators - Examples
Calculator functions are in bold.
1
2
Example 1: f ( x ) = x 3 − 2( x + 2 ) − 2
3
a. Start with Zoom: ZStandard. Adjust to give appropriate window to identify “hills” and “valleys.”
b. Give the coordinates for the “hills” (maximum):
XMin: -10
XMax: 10
Xscl: 1
YMin: -70
YMax: 70
Yscl: 10
c. Give the coordinates for the “valleys” (minimum):
d. Give the x-intercept(s) ( y 2 = 0 , intersect ):
e. Give the y-intercept (value):
f. Find f (− 6) . (value)
g. Find x when f ( x ) = 8 to the nearest hundredth.
( y 2 = 8 , intersect)
h. Sketch a graph of the function. Label axes.
Example 2: h(t ) = −16t 2 + 320t − 1200 gives the height of a rocket as a function of time.
a. Start with Zoom: ZStandard. Adjust to give an appropriate window to see the vertex.
b. Identify the variables:
x:
y:
XMin: -10
XMax: 25
Xscl: 5
YMin: -50
YMax: 500
Yscl: 50
c. Find t that makes the height a maximum:
d. What is the maximum height?
e. At what time t does the rocket hit the ground?
f. Sketch a graph of the function. Label axes.
A.3 – Part B – Graphing Calculators
4
1⎞
⎛
1. f ( x ) = ⎜ x − ⎟ + 5 x 3 − 3.7 x 2 − x
2⎠
⎝
a. Start with Zoom: ZStandard. Adjust to give an appropriate window to idenfity “hills” and
“valleys.” You may need to adjust the window more than once during the problem.
b. Give the coordinates for the “hills”:
XMin:
XMax:
Xscl:
YMin:
YMax:
Yscl:
c. Give the coordinates for the “valleys”:
d. Give the x-intercept(s):
e. Give the y-intercept:
f. Find f (− 6) .
g. Find both x where f ( x ) = −11 . Round to nearest hundredth.
h. Sketch a graph of the function. Label axes.
2. h(t ) = −16t 2 + 273t − 650 gives the height of a rocket as a function of time.
a. Start with Zoom: ZStandard. Adjust to give an appropriate window to see the vertex.
b. Identify the variables:
x:
XMin:
XMax:
Xscl:
YMin:
YMax:
Yscl:
y:
c. Find t that makes the height a maximum:
d. What is the maximum height?
e. At what time t does the rocket hit the ground?
f. Sketch a graph of the function. Label axes.
A.3 Exercises
Solution #16:
(a) 25 y 2 − y + 1
(b) Degree: 2
L. C. : 25
(c) Trinomial
Solution #34:
1.3 x 4 − 8.4 x − 34.1
What if you are
asked to simplify
and give only the
linear term?
Solution #38:
4 y 4 + 2 y3 − 3y 2
Simplify and give
only the cubic term:
A.3 Exercises
Solution #58:
(4 x + 5)(4 x + 5)
= 16 x 2 + 20 x + 20 x + 25
= 16 x 2 + 40 x + 25
Solution #104:
(x − 7 )(x + 7 )
Solution #112:
(5 x − 4 y )(5 x + 4 y )
A.3 Exercises
Solution #114:
(x + 5)(x + 5)
or (x + 5)
2
Solution #138:
(x − 6)(x − 7 )
Solution #144:
− (5u 2 + 13u − 6 )
= −(5u − 2)(u + 3)
Solution #148:
5 x 3 − 10 x 2 + (3 x − 6)
(
)
= 5 x 2 ( x − 2) + 3( x − 2)
(
)
= 5 x 2 + 3 (x − 2)
A.3 Exercises
Solution #160:
12(x 2 − 4)
= 12( x − 2 )( x + 2)
Solution #168:
y (2 y 2 − 7 y − 15)
= y (2 y + 3)( y − 5)
Solution #172:
12 2 1
6
x −
x−
96
96
96
1
(
=
12 x 2 − x − 6)
96
1
(4 x − 3)(3x + 2)
=
96
Precalculus
Appendix A
Review of Fundamental Concepts of Algebra
A.4 Rational Expressions
The quotient of two algebraic expressions is a fractional expression. Moreover, the quotient of two
polynomials is a rational expression. Recall that a fraction is in simplest form if its numerator and
denominator have no factors in common aside from ± 1 . To write a fraction in simplest form, divide
out common factors.
a ⋅c a
= , c≠0
b⋅c b
The key to success in simplifying rational expressions lies in your ability to factor polynomials.
