Algebra 2b Review
Short Answer
1. Rewrite the polynomial –2x2 + 12 – x4 – 3x3 – 7x5 – 20x in standard form. Then, identify the leading
coefficient, degree, and number of terms. Name the polynomial.
2. Add. Write your answer in standard form.
3. Find the product
.
4. Divide by using long division:
.
5. Write an expression that represents the width of a rectangle with length
6. Factor
.
7. Solve the polynomial equation
by factoring.
8. Identify the roots of
. State the multiplicity of each root.
9. Graph the function
.
10. Tell whether the function
shows growth or decay. Then graph the function.
11. Write the exponential equation
12. Write the logarithmic equation
13. Evaluate
as a single logarithm. Simplify, if possible.
16. Express
as a product. Simplify, if possible.
17. Simplify the expression
18. Evaluate
21. Simplify
22. Multiply
in exponential from.
as a single logarithm. Simplify, if possible.
15. Express
20. Solve
in logarithmic form.
16 by using mental math.
14. Express
19. Solve
and area
.
. If necessary, round your answer to the nearest tenth.
.
.
. Identify any t-values for which the expression is undefined.
. Assume that all expressions are defined.
.
23. Divide
. Assume that all expressions are defined.
24. Solve
. Check your answer.
25. Add
.
26. Simplify the expression
. Assume that all variables are positive.
27. Write the expression
in radical form, and simplify. Round to the nearest whole number if necessary.
28. Write the expression
by using rational exponents.
29. Simplify the expression
.
30. Solve the equation
.
31. Solve
.
32. Solve
.
33. Simplify
.
34. Solve the equation
35. Simplify
.
by rationalizing the denominator.
36. Find the center and radius of a circle that has a diameter with endpoints
37. Graph the equation
.
38. Write an equation in standard form for the ellipse shown with center (0, 0).
and
.
y
Focus
(0, 4)
x
(0, –5)
39. Graph the ellipse
.
40. Identify the conic section the equation
represents.
41. Identify the conic section that the equation
represents.
42. Find the vertices, co-vertices, and asymptotes of the hyperbola
, and then graph.
43. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola
. Then,
graph the parabola.
44. Simplify
. Identify any x-values for which the expression is undefined.
45. The area of a rectangle is equal to
units, what expression represents its width?
46. Identify the zeros and vertical asymptotes of
47. Identify the zeros and asymptotes of
48. Simplify
by rationalizing the denominator.
square units. If the length of the rectangle is equal to
. Then graph.
. Then graph.
49. Write the equation of a circle with center
and radius
.
50. Write an equation in standard form for the hyperbola with center
, vertex
, and focus
.
Algebra 2b Review
Answer Section
SHORT ANSWER
1. ANS:
leading coefficient: –7; degree: 5; number of terms: 6; name: quintic polynomial
The standard form is written with the terms in order from highest to lowest degree.
In standard form, the degree of the first term is the degree of the polynomial.
The polynomial has 6 terms. It is a quintic polynomial.
PTS: 1
DIF: Average
TOP: 6-1 Polynomials
2. ANS:
=
=
REF: Page 407
OBJ: 6-1.2 Classifying Polynomials
Identify like terms. Rearrange terms to get like terms
together.
Combine like terms.
PTS: 1
DIF: Basic
REF: Page 407
OBJ: 6-1.3 Adding and Subtracting Polynomials
STA: 2A.2.A
TOP: 6-1 Polynomials
3. ANS:
NAT: 12.5.3.c
Use the Distributive Property to multiply the monomial by each term inside the parentheses. Group terms to
get like bases together, and then multiply.
PTS: 1
DIF: Basic
REF: Page 414
OBJ: 6-2.1 Multiplying a Monomial and a Polynomial
STA: 2A.2.A
TOP: 6-2 Multiplying Polynomials
4. ANS:
NAT: 12.5.3.c
To divide, first write the dividend in standard form. Include missing terms with a coefficient of 0.
