REVISION CHAPTER ALGEBRA
Transcript of REVISION CHAPTER ALGEBRA
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REVISION CHAPTER
ALGEBRA
LEARNING OUTCOMES
Upon completing this topic, the students will be able to:
1. Differentiate between the types of mathematic function (PLO1-K-C3)
2. Manipulate algebraic expressions by using all of algebra rules (PLO1-K-
C3)
3. Separate an algebraic fraction into its partial fractions (PLO1-K-C3)
1.0 INTRODUCTION
Chapter 1 is started with section 1.1 which is about introduction to the types
of function in mathematics. Then, followed by the review of algebra rules in
section 1.2. Finally, in section 1.3, partial fraction will be discussed. A set
of tutorial for chapter 1 is available in section 1.4
1.1 TYPES OF FUNCTION
The purpose of this reference section is to show you graphs of various
types of functions in order that you can become familiar with the types.
You will discover that each type has its own distinctive graph. By showing
several graphs on one plot you will be able to see their common features.
Examples of the following types of functions are shown in this chapter:
linear
quadratic
power
polynomial
rational
exponential
logarithmic
sinusoidal
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In each case the argument (input) of the function is called x and the value
(output) of the function is called y.
1.1.1 LINEAR FUNCTION
These are functions of the form:
y = m x + b,
where m and b are constants. A typical use for linear functions is converting
from one quantity or set of units to another. Graphs of these functions are
straight lines (see Figure 1.1). m is the slope and b is the y intercept. If m
is positive then the line rises to the right and if m is negative then the line
falls to the right.
Figure 1.1 Linear graph
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1.1.2 QUADRATIC FUNCTION
These are functions of the form:
y = a x 2 + b x + c,
where a, b and c are constants. Their graphs are called parabolas (see
Figure 1.2). This is the next simplest type of function after the linear
function. Falling objects move along parabolic paths. If a is a positive
number then the parabola opens upward and if a is a negative number
then the parabola opens downward.
Figure 1.2 Quadratic graph
1.1.3 POWER FUNCTION
These are functions of the form:
y = a x b,
where a and b are constants. They get their name from the fact that the
variable x is raised to some power. Many physical laws (e.g. the
gravitational force as a function of distance between two objects, or the
bending of a beam as a function of the load on it) are in the form of power
functions. We will assume that a = 1 and look at several cases for b:
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The power b is a positive integer. (See Figure 1.3). When x = 0 these
functions are all zero. When x is big and positive they are all big and
positive. When x is big and negative then the ones with even powers are
big and positive while the ones with odd powers are big and negative.
Figure 1.3 Power graph for positive integer
The power b is a negative integer. (See Figure 1.4). When x = 0 these
functions suffer a division by zero and therefore are all infinite. When x is
big and positive they are small and positive. When x is big and negative
then the ones with even powers are small and positive while the ones with
odd powers are small and negative.
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Figure 1.4 Power graph for negative integer
The power b is a fraction between 0 and 1. (See Figure 1.5). When x =
0 these functions are all zero. The curves are vertical at the origin and as x
increases they increase but curve toward the x axis.
Figure 1.5 Power graph for integer between 0 and 1
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1.1.4 POLYNOMIAL FUNCTION
These are functions of the form:
y = an Β· x n + an β1 Β· x n β1 + β¦ + a2 Β· x 2 + a1 Β· x + a0,
where an, an β1, β¦ , a2, a1, a0 are constants. Only whole number powers of
x are allowed. The highest power of x that occurs is called the degree of
the polynomial. The graph in Figure 1.6 shows examples of degree 4 and
degree 5 polynomials. The degree gives the maximum number of βups and
downsβ that the polynomial can have and also the maximum number of
crossings of the x axis that it can have.
Polynomials are useful for generating smooth curves in computer graphics
applications and for approximating other types of functions.
