section F1 214FactoringFactoring is the reverse process of multiplication. Factoring polynomials inalgebra has similar role as factoring numbers in arithmetic. Any number canbe expressed as a product of prime numbers. For example, 6 2 3.Similarly, any polynomial can be expressed as a product of primepolynomials, which are polynomials that cannot be factored any further. Forexample, π₯π₯ 2 5π₯π₯ 6 (π₯π₯ 2)(π₯π₯ 3). Just as factoring numbers helps insimplifying or adding fractions, factoring polynomials is very useful insimplifying or adding algebraic fractions. In addition, it helps identify zeros of polynomials, which in turn allowsfor solving higher degree polynomial equations.In this chapter, we will examine the most commonly used factoring strategies with particular attention to specialfactoring. Then, we will apply these strategies in solving polynomial equations.F.1Greatest Common Factor and Factoring by GroupingPrime FactorsWhen working with integers, we are often interested in their factors, particularly primefactors. Likewise, we might be interested in factors of polynomials.Definition 1.1To factor a polynomial means to write the polynomial as a product of βsimplerβpolynomials. For example,5π₯π₯ 10 5(π₯π₯ 2), or π₯π₯ 2 9 (π₯π₯ 3)(π₯π₯ 3).In the above definition, βsimplerβ means polynomials of lower degrees or polynomials withcoefficients that do not contain common factors other than 1 or 1. If possible, we wouldlike to see the polynomial factors, other than monomials, having integral coefficients anda positive leading term.When is a polynomial factorization complete?In the case of natural numbers, the complete factorization means a factorization into primenumbers, which are numbers divisible only by their own selves and 1. We would expectthat similar situation is possible for polynomials. So, which polynomials should weconsider as prime?Observe that a polynomial such as 4π₯π₯ 12 can be written as a product in many differentways, for instance13 (4π₯π₯ 12), 2( 2π₯π₯ 6), 4( π₯π₯ 3), 4(π₯π₯ 3), 12 π₯π₯ 1 , etc.Since the terms of 4π₯π₯ 12 and 2π₯π₯ 6 still contain common factors different than 1 or 1, these polynomials are not considered to be factored completely, which means that theyshould not be called prime. The next two factorizations, 4( π₯π₯ 3) and 4(π₯π₯ 3) areboth complete, so both polynomials π₯π₯ 3 and π₯π₯ 3 should be considered as prime. But11what about the last factorization, 12 3 π₯π₯ 1 ? Since the remaining binomial 3 π₯π₯ 1does not have integral coefficients, such a factorization is not always desirable.Factoring
section F1 215Here are some examples of prime polynomials: 13any monomials such as 2π₯π₯ 2, ππππ 2 , or π₯π₯π₯π₯;any linear polynomials with integral coefficients that have no common factors otherthan 1 or 1, such as π₯π₯ 1 or 2π₯π₯ 5;some quadratic polynomials with integral coefficients that cannot be factored into anylower degree polynomials with integral coefficients, such as π₯π₯ 2 1 or π₯π₯ 2 π₯π₯ 1.For the purposes of this course, we will assume the following definition of a primepolynomial.Definition 1.2A polynomial with integral coefficients is called prime if one of the following conditionsis true- it is a monomial, or- the only common factors of its terms are ππ or ππ and it cannot be factored into anylower degree polynomials with integral coefficients.Definition 1.3A factorization of a polynomial with integral coefficients is complete if all of its factorsare prime.Here is an example of a polynomial factored completely: 6π₯π₯ 3 10π₯π₯ 2 4π₯π₯ 2π₯π₯(3π₯π₯ 1)(π₯π₯ 2)In the next few sections, we will study several factoring strategies that will be helpful infinding complete factorizations of various polynomials.Greatest Common FactorThe