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₯β΅π¦π§² 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 Learning Material 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 Learning Material 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 Learning Material 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 π parts. Fractions are of the form πwhere 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 Learning Material 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 Learning Material 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 ππ denominator and using the property ππ = π π if c ≠ 0, divide both their greatest factor. Examples: Reduce the following expressions to lowest terms: 12π₯²π¦ 1. 2. 3. 4. 4π₯𦳠= 8π’−8π£ 6π£−6π’ 3π₯²(4π₯π¦) π¦²(4π₯π¦) = π₯²−π₯−6 π₯2+2π₯−15 = 8(π’−π£) 6(π£−π’) = π¦2 = 4(π’−π£) −3(π’−π£) (π₯−3)(π₯+2) (π₯−3)(π₯+5) 2(π₯ 2 − π¦ 2 )π₯π¦+π₯ 4 −π¦β΄ π₯ 4 −π¦β΄ 3π₯² = = = - 4 3 π₯+2 π₯+5 2π₯π¦(π₯ 2 −π¦ 2 )(π₯ 2 +π¦ 2 )(π₯ 2 −π¦ 2 ) (π₯ 2 +π¦ 2 )(π₯ 2 −𦲠ISHRM 2014 Learning Material 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 10π§−4𧲠3. 6ππ§ π−3π² 4. 5π−15π² 5. π₯−5 π₯²−110π₯+25 π₯+4 6. π₯²+π₯−12 7. 2π−8π² 3π²−12π³ 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 π π₯ π π = ππ ππ To divide rational expressions, the property to be used is the π π π π π π ÷ = π₯ π π = ππ ππ ISHRM 2014 Learning Material College Algebra MT 111/MT 211 Module III if π π is not ≠ 0 Examples: Perform the indicated operation and reduce to lowest terms. 3π₯²π¦ 1. 2𧳠4𧲠• π₯² π₯²−𦲠2. ÷ π₯−π¦ (3π₯ 2 π¦)(4π§ 2 ) = (2π§ 3 )(π₯ 2 ) π₯+π¦ π₯ = π₯²−𦲠π₯−π¦ • = 12π₯²π¦π§² 2π₯²π§³ π₯ π₯+π¦ = = 6π¦ π§ (π₯−π¦)(π₯+π¦)π₯ (π₯−π¦)(π₯+π¦) = x Perform the following operations and reduce to lowest terms. 4−π₯² 1. π₯³+ 5π₯² π₯³π¦ 2. 2π₯π§ 3. 4. 5. • 5π₯²+15π₯ π₯²+5π₯+6 ÷ 3y2z π’π£+π’π€ π’π£−π’π€ π₯ π₯+1 2+π’ π’−4 • • • π£ π£+π€ ÷ π£ π£−π€ π₯²−1 π₯² π’²−16 8+π’³ Addition and Subtraction of rational expressions To add or subtract rational expressions, the following properties of fractions are used: π π + π π = π+π π and π π − π π = π−π ISHRM 2014 Learning Material College Algebra MT 111/MT 211 Module III π To add or subtract similar rational expressions, expressions having the same denominator, add or subtract their corresponding numerators. Examples: 1. 2π₯ 2. π¦ π₯−4 + π¦ 1 π+π − = 2π₯+π₯−4 π²−2 π+π π¦ = 3π₯−4 π¦ 1−(π2 −2) = π+π = 1−π2 +2 π+π = 3−π² π+π 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 ISHRM 2014 Learning Material 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 π’ 2 + − 2π π+π π₯²+6π₯+9 2 π’+3 − + 4 π’²+3π’ π π ISHRM 2014 Learning Material 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 ISHRM 2014 Learning Material 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(π¦+4) 2. 3(π¦2 −10)− (3π¦+30) π−3π² 3. 5π−15π² (π₯+π)(π₯ 2 −ππ₯+π2 ) 4. π₯ 4 +π³π₯ 5π§ 2 5. 10π§ 2π₯+10 6. π₯+5 π’³−π£³ 7. π’²−π£² 6ππ−3π² 8. 4ππ+π³−2π² π₯²−6π₯+5 9. 10. π₯²−7π₯+10 ISHRM 2014 Learning Material 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+1)² π²+3π+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−1 π2 −1 • π+2 4π+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+2 π−3 π₯²−16 π₯+5 π¦²−9 π¦ π₯²−25 π−3 π+2 ÷ ÷ π₯2 +2π₯−35 ) π₯2 +π₯−6 π₯²−16 ÷ ÷ ÷ ÷ π²−9 π²−4 π₯²−8π₯+16 π₯²+10π₯+25 π¦+3 π¦+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 + 2π π+π π₯+π¦ π₯−π¦ 3ππ 4 2π’ π€ 3 π₯ 3π₯+2 + π₯²+1 4 π − π π₯−π¦ + − π₯+π¦ 5 π€ π’ −π’ π₯²−7π₯+10 + π£ 1−π£ −2 2ππ π₯+1 π¦+2 π₯²+1 π₯ + 4 5π₯²−3 + + π¦+1 π¦²−4 + 3π£ π£+1 3 π₯²−π₯−2 − − 3 π¦−2 5 π£²−1 ISHRM 2014