Figure 1.6 Polynomial graph
1.1.5 RATIONAL FUNCTION
These functions are the ratio of two polynomials. One field of study where
they are important is in stability analysis of mechanical and electrical
systems (which uses Laplace transforms).
When the polynomial in the denominator is zero then the rational function
becomes infinite as indicated by a vertical dotted line (called an asymptote)
in its graph. For the example (see Figure 1.7) this happens when x= β2 and
7
when x = 7. When x becomes very large the curve may level off. The curve
to the right levels off at y = 5.
Figure 1.7 Rational graph with horizontal asymptote at y=5 and
vertical asymptote at x= β2 and x = 7
The graph in Figure 1.8 shows another example of a rational function. This
one has a division by zero at x = 0. It doesn't level off but does approach
the straight line y = x when x is large, as indicated by the dotted line (another
asymptote).
Figure 1.8 Rational graph
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1.1.6 EXPONENTIAL FUNCTION
These are functions of the form:
y = a b x,
where x is in an exponent (not in the base as was the case for power
functions) and a and b are constants. (Note that only b is raised to the power
x; not a.) If the base b is greater than 1 then the result is exponential growth
(see Figure 1.9). Many physical quantities grow exponentially (e.g. animal
populations and cash in an interest-bearing account).
Figure 1.9 Exponential graph for base greater than 1
If the base b is smaller than 1 then the result is exponential decay (see
Figure 1.10). Many quantities decay exponentially (e.g. the sunlight
reaching a given depth of the ocean and the speed of an object slowing
down due to friction).
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Figure 1.10 Exponential graph for base smaller than 1
1.1.7 LOGARITHMIC FUNCTION
There are many equivalent ways to define logarithmic functions. We will
define them to be of the form:
y = a ln (x) + b,
where x is in the natural logarithm and a and b are constants. They are
only defined for positive x. For small x they are negative and for large x
they are positive but stay small (See Figure 1.11). Logarithmic functions
accurately describe the response of the human ear to sounds of varying
loudness and the response of the human eye to light of varying
brightness.
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Figure 1.11 Logarithmic graph
1.1.8 SINUSOIDAL FUNCTION
These are functions of the form:
y = a sin (b x + c),
where a, b and c are constants. Sinusoidal functions are useful for
describing anything that has a wave shape with respect to position or time.
Examples are waves on the water, the height of the tide during the course
of the day and alternating current in electricity. Parameter a (called the
amplitude) affects the height of the wave, b (the angular velocity) affects
the width of the wave and c (the phase angle) shifts the wave left or right.
The sinusoidal graph is shown in Figure 1.12
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1.2 REVIEW OF ALGEBRA
Here we review the basic rules and procedures of algebra that you need
to know in order to be successful in calculus.
1.2.1 ARITHMETIC OPERATIONS
The real numbers have the following properties:
π + π = π + π ππ = ππ (Commutative Law)
(π + π) + π = π + (π + π) (ππ)π = π(ππ) (Associative Law)
π(π + π) = ππ + ππ (Distributive law)
In particular, putting π = β1 in the Distributive Law, we get
β(π + π) = (β1)(π + π) = (β1)π + (β1)π
and so
β(π + π) = βπ β π
Example 1.1
i. (3π₯π¦)(β4π₯) = 3(β4)π₯2π¦ = β12π₯2π¦
ii. 2π‘(7π₯ + 2π‘π₯ β 11) = 14π‘π₯ + 4π‘2 β 22π‘
iii. 4 β 3(π₯ β 2) = 4 β 3π₯ + 6 = 10 β 3π₯
If we use the Distributive Law three times, we get
(π + π)(π + π) = (π + π)π + (π + π)π = ππ + ππ + ππ + ππ
This says that we multiply two factors by multiplying each term in one
factor by each term in the other factor and adding the products.