Then write out in long division form, and divide.
50
Write out the answer with the remainder to get
PTS: 1
DIF: Average
REF: Page 422
OBJ: 6-3.1 Using Long Division to Divide Polynomials
STA: 2A.2.A
TOP: 6-3 Dividing Polynomials
5. ANS:
.
NAT: 12.5.3.c
.
Substitute.
Use synthetic division.
–5
1
12 47 60
–5 –35 –60
1
7
12
0
The width can be represented by
PTS: 1
NAT: 12.5.3.c
6. ANS:
=
=
=
.
DIF: Average
REF: Page 425
TOP: 6-3 Dividing Polynomials
OBJ: 6-3.4 Application
Group terms.
Factor common monomials from each group.
Factor out the common binomial.
Factor the difference of squares.
PTS: 1
DIF: Average
NAT: 12.5.3.d
STA: 2A.2.A
7. ANS:
The roots are 0, –6, and –1.
REF: Page 431
OBJ: 6-4.2 Factoring by Grouping
TOP: 6-4 Factoring Polynomials
Factor out the GCF, 3x3.
Factor the quadratic.
Set each factor equal to 0.
Solve for x.
PTS: 1
DIF: Average
REF: Page 438
OBJ: 6-5.1 Using Factoring to Solve Polynomial Equations
STA: 2A.2.A
TOP: 6-5 Finding Real Roots of Polynomial Equations
8. ANS:
is a factor once, and
is a factor twice.
The root 3 has a multiplicity of 1, and the root has a multiplicity of 2.
is a factor once, and
is a factor twice.
The root 3 has a multiplicity of 1.
The root
has a multiplicity of 2.
PTS: 1
DIF: Average
REF: Page 439
TOP: 6-5 Finding Real Roots of Polynomial Equations
9. ANS:
OBJ: 6-5.2 Identifying Multiplicity
y
25
20
15
10
5
–10 –8
–6
–4
–2
–5
2
4
6
8
10
x
–10
–15
–20
–25
Step 1: Identify the possible rational roots by using the Rational Root Theorem. p = 4 and q = 1, so roots are
positive and negative values in multiples of 2 from 1 to 4.
Step 2: Test possible rational zeros until a zero is identified.
Test x = –1.
Test x = 1.
is a zero, and
Step 3: Factor:
The zeros are –1, –2, and –2.
.
.
Step 4: Plot other points as guidelines.
so the y-intercept is + 4. Plot points between the zeros.
and
Step 5: Identify end behavior.
The degree is odd and the leading coefficient is positive, so as
.
and as
Step 6: Sketch the graph by using all of the information about f(x).
PTS: 1
DIF: Average
REF: Page 455
OBJ: 6-7.3 Graphing Polynomial Functions
TOP: 6-7 Investigating Graphs of Polynomial Functions
10. ANS:
This is an exponential growth function.
y
9
8
7
6
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
Step 1 Find the value of the base: 2.
The base is greater than 1. So, this is an exponential growth function.
Step 2 Choose several values of x and generate ordered pairs. Then, graph the ordered pairs and connect with
a smooth curve.
PTS:
OBJ:
STA:
KEY:
11. ANS:
1
DIF: Basic
REF: Page 491
7-1.1 Graphing Exponential Functions
NAT: 12.5.1.h
2A.4.A
TOP: 7-1 Exponential Functions Growth and Decay
exponential function | graph
The base of the exponent becomes the base of the logarithm.
The exponent is the logarithm.
becomes
.
PTS: 1
DIF: Basic
REF: Page 505
OBJ: 7-3.1 Converting from Exponential to Logarithmic Form
TOP: 7-3 Logarithmic Functions
12. ANS:
STA: 2A.11.A
Logarithmic form:
Exponential form:
The base of the logarithm becomes the base of the power, and the logarithm is the exponent.
PTS: 1
DIF: Basic
REF: Page 506
OBJ: 7-3.2 Converting from Logarithmic to Exponential Form STA: 2A.11.A
TOP: 7-3 Logarithmic Functions
13. ANS:
4
The log is the exponent.
