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MODULE III
FOR PRE – FINAL PERIOD
Learning Objectives
At the end of Module III, the students are expected to be able to:
1. Enumerate and discuss the methods of factoring polynomials;
2. Factor polynomials;
3. Review further on operation of fractions;
4. Explain the concept of rational expressions;
5. Reduce rational expressions to the lowest terms; and,
6. Operate on rational exponent and reduce to lowest terms.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
MODULE III
3.1 FACTORING POLYNOMIALS
Factoring is the process of finding two or more factors whose product is the
original expression. Each of the polynomials whose product is equal to the given are said
to be factors of the originals. For example, since (x2 + 1) (x2 – 1) = x4 – 1, then x2 – 1
and x2 + 1 are factors of x4 – 1. However, (x2 + 1) (x + 1) (x – 1) are other factors of x4
– 1.
If the only factors of a polynomial are itself, its negative, and ±1, then the
polynomial is said to be prime. A polynomial that is expressed as a product of primes is
said to be completely factored.
The following are some techniques in factoring:
1. Common Factor
In factoring, the initial step is to check for common factors and
bring these common factors out.
Illustration:
Factor 8x – 6y. The common factor of 8 and 6 is 2. Both 8 and 6
can be divided by the common factor 2. Therefore:
8x – 6y = 2(4x – 3y)
Factor 3x5yz2 – 6x3y2z + 12x2y3. The GCF in each term is the 3x2y.
Rewriting:
3π‘₯⁡𝑦𝑧²
3π‘₯²π‘¦
-
6π‘₯³π‘¦²π‘§
3π‘₯²π‘¦
+
12π‘₯²π‘¦³
3π‘₯²π‘¦
= x3z2 – 2xyz + 4y2
Or: 3x2y (3x5yz2 - 6x3y2z + 12x2y3) = x3z2 – 2xyz + 4y2
Factor the following:
1. 2x4y2z3 + 8x2yz2 - 6x3y5z44. – 3xy + 4x2y – 5xy2
2. x3n - 2x2n5. 2m2 - 3m2n2 + 5m6
3. 15xy + 3xz
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
2. Factor by grouping
There are instances when polynomials, especially those containing
more than three terms, do not have a common factor. However, if the
terms are grouped in some way, other factoring technique could be used
to factor the resulting expression.
Illustration:
Factor am + bm + an + bn, grouping (am + an) + (bm + bn).
Take out common factors a(m + n) + b(m + n)
Finally, (a + b) (m + n)
Factor the following:
1. 6x2 - 4ax - 9bx + 6ab
2. (2x – 3a) (2x – 3b)
3. m2 + 2mn + n2 - 4
4. mx + my - nx - ny
5. ax + ay + bx + by + cx + cy
3. Factoring the difference of two squares
From the product of the sum and difference of two binomials, a
polynomial which is a difference of two squares is obtained. From the
expression
(x + y) (x – y) = x2 - y2
And given x2 - y2 = (x + y) (x – y)
Factor the following completely:
1. x2 - 9
2. 25x2 - 16
3. 4u4v2 - w2
4. x4 - 16y4
ISHRM 2014
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College Algebra
MT 111/MT 211 Module III
5. 4x2 - 25y4
4. Factoring the Sum and Difference of Two Cubes
For the sum and difference of two cubes the formulas are:
x3 + y3 = (x + y) (x2 - xy + y2)
x3 - y3 = (x - y) (x2 + xy + y2)
Illustrations:
1. 125 - y6 = (5 - y2) (25 + 5y2 + y4)
2. 8x3 + 27y3 = (2x + 3y) (4x2 - 6xy + 9y2)
Factor the following completely
1. z9 - 1
2. w3 + 125
3. 27s3 + 64t3
4. 8 - 216z3
5. m3 - z3
5. Factoring Perfect Square Trinomials
The square of binomial is called the perfect square trinomial. To
recognize this form, the terms have to be arranged in descending powers
of the variable. Then, determine if it satisfies the following properties:
1. It is a trinomial.
2. The last term is positive
3. The first and the last terms are perfect squares.
4. The middle term is twice the product of the square roots of the
first and the last terms.
Given a square of a binomial: (x + y)2 = x2 + 2xy + y2.