In the case where π = π and π = π, we have
(π + π)2 = π2 + ππ + ππ + π2 = π2 + 2ππ + π2
Similarly, we obtain
(π β π)2 = π2 β 2ππ + π2
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Exercise 1.1
Solve these using appropriate law.
i. (2π₯ + 1)(3π₯ β 5)
ii. (π₯ + 6)2
iii. 3(π₯ β 1)(4π₯ + 3) β 2(π₯ + 6)
1.2.2 FRACTIONS
To add two fractions with the same denominator, we use the Distributive
Law:
π
π+
π
π=
1
πΓ π +
1
πΓ π =
1
π(π + π) =
π + π
π
Thus, it is true that
π + π
π=
π
π+
π
π
But remember to avoid the following common error:
π
π + πβ
π
π+
π
π
(For instance, take π = π = π = 1 to see the error.)
To add two fractions with different denominators, we use a common
denominator:
π
π+
π
π=
ππ + ππ
ππ
We multiply such fractions as follows:
π
πβ
π
π=
ππ
ππ
In particular, it is true that
βπ
π= β
π
π=
π
βπ
To divide two fractions, we invert and multiply:
ππππ
=π
πΓ
π
π=
ππ
ππ
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Example 1.2
i. π₯+3
π₯=
π₯
π₯+
3
π₯= 1 +
3
π₯
ii. π 2π‘
π’β
π’π‘
β2=
π 2π‘2π’
β2π’= β
π 2π‘2
2
Exercise 2.2
i. 3
π₯β1+
π₯
π₯+2
ii.
π₯
π¦+1
1βπ¦
π₯
1.2.3 FACTORING
We have used the Distributive Law to expand certain algebraic expressions.
We sometimes need to reverse this process (again using the Distributive
Law) by factoring an expression as a product of simpler ones. The easiest
situation occurs when the expression has a common factor as follows:
To factor a quadratic of the form π₯2 + ππ₯ + π we note that
(π₯ + π)(π₯ + π ) = π₯2 + (π + π )π₯ + ππ
so we need to choose numbers π and π so that π + π = π and ππ = π.
Example 1.3
Factor π₯2 + 5π₯ β 24.
Solution:
The two integers that add to give 5 and multiply to give -24 are -3
and 8.
Therefore, π₯2 + 5π₯ β 24 = (π₯ β 3)(π₯ + 8)
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Example 1.4
Factor 2π₯2 β 7π₯ β 4.
Solution:
Even though the coefficient of is not 1, we can still look for factors of
the form 2π₯ + π and π₯ + π , where ππ = β4. Experimentation reveals
that
2π₯2 β 7π₯ β 4 = (2π₯ + 1)(π₯ β 4)
Some special quadratics can be factored by using the formula for a
difference of squares:
a. a2 β b2 = (a β b)(a + b)
b. a3 β b3 = (a β b)(a2 + ab + b2)
c. a3 + b3 = (a + b)(a2 β ab + b2)
Exercise 1.3
i. π₯2 β 6π₯ + 9
ii. 4π₯2 β 25
iii. π₯3 + 8
Example 1.5
Simplify π₯2β16
π₯2β2π₯β8
Solution:
Factoring numerator and denominator, we have
π₯2 β 16
π₯2 β 2π₯ β 8=
(π₯ β 4)(π₯ + 4)
(π₯ β 4)(π₯ + 2)=
π₯ + 4
π₯ + 2
To factor polynomials of degree 3 or more, we sometimes use the
following fact:
THE FACTOR THEOREM: If π is a polynomial and π(π) = 0, then
π₯ β π is a factor of π(π₯)
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Example 2.6
Factor π₯3 β 3π₯2 β 10π₯ + 24
Solution:
Let π(π₯) = π₯3 β 3π₯2 β 10π₯ + 24. If π(π) = 0, where π is an integer,
then π is a factor of 24. Thus, the possibilities for π are Β±1, Β±2, Β±3,
Β±4, Β±6, Β±8, Β±12 and Β±24. We find that π(1) = 12, π(β1) = 30, π(2) =
0. By the Factor Theorem, π₯ β 2 is a factor. Instead of substituting
further, we use long division as follows:
Therefore, π₯3 β 3π₯2 β 10π₯ + 24 = (π₯ β 2)(π₯2 β π₯ β 12)
= (π₯ β 2)(π₯ + 3)(π₯ β 4)
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1.2.4 COMPLETING THE SQUARE
Completing the square is a useful technique for graphing parabolas or
integrating rational functions. Completing the square means rewriting a
quadratic ππ₯2 + ππ₯ + π in the form π(π₯ + π)2 + π and can be accomplished
by:
1. Factoring the number from the terms involving π₯.
2. Adding and subtracting the square of half the coefficient of π₯.
In general, we have
ππ₯2 + ππ₯ + π = π [π₯2 +π
ππ₯] + π
= π [π₯2 +π
ππ₯ + (
π
2π)2 β (
π
2π)2] + π
= π(π₯ +π
2π)2 + (π β
π2
4π)
Example 1.7
Rewrite π₯2 + π₯ + 1 and 2π₯2 β 12π + 11 by completing the square.
Solutions:
The square of half the coefficient of is 1
4. Thus
π₯2 + π₯ + 1 = π₯2 + π₯ +1
4β
1
4+ 1 = (π₯ +
1
2)2 +
3
4
2π₯2 β 12π + 11 = 2[π₯2 β 6π₯] + 11 = 2[π₯2 β 6π₯ + 9 β 9] + 11
= 2[(π₯ β 3)2 β 9] + 11 = 2(π₯ β 3)2 β 7
Exercise 1.4
i. 3π₯2 + 4π₯ β 2
ii. π₯2 β 6π₯ β 4
iii. π₯2 + 8π₯ + 5
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1.2.5 RADICALS
The most commonly occurring radicals are square roots. The symbol β
means βthe positive square root of.β Thus
π₯ = βπ πππππ π₯2 = π πππ π₯ β₯ 0
Since π = π₯2 β₯ 0, the symbol βπ makes sense only when π β₯ 0 . Here are
two rules for working with square roots:
a. βππ = βπβπ
b. βπ
π=
βπ
βπ
However, there is no similar rule for the square root of a sum. In fact, you
should remember to avoid the following common error:
βπ + π β βπ + βπ
(For instance, take π = 9 and π = 16 to see the error.)
Example 1.8
i. β18
β2= β
18
2= β9 = 3
ii. βπ₯2π¦ = βπ₯2 βπ¦ = π₯βπ¦
In general, if π is a positive integer,
π₯ = βπ π
πππππ π₯π = π
βπππ
= βππ βπ
π & β
π
π
π=
βππ
βππ
To rationalize a numerator or denominator that contains an expression
such as βπ β βπ , we multiply both the numerator and the denominator by
the conjugate radical βπ + βπ. Then we can take advantage of the formula
for a difference of squares:
(βπ β βπ)(βπ + βπ) = (βπ)2 β (βπ)2 = π β π
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Example 1.9
Rationalize the numerator in the expression βπ₯+4β2
π₯.
Solution:
We multiply the numerator and the denominator by the conjugate
radical βπ₯ + 4 + 2
βπ₯ + 4 β 2
π₯= (
βπ₯ + 4 β 2
π₯) (
βπ₯ + 4 + 2
βπ₯ + 4 + 2) =
(π₯ + 4) β 4
π₯(βπ₯ + 4 + 2)
=π₯
π₯(βπ₯ + 4 + 2)=
1
βπ₯ + 4 + 2
Exercise 1.5
i. βπ₯43
ii. βπ₯+1β3
π₯β8
iii. β3 β β12
1.2.6 EXPONENTS
Let π be any positive number and let π be a positive integer. Then, by
definition,
a. ππ = (π β π β π β β π)} π ππππ‘πππ
b. π0 = 1
c. πβπ =1
ππ
d. π1 πβ = βππ
e. ππ πβ = βπππ= ( βπ
π)π π ππ πππ¦ πππ‘ππππ
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Laws of Exponents: Let π and π be positive numbers and let π and π be
any rational numbers (that is, ratios of integers). Then,
a. ππ Γ ππ = ππ+π
b. ππ
ππ = ππβπ
c. (ππ)π = πππ
d. (ππ)π = ππππ
e. (π
π)π =
ππ
ππ π β 0
In words, these five laws can be stated as follows:
1. To multiply two powers of the same number, we add the exponents.
2. To divide two powers of the same number, we subtract the exponents.
3. To raise a power to a new power, we multiply the exponents.
4. To raise a product to a power, we raise each factor to the power.
5. To raise a quotient to a power, we raise both numerator and denominator
to the power
Example 1.10
(π₯
π¦)3(
π¦2π₯
π§)4 =
π₯3
π¦3β
π¦8π₯4
π§4= π₯7π¦5π§β4
Exercise 1.6
i. 28 Γ 82
ii. π₯β2βπ¦β2
π₯β1+π¦β1
iii. 1
βπ₯43
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1.3 PARTIAL FRACTION
Given a rational function, if the degree of the polynomial in the numerator
is less than the degree of the polynomial in the denominator, then the
rational function can be expressed as the sum of rational functions whose
denominators are powers of linear polynomials and powers of irreducible
quadratic functions. This sum is called the partial fraction decomposition of
the rational function.
Given the rational function π(π₯) =π(π₯)
π(π₯).
If the factorization of the polynomial π(π₯) contains π identical linear factors,
ππ₯ + π then the partial fraction decomposition of π(π₯) contains a sum of the
form:
π΄1
ππ₯ + π+
π΄2
(ππ₯ + π)2+
π΄3
(ππ₯ + π)3+βββββ +
π΄π
(ππ₯ + π)π
Where π΄1, π΄2, π΄3, β¦ . . , π΄π are constants to be determined.
If the factorization of the polynomial π(π₯) contains π identical irreducible
quadratic factors ππ₯2 + ππ₯ + π, then the partial fraction decomposition of
π(π₯) contains a sum of the form
π΄1π₯ + π΅1
ππ₯2 + ππ₯ + π+
π΄2π₯ + π΅2
(ππ₯2 + ππ₯ + π)2+βββββ +
π΄ππ₯ + π΅π
(ππ₯2 + ππ₯ + π)π
where π΄1, π΄2, β¦ . . , π΄π and π΅1, π΅2, β¦ . . , π΅π are constants to be determined.
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Example 1.11
Evaluate the following function
6
π₯2 β 9π₯
Solution:
= 6
π₯(π₯ β 9)=
π΄
π₯+
π΅
π₯ β 9
We need to solve for A and B.
Multiplying both sides of the equation by π₯(π₯ β 9), we obtain the
equation 6 = π΄(π₯ β 9) + π΅π₯. From here, there are two methods for
solving for A and B.
Method 1
The first method is to solve a system of equations obtained from
equating the coefficients of the terms on each side of the equation.
6 = π΄(π₯ β 9) + π΅π₯ β 6 = π΄π₯ β 9π΄ + π΅π₯ β 6 = (π΄ + π΅)π₯ β 9π΄
The coefficient of the x term on the right side of the equation is π΄ +
π΅. Since there is not an x term on the left side of the equation, then
its coefficient is zero. Equating the coefficients of the x terms on each
side of the equation, we obtain that. π΄ + π΅ = 0. The constant term
on the right side of the equation is β9π΄. The constant term on the
left side of the equation is 6. Equating the constant terms on each
side of the equation, we obtain that β9π΄ = 6. Thus, to solve for A
and B, we will solve the system of equations π΄ + π΅ = 0 and β9π΄ =
6. The second equation gives us that = β2
3 . The first equation
gives us that π΅ = βπ΄ =2
3.