16
Think: What power of the base is the number?
16
16 = 4
PTS: 1
DIF: Average
REF: Page 506
OBJ: 7-3.3 Evaluating Logarithms by Using Mental Math
TOP: 7-3 Logarithmic Functions
14. ANS:
4
STA: 2A.11.A
To add the numbers, multiply the logarithms
4
Think. What exponent on base 7 gives 2401?
PTS: 1
DIF: Basic
REF: Page 514
STA: 2A.2.A
TOP: 7-4 Properties of Logarithms
15. ANS:
3
To subtract the logarithms divide the numbers.
=
=
=3
OBJ: 7-4.1 Adding Logarithms
PTS: 1
STA: 2A.2.A
16. ANS:
–8
OBJ: 7-4.2 Subtracting Logarithms
DIF: Average
REF: Page 513
TOP: 7-4 Properties of Logarithms
=
The Power Property of Logarithms says
=
Because
,
.
.
= –8
PTS: 1
DIF: Average
REF: Page 513
OBJ: 7-4.3 Simplifying Logarithms with Exponents
STA: 2A.2.A
TOP: 7-4 Properties of Logarithms
17. ANS:
3
Factor 8. Then write it in the form of , and apply the Inverse Properties of Logarithms and Exponents.
PTS: 1
STA: 2A.2.A
DIF: Basic
REF: Page 514
TOP: 7-4 Properties of Logarithms
OBJ: 7-4.4 Recognizing Inverses
18. ANS:
1.3
Method 1 Change to base 10.

1.3
Use a calculator.
Divide. Round to the nearest tenth.
Method 2 Change to base 5, because both 125 and 625 are powers of 5.
»1.3
PTS: 1
DIF: Average
REF: Page 514
OBJ: 7-4.5 Changing the Base of a Logarithm
TOP: 7-4 Properties of Logarithms
19. ANS:
x=4
STA: 2A.2.A
Rewrite each side as powers of the same base.
To raise a power to a power, multiply the exponents.
The bases are the same, so the exponents must be equal.
The solution is
.
PTS: 1
DIF: Average
REF: Page 522
OBJ: 7-5.1 Solving Exponential Equations
STA: 2A.11.D
TOP: 7-5 Exponential and Logarithmic Equations and Inequalities
20. ANS:
Apply the Quotient Property.
Simplify.
Use the Power Property.
Divide.
Use 5 as the base for both sides.
Use inverse properties.
PTS:
OBJ:
TOP:
21. ANS:
6t; t
1
DIF: Advanced
REF: Page 523
7-5.3 Solving Logarithmic Equations
STA: 2A.11.D
7-5 Exponential and Logarithmic Equations and Inequalities
2 or 0
Factor common factors out of the numerator and/or denominator. Divide out the common factors to simplify
the expression. Finally, use the original denominator to determine any t-values for which the expression is
undefined.
PTS: 1
DIF: Average
REF: Page 577
OBJ: 8-2.1 Simplifying Rational Expressions
NAT: 12.5.3.c
STA: 2A.2.A
TOP: 8-2 Multiplying and Dividing Rational Expressions
22. ANS:
Arrange the expressions so like terms are together:
.
Multiply the numerators and denominators, remembering to add exponents when multiplying:
Divide, remembering to subtract exponents:
Since
, this expression simplifies to
.
.
PTS: 1
DIF: Basic
REF: Page 578
OBJ: 8-2.3 Multiplying Rational Expressions
NAT: 12.5.3.c
STA: 2A.2.A
TOP: 8-2 Multiplying and Dividing Rational Expressions
23. ANS:
Rewrite as multiplication by the reciprocal.
=
Simplify by canceling common factors.
=
PTS: 1
DIF: Basic
REF: Page 579
OBJ: 8-2.4 Dividing Rational Expressions
NAT: 12.5.3.c
STA: 2A.2.A
TOP: 8-2 Multiplying and Dividing Rational Expressions
24. ANS:
There is no solution because the original equation is undefined at
.