Factor this perfect square trinomial: x2 + 2xy + y2 = (x + y)2.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
Illustrations:
1. m4 – 2m2n + n2 = (m2 - n)2
2. x2y2 + 4xy + 4 = (xy + 2)2
Factor the following completely:
1. 4x2 + 12x + 9
2. m2n2 + 12mn + 36
3. 9x2 + 30xy + 25y2
4. x2 + 4xy + 4y2
5. 25a2 - 40ab + 16b2
6. Factoring other Quadratic Trinomials
These are the quadratic trinomials which are not perfect square. In
such case, the special products of the forms (x + a) (x + b) shall be used as
reference. To factor trinomials which are not perfect squares, use either
x2 + (a + b)x + ab = (x + a) (x + b) or
acx2 + (ad + bc)xy + bdy2 = (ax + by) (cx + dy)
Illustrations:
1. 2x2 -19 + 42 = (x – 6) (2x – 7)
2. 28x2 – 2x – 6 = (4x – 2) (7x + 3)
Factor the following completely:
1. 6x2 – 13x – 28
2. 12x2+ 2x – 2
3. 10x2 - 29x - 21
4. x2 + 5x + 6
5. 15a2 +ab – 6b2
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO.1, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
I. Factoring Polynomials.
A. Common Factor
1. 27xy + 9xz
2. 4x4y2z3 + 28x2yz2 - 36x3y5z4
3. 21x2 - 12x + 27x3 + 6x4
4. – 9xy + 3x2y – 18xy2
5. 2m2 - 10m2n2 + 20m6
B. Factor by grouping
1. 3px + 4py - 18x - 24y
2. 15mx – 10my + 12nx – 8ny
3. a2 + 2ab + b2 - 4
4. ma + mb - na - nb
5. amx + ay – bmx – by
ISHRM 2014
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College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO. 2 , MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
C. Factoring the difference of two squares
Factor the following completely:
1. x2 - 100
2. 49x2 - 36
3. 25s4t2 - u2
4. m4 - 9n4
5. 4x2 - 36y4
D. Factoring the Sum and Difference of Two Cubes
Factor the following completely:
1. z9 - 1
2. a3 + 27
3. 8s3 + 125t3
4. 64 - 216z3
5. x3 - y3
ISHRM 2014
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College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO.3, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
E. Factoring Perfect Square Trinomials
Factor the following completely:
1. a2x2 - 2abxy + b2y2
2. 4x2 - 28xy + 49y2
3. 16a2 – 56ab + 49b2
4. 64a2 + 144ab + 81b2
5. 49m2 + 56mn + 16n2
F. Factoring other Quadratic Trinomials
Factor the following completely:
1. 6x2 + 9xy – 10y2
2. 15a2 + 26ab + 8b2
3. 24m2 – 18mn – 15n2
4. 8x2 + 2xy – 15y2
5. 28a2b2 – 3abmn – 18m2n2
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
3.2 FRACTION
Review on operation of fraction.
A fraction represents a part of a whole or, more generally, any number of equal
𝒂
parts. Fractions are of the form 𝒃where aand b are both real numbers and b≠ 0. The
numbera is called the numerator and b is the denominator. The result of fraction is
called a quotient.
Similar fractions are fractions whose denominators are the same or equal.
Examples are:
1
7
2 3
+
3
4
+
7 7
1
-
=
4
6
=
(Addition of similar fractions)
7
2
1
4
2
=
(Subtraction of similar fraction)
Dissimilar fractions are fractions whose denominators have different values.
Examples are:
2
3
3
7
+
1
2
2
-
=
5
4+3
=
=
6
15−14
35
=
7
=1
6
1
(Addition of dissimilar fractions)
6
1
(Subtraction of dissimilar fractions)
35
Proper fraction is a fraction whose denominator is greater than the numerator.
7
The fractions written above are all proper fractions, except the quotient 6 and the
1
equivalent 1 6 .
Improper fraction is a fraction whose numerator is greater than the denominator.
Examples are:
4
3
+
3
=
2
8+9
6
=
17
=
6
5
2 (Always change improper fraction to
6
mixed number)
Or using mixed number to solve improper fractions,
Change the given improper fractions above to mixed numbers.
4
1
3
3
=1
1
1
3
and
+ 1
1
2
3
2
1
2
= 1 adding the two mixed fractions
= (1 + 1)
1
3
+
1
2
=
ISHRM 2014
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College Algebra
MT 111/MT 211 Module III
2+3
2
6
=
5
2
6
Multiplication of fractions
1.
2.
3.
5
7
1
4
6
6
3
2
8
5
x
x
4. 1
5.
x
1
2
3
4
1
= (Always reduce the answer to the lowest term)
36
6
=
40
1
9
2
20
3
7
2
3
x
=
5
3
=
=
3
x 3
7
21
4
=
x 2
2
10
=
=
4
17
7
5
x
=
21
6
68
35
=
1
3
= 1
2
33
35
Division of fractions
2.