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Method 2
The second method involves choosing a value for x that will leave A
but will eliminate B in the equation and choosing a value for x that
will leave B but will eliminate A.
To solve for A, choose:
π₯ = 0, 6 = π΄(0 β 9) + π΅(0) β 6 = β9π΄ β΄ π΄ = β2
3
To solve for B, choose:
π₯ = 9, 6 = π΄(9 β 9) + π΅(9) β 6 = 9π΅ β΄ π΅ = 2
3
β΄ 6
π₯(π₯ β 9)=
β23
π₯+
23
π₯ β 9=
2
3(
1
π₯ β 9β
1
π₯)
If the degree of the polynomial in the numerator is greater or equal
than the degree of the polynomial in the denominator, then we need
to use long division first before decompose them into simpler parts.
Example 1.12
Express 242
5054322
23
xx
xxxin partial fractions
Solutions:
First use long division because the degree of the polynomial in the
numerator is more than the degree of the polynomial in the
denominator
264....................................
242.............................
506.............................
4842.....................
12
505432242
2
2
23
232
x
xx
xx
xxx
x
xxxxx
Then, 242
26412
242
50543222
23
xx
xx
xx
xxx
24
64)6)(4(
264
x
B
x
A
xx
x )4()6(264 xBxAx
Choose x = 4, then A = -1 and choose x = -6, then B = 5
Final answer:
6
5
4
112
242
5054322
23
xxx
xx
xxx
Example 1.13
Express )362)(4(
71872
2
xxx
xx in partial fractions.
Solutions:
The factor 12)4(),362( 22 acbxx which is not a perfect
square. Therefore, )362( 2 xx is irreducible.
The partial fractions of )362)(4(
71872
2
xxx
xx will be of the form of
362)4( 2
xx
CBx
x
A.
Multiplying throughout by the complete denominator:
)4)(()362(7187 22 xCBxxxAxx
Then multiply out and collect up like terms, and that gives:
CAxCBAxBAxx 43)46()2(7187 22
Now you can equate coefficients of like terms on each side and finish
it.
[x2] BA 27 AB 27 (1)
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[Constant] CA 437 4
73
AC (2)
[x] )4
738286(18
AAA A = 3
Substitution in (1) and (2) gives B = 1 and C = 4
362
4
4
3
)362)(4(
718722
2
xx
x
xxxx
xx
Example 1.12
Express 2)27(
1435
x
x in partial fraction
Solutions:
There is a rule that applies:
Repeated factors in the denominator of the algebraic expessionn of
the form 2)( bax give partial fractions of the form 2)( bax
B
bax
A
Consequently, we write:
22 )27(27)27(
1435
x
B
x
A
x
x Then we multiply throughout as usual
by the original denominator.
BAAxx 271435
Now we simply equate coefficients and A and B are found:
22 )27(
4
27
5
)27(
1435
xxx
x
Exercise 1.7
i. 1
π₯2+3π₯
ii. π₯β5
π₯2β2π₯β8
iii. 2π₯2βπ₯+20
(π₯β2)(π₯2+9)
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1.4 TUTORIAL
a. Expand and simplify.
i. (β6ππ)(0.5ππ) ii. β(2π₯2π¦)(βπ₯π¦4)
iii. 2π₯(π₯ β 5) iv. (4 β 3π₯)π₯
v. β2(4 β 3π) vi. 8 β (4 + π₯)
vii. 4(π₯2 β π₯ + 2) β 5(π₯2 β 2π₯ +
1)
viii. 5(3π‘ β 4) β (π‘2 + 2) β
2π‘(π‘ β 3)
ix. (4π₯ β 1)(3π₯ + 7) x. π₯(π₯ β 1)(π₯ + 2)
xi. (2π₯ β 1)2 xii. (2 + 3π₯)2
xiii. π¦4(6 β π¦)(5 + π¦) xiv. (π‘ β 5)2 β 2(π‘ + 3)(8π‘ β 1)
xv. (π‘ β 5)2 β 2(π‘ + 3)(8π‘ β 1) xvi. (1 + 2π₯)(π₯2 β 3π₯ + 1)
b. Perform the indicated operations and simplify.
i. 2+8π₯
2 ii.