Note that
.
Factor.
The factor
cancels.
Because the left side of the original equation is undefined when
PTS: 1
DIF: Average
REF: Page 579
, there is no solution.
.
OBJ: 8-2.5 Solving Simple Rational Equations
NAT: 12.5.3.c
STA: 2A.10.D
TOP: 8-2 Multiplying and Dividing Rational Expressions
25. ANS:
Factor the denominators. The LCD is
=
Multiply by
.
.
=
=
Add the numerators.
=
Factor the numerator.
=
Divide the common factor.
PTS: 1
NAT: 12.5.3.c
26. ANS:
=
=
=
DIF: Average
STA: 2A.2.A
REF: Page 584
OBJ: 8-3.3 Adding Rational Expressions
TOP: 8-3 Adding and Subtracting Rational Expressions
Factor into perfect powers of four.
Use the Product Property of Roots.
Simplify.
PTS: 1
DIF: Basic
REF: Page 611
OBJ: 8-6.2 Simplifying Radical Expressions
NAT: 12.5.3.c
STA: 2A.2.A
TOP: 8-6 Radical Expressions and Rational Exponents
27. ANS:
; 25
Write with a radical.
Evaluate the root.
Evaluate the power.
PTS: 1
DIF: Average
REF: Page 612
OBJ: 8-6.3 Writing Expressions in Radical Form
NAT: 12.5.3.b
STA: 2A.2.A
TOP: 8-6 Radical Expressions and Rational Exponents
28. ANS:
PTS: 1
DIF: Average
REF: Page 612
OBJ: 8-6.4 Writing Expressions by Using Rational Exponents NAT: 12.5.3.b
STA: 2A.2.A
TOP: 8-6 Radical Expressions and Rational Exponents
29. ANS:
27
Product of Powers
Simplify.
PTS: 1
DIF: Basic
REF: Page 613
OBJ: 8-6.5 Simplifying Expressions with Rational Exponents NAT: 12.5.3.c
STA: 2A.2.A
TOP: 8-6 Radical Expressions and Rational Exponents
30. ANS:
Subtract –5 from both sides.
Square both sides.
Simplify.
Check
OK
PTS: 1
DIF: Basic
REF: Page 628
OBJ: 8-8.1 Solving Equations Containing One Radical
TOP: 8-8 Solving Radical Equations and Inequalities
31. ANS:
x=3
Square both sides.
Simplify.
Distribute 4.
Solve for x.
STA: 2A.9.D
PTS: 1
DIF: Average
REF: Page 629
OBJ: 8-8.2 Solving Equations Containing Two Radicals
TOP: 8-8 Solving Radical Equations and Inequalities
32. ANS:
x=6
Step 1 Solve for x.
STA: 2A.9.D
Raise both sides to the reciprocal power.
or
or
Simplify.
Write in standard form.
Factor.
Use the Zero-Product Property.
Solve for x.
Step 2 Use substitution to check for extraneous solutions.
Because
does not satisfy the original equation, it is extraneous. The only solution is
PTS: 1
DIF: Average
REF: Page 630
OBJ: 8-8.4 Solving Equations with Rational Exponents
TOP: 8-8 Solving Radical Equations and Inequalities
33. ANS:
.
STA: 2A.9.D
Use the Power Property of Logarithms.
Simplify.
PTS: 1
DIF: Advanced
34. ANS:
There is no solution.
STA: 2A.2.A
TOP: 7-4 Properties of Logarithms
Multiply each term by the LCD, (x – 7).
Simplify. Note that x g 7.
Solve for x.
The solution
is extraneous because it makes the denominators of the original equation equal to 0.
Therefore the equation has no solution.