5
9
÷
2
3
=
1.
4
7
3
5
5
9
x
4
5
7
3
3
15
5
2
18
6
=
=
x
=
20
=
21
Solve the following:
1.
2.
3
2
+
2
4
3.
4
5
6
+
-
5
7
9
6
10
5
-
3
4. 2
1
5
8
9
+
+
2
6
2
3
3
5
1 +
ISHRM 2014
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College Algebra
MT 111/MT 211 Module III
3
5.
-
7
2
6. ( 5
+
3
1
4
2
7
+ )(
3
1
4
3
( -
7.
2
1
)÷
-
4
5
3
4
)
2
5
8.
2
3
5
9. 2
4
7
10. 5 x
÷
2
3
2
5
3. 3. RATIONAL EXPRESSION
Rational expression is defined as the quotient or ratio of two integers where the
denominator or the divisor is never zero. If the numerator and denominator have no
common factor except 1 and -1, then the rational expression is said to be in lowest terms.
If a given rational expression is not in lowest terms, factor both the numerator and
π‘Žπ‘
denominator and using the property 𝑏𝑐 =
π‘Ž
𝑏
if c ≠ 0, divide both their greatest factor.
Examples:
Reduce the following expressions to lowest terms:
12π‘₯²π‘¦
1.
2.
3.
4.
4π‘₯𝑦³
=
8𝑒−8𝑣
6𝑣−6𝑒
3π‘₯²(4π‘₯𝑦)
𝑦²(4π‘₯𝑦)
=
π‘₯²−π‘₯−6
π‘₯2+2π‘₯−15
=
8(𝑒−𝑣)
6(𝑣−𝑒)
=
𝑦2
=
4(𝑒−𝑣)
−3(𝑒−𝑣)
(π‘₯−3)(π‘₯+2)
(π‘₯−3)(π‘₯+5)
2(π‘₯ 2 − 𝑦 2 )π‘₯𝑦+π‘₯ 4 −𝑦⁴
π‘₯ 4 −𝑦⁴
3π‘₯²
=
=
= -
4
3
π‘₯+2
π‘₯+5
2π‘₯𝑦(π‘₯ 2 −𝑦 2 )(π‘₯ 2 +𝑦 2 )(π‘₯ 2 −𝑦 2 )
(π‘₯ 2 +𝑦 2 )(π‘₯ 2 −𝑦²
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College Algebra
MT 111/MT 211 Module III
=
?
Reduce the following rational expression to its lowest terms:
6
1.
8π‘₯
π‘₯−2
2.
3π‘₯−6
10𝑧−4𝑧²
3.
6π‘Žπ‘§
π‘Ž−3𝑏²
4.
5π‘Ž−15𝑏²
5.
π‘₯−5
π‘₯²−110π‘₯+25
π‘₯+4
6.
π‘₯²+π‘₯−12
7.
2π‘š−8π‘š²
3π‘š²−12π‘š³
π‘₯²−10π‘₯+9
8.
π‘₯²−8π‘₯−9
Multiplication and division of rational expressions
Among the basic operations on rational expressions, multiplication and
division will be studies first because they are simpler than addition and
subtraction.
The property of fractions used to multiply rational expression is the
π‘Ž
𝑏
π‘₯
𝑐
𝑑
=
π‘Žπ‘
𝑏𝑑
To divide rational expressions, the property to be used is the
π‘Ž
𝑏
𝑐
π‘Ž
𝑑
𝑏
÷ =
π‘₯
𝑑
𝑐
=
π‘Žπ‘‘
𝑏𝑐
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College Algebra
MT 111/MT 211 Module III
if
𝑐
𝑑
is not ≠ 0
Examples:
Perform the indicated operation and reduce to lowest terms.
3π‘₯²π‘¦
1.
2𝑧³
4𝑧²
•
π‘₯²
π‘₯²−𝑦²
2.
÷
π‘₯−𝑦
(3π‘₯ 2 𝑦)(4𝑧 2 )
=
(2𝑧 3 )(π‘₯ 2 )
π‘₯+𝑦
π‘₯
=
π‘₯²−𝑦²
π‘₯−𝑦
•
=
12π‘₯²π‘¦π‘§²
2π‘₯²π‘§³
π‘₯
π‘₯+𝑦
=
=
6𝑦
𝑧
(π‘₯−𝑦)(π‘₯+𝑦)π‘₯
(π‘₯−𝑦)(π‘₯+𝑦)
= x
Perform the following operations and reduce to lowest terms.
4−π‘₯²
1.
π‘₯³+ 5π‘₯²
π‘₯³π‘¦
2.