9πβ6
3π
iii. 1
π₯+5+
2
π₯β3 iv.
1
π₯+1+
1
π₯β1
v. π’ + 1 +π’
π’+1 vi.
2
π2β
3
ππ+
4
π2
vii. π₯ π¦β
π§ viii.
π₯
π¦ π§β
ix. (β2π
π ) (
π 2
β6π‘) x.
π
ππΓ·
π
ππ
xi. 1+
1
πβ1
1β1
πβ1
xii. 1 +
1
1+1
1+π₯
27
c. Factor the expression.
i. 6π₯2 β 5π₯ β 6 ii. π₯2 + 10π₯ + 25
iii. π‘3 + 1 iv. 4π‘2 β 9π 2
v. 4π‘2 β 12π‘ + 9 vi. π₯3 β 27
vii. π₯3 + 2π₯2 + π₯ viii. π₯3 β 4π₯2 + 5π₯ β 2
ix. π₯3 + 3π₯2 β π₯ β 3 x. π₯3 β 2π₯2 β 23π₯ + 60
xi. π₯3 + 5π₯2 β 2π₯ β 24 xii. π₯3 β 3π₯2 β 4π₯ + 12
d. Simplify the expression.
i. π₯2+π₯β2
π₯2β3π₯+2 ii.
2π₯2β3π₯β2
π₯2β4
iii. π₯2β1
π₯2β9π₯+8 iv.
π₯3+5π₯2+6π₯
π₯2βπ₯β12
v. 1
π₯+3+
1
π₯2β9 vi.
π₯
π₯2+π₯β2β
2
π₯2β5π₯+4
e. Complete the square.
i. π₯2 + 2π₯ + 5 ii. π₯2 β 16π₯ + 80
iii. π₯2 β 5π₯ + 10 iv. π₯2 + 3π₯ + 1
v. 4π₯2 + 4π₯ β 2 vi. 3π₯2 β 24π₯ + 50
f. Simplify the radicals.
i. β32β2 ii. βπ₯π¦βπ₯3π¦
iii. ββ23
β543
iv. β16π4π3
v. β32π₯44
β24 vi.
β96π65
β3π5
28
g. Use the Laws of Exponents to rewrite and simplify the
expression.
i. 3β1 2β ii. 961 5β
iii. πβ3π4
πβ5π5 iv.
π₯β1+π¦β1
(π₯+π¦)β1
v. (2π₯2π¦4)3 2β vi. (π₯β5π¦3π§10)β3 5β
vii. βπ¦65 viii. (βπ
4)3
ix. 1
(βπ‘)5 x.
βπ₯58
βπ₯34
xi. ββπ π‘ π‘1 2β
π 2 3β
4
xii. βπ2π+14
Γ βπβ14
h. Rationalize the expression.
i. βπ₯β3
π₯β9 ii.
(1 βπ₯β )β1
π₯β1
iii. π₯βπ₯β8
π₯β4 iv.
β2+β+β2ββ
β
v. 2
3ββ5 vi.
1
βπ₯ββπ¦
vii. βπ₯2 + 3π₯ + 4 β π₯ viii. βπ₯2 + π₯ β βπ₯2 β π₯
i. Express each of the following in partial fraction form.
i. 3π₯+9
π₯2+8π₯+12 ii.
π₯2+π₯+1
π₯2+3π₯+2
iii. 7π₯2+6π₯+5
(π₯+1)(π₯2+π₯+1) iv.
5π₯+6
(π₯β1)2
v. 2π₯3β5π₯+13
(π₯+4)2