PTS: 1
NAT: 12.5.4.a
35. ANS:
DIF: Average
STA: 2A.10.D
REF: Page 601
OBJ: 8-5.2 Extraneous Solutions
TOP: 8-5 Solving Rational Equations and Inequalities
Multiply by a form of 1 to get a perfect-square radicand in
the denominator.
Simplify the denominator.
PTS: 1
DIF: Average
REF: Page 23
OBJ: 1-3.3 Rationalizing the Denominator
STA: 2A.2.A
TOP: 1-3 Square Roots
36. ANS:
center
; radius 5
Step 1 Use the midpoint formula to find the center.
The center is
NAT: 12.5.3.c
.
Step 2 Use the Distance Formula with
and
to find the radius.
The radius is 5.
PTS: 1
DIF: Basic
REF: Page 725
OBJ: 10-1.3 Finding the Center and Radius of a Circle
37. ANS:
TOP: 10-1 Introduction to Conic Sections
y
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
Use a graphing calculator to graph the equation. Use a square window on your graphing calculator for an
accurate graph. The graph is an ellipse with center (0, 0) and intercepts (–4, 0), (4, 0), (0, –2), and (0, 2).
PTS: 1
38. ANS:
DIF: Advanced
STA: 2A.5.B
TOP: 10-1 Introduction to Conic Sections
Step 1 Choose the appropriate form of the equation.
. Because the vertical axis is longer.
Step 2 Identify the values of a and c.
; The vertex (0, –5) gives the value of a.
; The focus
gives the value of c.
Step 3 Use the equation
to find the value of
.
Step 4 Write the equation
PTS: 1
DIF: Average
REF: Page 737
OBJ: 10-3.2 Using Standard Form to Write an Equation for an Ellipse
STA: 2A.2.A
TOP: 10-3 Ellipses
39. ANS:
y
18
12
6
–18
–12
–6
6
–6
12
18
x
(6, –5)
–12
–18
Step 1 Rewrite the equation as
Step 2 Identify the values of h, k, a, and b.
and
, so the center is
.
and
; because
the major axis is horizontal.
Step 3 The vertices are
and the co-vertices are
, or
, or
and
and
,
.
PTS: 1
DIF: Basic
REF: Page 738
STA: 2A.5.B
TOP: 10-3 Ellipses
40. ANS:
ellipse
The standard form of an ellipse, where
, can be written as
OBJ: 10-3.3 Graphing Ellipses
, where the major axis is horizontal, or
, where the major axis is vertical.
is an ellipse with a vertical axis.
PTS: 1
DIF: Advanced
REF: Page 760
OBJ: 10-6.1 Identifying Conic Sections in Standard Form
TOP: 10-6 Identifying Conic Sections
41. ANS:
hyperbola
The general form for a conic section is
,
,
STA: 2A.5.D
Identify the values for A, B, and C.
Substitute into
.
Simplify.
.
Because
, the equation represents a hyperbola.
PTS: 1
DIF: Average
REF: Page 761
OBJ: 10-6.2 Identifying Conic Sections in General Form
TOP: 10-6 Identifying Conic Sections
42. ANS:
Vertices:
and
Co-vertices:
and
Asymptotes:
STA: 2A.5.D
and
y
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
Step 1 The equation is of the form
, so the transverse axis is vertical with center
.
Step 2 Because
.
and
, the vertices are
Step 3 The equations of the asymptotes are
Step 4 Draw a box using the vertices and covertices.Draw the asymptotes through the corners of
the box.
Step 5 Draw the hyperbola using the vertices and the
asymptotes.
and
and the co-vertices are
and
and
.
y
11
10
9
8
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2–1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
1 2 3 4 5 6 7 8 9
x
PTS: 1
DIF: Average
REF: Page 746
OBJ: 10-4.3 Graphing a Hyperbola
STA: 2A.5.B
TOP: 10-4 Hyperbolas
43. ANS:
Vertex
, focus
,
, axis of symmetry
, and directrix
.
y
5
4
3
2
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
–4
–5
The standard form for a parabola with a vertical axis of symmetry is
Step 1 The vertex is (h, k) or
Step 2
, so
.