2π‘₯𝑧
3.
4.
5.
•
5π‘₯²+15π‘₯
π‘₯²+5π‘₯+6
÷ 3y2z
𝑒𝑣+𝑒𝑀
𝑒𝑣−𝑒𝑀
π‘₯
π‘₯+1
2+𝑒
𝑒−4
•
•
•
𝑣
𝑣+𝑀
÷
𝑣
𝑣−𝑀
π‘₯²−1
π‘₯²
𝑒²−16
8+𝑒³
Addition and Subtraction of rational expressions
To add or subtract rational expressions, the following properties of
fractions are used:
π‘Ž
𝑐
+
𝑏
𝑐
=
π‘Ž+𝑏
𝑐
and
π‘Ž
𝑐
−
𝑏
𝑐
=
π‘Ž−𝑏
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
𝑐
To add or subtract similar rational expressions, expressions having the
same denominator, add or subtract their corresponding numerators.
Examples:
1.
2π‘₯
2.
𝑦
π‘₯−4
+
𝑦
1
π‘Ž+𝑏
−
=
2π‘₯+π‘₯−4
π‘Ž²−2
π‘Ž+𝑏
𝑦
=
3π‘₯−4
𝑦
1−(π‘Ž2 −2)
=
π‘Ž+𝑏
=
1−π‘Ž2 +2
π‘Ž+𝑏
=
3−π‘Ž²
π‘Ž+𝑏
To add or subtract dissimilar rational expressions, having different
denominators, replace them first with equivalent fractions having their least
common denominator or LCD as the denominator.
Examples:
1.
=
2
π‘₯−4
2
π‘₯−4
+
+
π‘₯
π‘₯²−8π‘₯+16
, where:
π‘₯
(x – 4)2 =
(π‘₯−4)²
x2 – 8x + 16 = (x – 4)2or (x – 4) (x – 4)
(π‘₯−4)(π‘₯−4)
π‘₯−4
=
=
2(π‘₯−4)+π‘₯
(π‘₯−4)²
=
2π‘₯−8+π‘₯
(π‘₯−4)²
=
• 2 = 2 (x – 4)
(π‘₯−4)(π‘₯−4)
(π‘₯−4)²
3π‘₯−8
(π‘₯−4)²
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College Algebra
MT 111/MT 211 Module III
•x =x
π‘₯+1
2.
π‘₯+2
π‘₯(π‘₯+1)
−
-
π‘₯+3
π‘₯
π‘₯+1
=
π‘₯+2
(π‘₯+3)(π‘₯+2
π‘₯(π‘₯+2)
π‘₯(π‘₯+2)
−4π‘₯−6
=
=
π‘₯
π‘₯+3
π‘₯
π‘₯
• −
•
π‘₯+2
π‘₯²+π‘₯ −(π‘₯ 2 +5π‘₯+6)
π‘₯(π‘₯+2)
=
π‘₯+2
=
π‘₯(π‘₯+2)
Perform the indicated operation and reduce to lowest terms.
1
1.
π‘₯²−9
2.
3.
1
𝑒
2
+
−
2𝑏
π‘Ž+𝑏
π‘₯²+6π‘₯+9
2
𝑒+3
−
+
4
𝑒²+3𝑒
𝑏
π‘Ž
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College Algebra
MT 111/MT 211 Module III
π‘₯²+π‘₯−π‘₯ 2 −5π‘₯−6
π‘₯(π‘₯+2)
WORK PROJECT / LEARNING ACTIVITY NO.4, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
A. Review on Fractions
Perform the indicated operation and reduced to lowest terms.
2
1.
3
3
−
+
2.
4
7
−
3.
8
3
+
4.
5
5. +
4
4
2
12. 2
5
1
5
4
7
1
8
12
π‘₯ 3 49
14.
5
1
(5 π‘₯ 3)
2
15. 1 2
5
( π‘₯ )
÷ 2
3
11
16.
1
2
5
÷
3
4
17. 7 ÷
3
2
7
4
5
20
3
18. 2
10
2
3
2
3
4
6
− )+( −
3
6
7
11
5 + 2 ) − (1
÷4
3
5
+1
4
1
2
1
10.(
5
π‘₯
1
5
13. 2
23
5
-
3
1
( + )−
4
9.(
7
1
4
3
6
6. 2
8.
11.
2
1
7.
1
)
+ 2)
19. 2
÷
3
6
x3
1
3
2
5
4
7
1 2
π‘₯
4 3
20. 1 2
÷
7 3
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College Algebra
MT 111/MT 211 Module III
2
3
4
7
( π‘₯ )
WORK PROJECT / LEARNING ACTIVITY NO.5, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
B. Reduce the following rational expression to its lowest terms.
12π‘₯³π‘¦²
1.