You can only cancel common factors, not individual terms in a sum or difference.
Example:
(
)
(x + 3) = x + 3
x 3 + 5x 2 + 6 x x x 2 + 5x + 6
x( x + 3)( x + 2)
=
= 2
=
4
2
2
2
x − 4x
x x −4
x ( x + 2)( x − 2) x( x − 2)
x 2 − 2x
(
)
Exercises
B. Write the rational expression in simplest form.
1.
15 x 2
10 x
3xy
2.
xy + x
3.
x−5
10 − 2 x
x 3 + 5x 2 + 6 x
4.
x2 − 4
5.
x2 − 9
x3 + x 2 − 9x − 9
6.
x 2 + 8 x − 20
x 2 + 11x + 10
Operations with Rational Expressions
To multiply or divide rational expressions, use the properties of fractions. Recall that to divide
fractions, you invert the divisor and multiply. To add or subtract rational expressions, you need a
common denominator.
C. Perform the operations and simplify. You may leave binomial factors in your answer.
5.
1.
Solution:
(r + 1)(r − 1) =
r (r − 1)
2.
r +1
r
6.
3.
7.
4.
8.
Complex Fractions
Fractional expressions with separate fractions in the numerator, denominator, or both are called
complex fractions. A complex fraction can be simplified by combining the fractions in its numerator
into a single fraction and then combining the fractions in its denominator into a single fraction. Then
invert the denominator and multiply.
Exercises
D. Simplify the Complex Fraction.
1.
Solution to D #2:
2.
(x − 4)
⎛ x x 4 4⎞
⎜ ⋅ − ⋅ ⎟
⎝4 x x 4⎠
4.
3.
=
(x − 4)
⎛ x − 16 ⎞
⎜⎜
⎟⎟
⎝ 4x ⎠
2
= (x − 4) ⋅
4x
4x
4x
= (x − 4) ⋅
=
(x − 4)(x + 4) x + 4
x − 16
2
Precalculus
Appendix A
Review of Fundamental Concepts of Algebra
A.5 Solving Equations
An equation in x is a statement that two algebraic expressions are equal. To solve an equation in x
means to find all values of x for which the equation is true. Such values are solutions.
When multiplying or dividing an equation by a variable quantity, it is possible to introduce an
extraneous solution. An extraneous solution is on that does not satisfy the original equation.
Therefore, it is essential that you check your solutions.
A. Solve the equation and check your solution. If not possible, explain why.
1. 2( x + 5) − 7 = 3( x − 2)
5.
100 − 4 x 5 x + 6
=
+6
3
4
2. x − 3( 2 x + 3) = 8 − 5 x
6. 10 −
3. 9 x − 10 = 5 x + 2( 2 x − 5)
3x 1
4.
+ ( x − 2) = 10
2 4
13
5
= 4+
x
x
7.
x
4
+
+2=0
x+4 x+4
8.
1
1
10
+
= 2
x −3 x+3 x −9
Solution: Multiply both sides by ( x − 3)( x + 3)
(x − 3)(x + 3) + (x − 3)(x + 3) = 10(x − 3)(x + 3)
(x − 3)
( x + 3)
(x − 3)(x + 3)
(x + 3) + (x − 3) = 10
2 x = 10
x=5
Since we multiplied by a variable, check the solution.
1
1
10
1 1 10
+
= 2
→ + = 3
5−3 5+3 5 −9
2 8 16
Quadratic Equations
A quadratic equation in x is an equation that can be written in the general form ax 2 + bx + c = 0 where
a, b, and c are real numbers with a ≠ 0 . A quadratic equation in x is also known as a second-degree
polynomial equation in x.
B. Solve the quadratic equation by factoring.
1. 6 x 2 + 3 x = 0
3. x 2 + 10 x + 25 = 0
Solution: 3 x(2 x + 1) = 0
3x = 0 2 x + 1 = 0
x = 0,− 12
2. x 3 − 2 x 2 − 8 x = 0
4. 3 + 5 x − 2 x 2 = 0
C. Solve the quadratic equation by extracting square roots.
1. x 2 = 32
3. ( x − 12) 2 = 16
Solution: x − 12 = ±4
x − 12 = 4 x − 12 = −4
x = 8, 16
2. 9 x 2 = 36
What if you were asked to solve and give the
product of the solutions?