.
and
.
Step 3 The graph has a vertical axis of symmetry, opens down, and
Step 4 The focus is (h, k + p). Substitute to get
Step 5 The directrix is a horizontal line
. Substitute to get
Step 6 Use the information to sketch the graph.
y
5
directrix y = 4
4
–5
vertex
(–2, 2)
3
focus
(–2, 0)
1
–4
–3
–2
axis of
symmetry
x = –2
p = –2
2
–1
–1
–2
–3
–4
–5
1
2
3
4
5
or
x
.
.
or
.
PTS: 1
STA: 2A.5.B
44. ANS:
DIF: Average
REF: Page 753
TOP: 10-5 Parabolas
The expression is undefined at
and
Factor
OBJ: 10-5.3 Graphing Parabolas
.
from the numerator and reorder the terms.
=
Factor the numerator and denominator.
=
Divide the common factors and simplify.
The expression is undefined at those x-values, 2 and
PTS: 1
NAT: 12.5.3.c
45. ANS:
DIF: Average
STA: 2A.2.A
, that make the original denominator 0.
REF: Page 578
OBJ: 8-2.2 Simplifying by Factoring -1
TOP: 8-2 Multiplying and Dividing Rational Expressions
Area of a rectangle equals length times width.
Substitute the area and length expressions given.
Factor and solve for w.
Simplify.
PTS: 1
DIF: Advanced
NAT: 12.5.3.c
TOP: 8-2 Multiplying and Dividing Rational Expressions
46. ANS:
Zeros at and .
Vertical asymptote:
y
STA: 2A.2.A
20
10
–20
–10
10
–10
–20
Factor the numerator.
20
x
The zeros are the values that make the numerator zero, x = and x = .
The vertical asymptote is where the denominator is zero, x =
Plot the zeros and draw the asymptote, then make a table of values to fill in missing points.
PTS: 1
DIF: Average
REF: Page 594
OBJ: 8-4.3 Graphing Rational Functions with Vertical Asymptotes
STA: 2A.10.A
TOP: 8-4 Rational Functions
47. ANS:
Zeros:
and 3
Vertical asymptotes:
,
Horizontal asymptote:
y
12
10
8
6
4
2
–12 –10 –8 –6 –4 –2–2
2
4
6
8 10 12
x
–4
–6
–8
–10
–12
Factor the numerator and denominator.
Zeros:
and 3
Vertical asymptotes:
Horizontal asymptote:
,
The numerator is 0 when
or
.
The denominator is 0 when
or
.
Both p and q have the same degree: 2. The
horizontal asymptote is
.
PTS: 1
DIF: Average
REF: Page 595
OBJ: 8-4.4 Graphing Rational Functions with Vertical and Horizontal Asymptotes
STA: 2A.10.A
TOP: 8-4 Rational Functions
48. ANS:
Multiply by a form of 1 to get a perfect-square radicand in
the denominator.
Simplify the denominator.
PTS: 1
DIF: Average
REF: Page 23
OBJ: 1-3.3 Rationalizing the Denominator
STA: 2A.2.A
TOP: 1-3 Square Roots
49. ANS:
Use the Distance Formula with
,
NAT: 12.5.3.c
, and distance equal to the radius, 3.
Use the Distance Formula.
Substitute.
Square both sides.
PTS: 1
DIF: Basic
REF: Page 729
OBJ: 10-2.1 Using the Distance Formula to Write the Equation of a Circle
STA: 2A.2.A
TOP: 10-2 Circles
50. ANS:
Step 1 The vertex and focus are on the vertical axis so the equation will be of the form:
.
Step 2 The vertex is
and the focus is
Step 3 The equation of the hyperbola is
, so
and
. Use
.
PTS: 1
DIF: Average
REF: Page 745
OBJ: 10-4.2 Writing Equations of Hyperbolas
TOP: 10-4 Hyperbolas
STA: 2A.5.C
to solve for
.