3π‘₯³π‘§²
3(𝑦+4)
2.
3(𝑦2 −10)− (3𝑦+30)
π‘Ž−3𝑏²
3.
5π‘Ž−15𝑏²
(π‘₯+π‘Ž)(π‘₯ 2 −π‘Žπ‘₯+π‘Ž2 )
4.
π‘₯ 4 +π‘Ž³π‘₯
5𝑧 2
5.
10𝑧
2π‘₯+10
6.
π‘₯+5
𝑒³−𝑣³
7.
𝑒²−𝑣²
6π‘Žπ‘−3𝑏²
8.
4π‘Žπ‘+𝑏³−2𝑏²
π‘₯²−6π‘₯+5
9.
10.
π‘₯²−7π‘₯+10
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π‘₯ 4 −π‘₯ 2 −6
π‘₯ 4 +π‘₯²−12
College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO.6, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
C. Multiplication of Rational Expressions
1. (
2.
3.
4.
5.
6.
7.
π‘₯2 +2π‘₯+1 π‘₯2 +2π‘₯+8
π‘₯+4 ) (π‘₯2 +3π‘₯+2 )
π‘₯²−π‘₯−6
π‘₯²+π‘₯−42
π‘₯²+2π‘₯+1
π‘₯²−1
π‘₯
3
•
π‘₯+4
π‘₯−5
4
π‘₯2
•
π‘₯ 2 −π‘₯−30
•
π‘₯ 2 −4
•
π‘₯+1
5
π‘₯−6
π‘₯+7
π‘₯²+9π‘₯+20
π‘₯²+8π‘₯+16
4(π‘š+1)²
π‘š²+3π‘š+2
π‘₯²−7π‘₯+12
π‘₯²−2π‘₯+1
π‘₯3
•
π‘₯²+6π‘₯−16
•
•
•
•
•
15
4π‘₯
π‘₯²−49
π‘₯²−16
π‘₯²+10π‘₯+25
π‘₯²−25
π‘š−1
π‘š2 −1
•
π‘š+2
4π‘š+8
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO.7, MODULE III
Name: ________________________________ Yr/Sec: ____________ Date: _______________
D. Division of Rational Expressions
1.
2.
12π‘₯²+4π‘₯
6π‘₯+4
35π‘₯ 5 +30π‘₯³
π‘₯²−π‘₯−132
3. (
4.
5.
6.
7.
÷
9π‘₯²−1
10π‘₯−4
÷
7π‘₯ 4 +6π‘₯²
π‘₯²−5π‘₯−84
π‘₯3 +6π‘₯2 −37π‘₯−210
π‘₯2 −3π‘₯−18
π‘₯+4
π‘₯−5
π‘š+2
π‘š−3
π‘₯²−16
π‘₯+5
𝑦²−9
𝑦
π‘₯²−25
π‘š−3
π‘š+2
÷
÷
π‘₯2 +2π‘₯−35
)
π‘₯2 +π‘₯−6
π‘₯²−16
÷
÷
÷
÷
π‘š²−9
π‘š²−4
π‘₯²−8π‘₯+16
π‘₯²+10π‘₯+25
𝑦+3
𝑦+2
÷
π‘₯²+9π‘₯+20
π‘₯²−25
÷
π‘₯+1
π‘₯²−3π‘₯−10
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module III
WORK PROJECT / LEARNING ACTIVITY NO.7, MODULE III
Name: ___________________________________ Yr/Sec: ____________ Date: ____________
E. Addition of Rational Expressions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
2
+
π‘₯²
1
π‘₯²+1
π‘₯
3
+
2𝑏
π‘Ž+𝑏
π‘₯+𝑦
π‘₯−𝑦
3π‘Žπ‘
4
2𝑒
𝑀
3
π‘₯
3π‘₯+2
+
π‘₯²+1
4
𝑏
−
π‘Ž
π‘₯−𝑦
+
−
π‘₯+𝑦
5
𝑀
𝑒
−𝑒
π‘₯²−7π‘₯+10
+
𝑣
1−𝑣
−2
2π‘Žπ‘
π‘₯+1
𝑦+2
π‘₯²+1
π‘₯
+
4
5π‘₯²−3
+
+
𝑦+1
𝑦²−4
+
3𝑣
𝑣+1
3
π‘₯²−π‘₯−2
−
−
3
𝑦−2
5
𝑣²−1
ISHRM 2014
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