4. ( x + 2) 2 = 14
D. Solve using a graphing calculator. (Set y1 as the left side, y 2 as the right side, use “Intersect”)
1. x 2 + 4 x − 32 = 0
3. 9 x 2 − 18 x = −3
2. x 2 + 12 x + 25 = 0
4. − x 2 + 4 x + 8 = 0
Solution: x = −9.32,−2.68
E. Solve the quadratic equation by using the quadratic formula.
1. 2 x 2 − x − 1 = 0
2. 2 + 2 x − x 2 = 0
3. 4 x 2 + 4 x = 7
4. 6 x = 4 − x 2
Solution:
4x 2 + 4x − 7 = 0
− 4 ± 4 2 − 4(4)(− 7 )
x=
2(4)
− 4 ± 128 − 4 ± 8 2
=
8
8
= − 12 ± 2
=
Polynomial Equations of Higher Degree
The methods used to solve quadratic equations can sometimes be extended to solve polynomial
equations of higher degree.
F. Find all solutions of the equation. Check your solutions in the original equation.
1. 4 x 4 − 18 x 2 = 0
3. x 3 + 2 x 2 + 3 x + 6 = 0
Solution:
(
)
2x 2 2x 2 − 9 = 0
2x 2 = 0 2x 2 − 9 = 0
x=0
x2 =
9
2
x=±
x = 0, ±
9
2
=±
3
2
= ± 3 22
3 2
2
2. 5 x 3 + 30 x 2 + 45 x = 0
4. x 4 + 5 x 2 − 36 = 0
Precalculus
Appendix A – Key to Exercises
A.1
A 2.(a) 5, 12 (b) 0, 5, 12
3. (a) 63 , 8
(b) 63 , 8
B 1. (− ∞, 5]
3. [− 2, ∞ )
(c) -7, -3, 0, 5, 12
(d) − 7, − 3, − 73 , 0, 54 , 3.12, 5, 12
(e)
(c) − 22, − 1, 63 , 8
(d) − 22, − 7.5, − 1, − 13 , .5, 63 , 8
(e) − π , 2
4. (3, ∞ )
5. x ≥ 4
5
7. (− 2, 2)
6. x < 2
A.2
A 1. − 125z 3
2. 5x 6
4. 3x 2
2. 3 y 2 6 x
B 1. 6 x 2 x
3
3
C 1.
3. 24 y 2
2.
3.
10
2
3.
5.
7
x
5x 3
D 1. 9 2 , 64 3 , 5 32 , − 144 , (− 216 ) 3 , 5 − 243 , 81 4 ,
E 1.
2
x
2. x 3 y
F 1.
A.4
3y
y +1
B 1.
3
2
C 1.
1
5 x − 10
2.
x+5
x −1
6.
5.
D 1.
x
2.
1
2
3
3. − 12
r +1
r
6 x + 13
x+3
4x
2.
x+4
1
r2
5.
2
4
6. 4 y
6.
2.
3 5 −3 3
x
x ( x + 3)
1
5.
x−2
x +1
t −3
3.
(t + 3)(t − 2)
4.
− x2 − 3
(x + 1)(x − 2)(x − 3)
3. x ( x + 1)
x−2
x +1
(x + 6)(x + 1)
4.
x2
6.
4x + 1
(x + 1)(x − 1)
x +1
4.
x −1
8.
C 1. x = ±4 2
D 1. x = −8, 4
2. x = ±2
3. x = 8, 16
2. x = −9.32, − 2.68 3. x = 0.18, 1.82
E 1. x = − 12 , 1
2. x = 1 ± 3
3. x = − 12 + 2
4. x = −3 ± 13
2. x = −3, 0
3. x = −2
4. x = ±2
3 2
2
2
5
A.5
A 1. x = 9
2. no solution
3. ℜ
4. x = 6
5. x = 10
6. x = 3
7. no solution (-4 is an extraneous solution)
8. x = 5
1
B 1. x = 0, − 2
2. x = −2, 0, 4
3. x = −5
4. x = − 12 , 3
F 1. 0,±
125 x 9
y 12
8.
( 16 )
3
7.
7.
5. 22 2
4. − 3 5 + 3 6
1
1
1
4
( x + y )2
3
4. 34 2
y2
5+ 3
11
6.
4. x = −2 ± 14
4. x = −1.46, 5.46
A.3