first strategy of factoring is to factor out the greatest common factor (GCF).Definition 1.4The greatest common factor (GCF) of two or more terms is the largest expression that isa factor of all these terms.In the above definition, the βlargest expressionβ refers to the expression with the mostfactors, disregarding their signs.To find the greatest common factor, we take the product of the least powers of each type ofcommon factor out of all the terms. For example, suppose we wish to find the GCF of theterms6π₯π₯ 2 π¦π¦ 3, 18π₯π₯ 5 π¦π¦, and 24π₯π₯ 4 π¦π¦ 2.First, we look for the GCF of 6, 18, and 24, which is 6. Then, we take the lowest powerout of π₯π₯ 2 , π₯π₯ 5 , and π₯π₯ 4 , which is π₯π₯ 2 . Finally, we take the lowest power out of π¦π¦ 3 , π¦π¦, and π¦π¦ 2 ,which is π¦π¦. Therefore,GCF(6π₯π₯ 2 π¦π¦ 3 , 18π₯π₯ 5 π¦π¦, 24π₯π₯ 4 π¦π¦ 2 ) 6π₯π₯ 2 π¦π¦This GCF can be used to factor the polynomial 6π₯π₯ 2 π¦π¦ 3 18π₯π₯ 5 π¦π¦ 24π₯π₯ 4 π¦π¦ 2 by first seeingit as6π₯π₯ 2 π¦π¦ π¦π¦ 2 6π₯π₯ 2 π¦π¦ 3π₯π₯ 3 6π₯π₯ 2 π¦π¦ 4π₯π₯ 2 π¦π¦,Greatest Common Factor and Factoring by Grouping
section F1 216and then, using the reverse distributing property, βpullingβ the 6π₯π₯ 2 π¦π¦ out of the bracket toobtain6π₯π₯ 2 π¦π¦(π¦π¦ 2 3π₯π₯ 3 4π₯π₯ 2 π¦π¦).Note 1: Notice that since 1 and 1 are factors of any expression, the GCF is defined upto the sign. Usually, we choose the positive GCF, but sometimes it may be convenient tochoose the negative GCF. For example, we can claim thatGCF( 2π₯π₯, 4π¦π¦) 2 or GCF( 2π₯π₯, 4π¦π¦) 2,depending on what expression we wish to leave after factoring the GCF out: 2π₯π₯ 4π¦π¦ οΏ½οΏ½οΏ½ππππΊπΊπΊπΊπΊπΊ( π₯π₯ 2π¦π¦) or 2π₯π₯ 4π¦π¦ οΏ½οΏ½ 2 (π₯π₯ 2π¦π¦) οΏ½οΏ½πππ οΏ½οΏ½π‘π‘π‘π‘π‘π‘Note 2: If the GCF of the terms of a polynomial is equal to 1, we often say that these termsdo not have any common factors. What we actually mean is that the terms do not have acommon factor other than 1, as factoring 1 out does not help in breaking the originalpolynomial into a product of simpler polynomials. See Definition 1.1.Finding the Greatest Common FactorFind the Greatest Common Factor for the given expressions.a.c.Solutiona.6π₯π₯ 4 (π₯π₯ 1)3 , 3π₯π₯ 3 (π₯π₯ 1), 9π₯π₯(π₯π₯ 1)2ππππ 2 , ππ2 ππ, ππ, ππb.d.4ππ(π¦π¦ π₯π₯), 8ππ(π₯π₯ π¦π¦)3π₯π₯ 1 π¦π¦ 3 , π₯π₯ 2 π¦π¦ 2 π§π§Since GCF(6, 3, 9) 3, the lowest power out of π₯π₯ 4 , π₯π₯ 3 , and π₯π₯ is π₯π₯, and the lowestpower out of (π₯π₯ 1)3 , (π₯π₯ 1), and (π₯π₯ 1)2 is (π₯π₯ 1), thenGCF(6π₯π₯ 4 (π₯π₯ 1)3 , 3π₯π₯ 3 (π₯π₯ 1), 9π₯π₯(π₯π₯ 1)2 ) ππππ(ππ ππ)b.Since π¦π¦ π₯π₯ is opposite to π₯π₯ π¦π¦, then π¦π¦ π₯π₯ can be written as (π₯π₯ π¦π¦). So 4, ππ, and(π₯π₯ π¦π¦) is common for both expressions. Thus,GCF 4ππ(π¦π¦ π₯π₯), 8ππ(π₯π₯ π¦π¦) ππππ(ππ ππ)Note: The Greatest Common Factor is unique up to the sign. Notice that in the aboveexample, we could write π₯π₯ π¦π¦ as (π¦π¦ π₯π₯) and choose the GCF to be 4ππ(π¦π¦ π₯π₯).c.The terms ππππ 2 , ππ2 ππ, ππ, and ππ have no common factor other than 1, soGCF(ππππ 2 , ππ2 ππ, ππ, ππ) ππFactoring
section F1 d.217The lowest power out of π₯π₯ 1 and π₯π₯ 2 is π₯π₯ 2 , and the lowest power out of π¦π¦ 3 andπ¦π¦ 2 is π¦π¦ 3 . Therefore,GCF(3π₯π₯ 1 π¦π¦ 3 , π₯π₯ 2 π¦π¦ 2 π§π§) ππ ππ ππ ππFactoring out the Greatest Common FactorFactor each expression by taking the greatest common factor out. Simplify the factors, ifpossible.a.c.Solutiona.54π₯π₯ 2 π¦π¦ 2 60π₯π₯π¦π¦ 3b. π₯π₯(π₯π₯ 5) π₯π₯ 2 (5 π₯π₯) (π₯π₯ 5)2d.ππππ ππ2 ππ(ππ 1)π₯π₯ 1 2π₯π₯ 2 π₯π₯ 3To find the greatest common factor of 54 and 60, we can use the method of dividingby any common factor, as presented below.all commonfactors are listedin this column2 54, 603 27, 309, 10no morecommon factorsfor 9 and 10So, GCF(54, 60) 2 3 6.Since GCF(54π₯π₯ 2 π¦π¦ 2 , 60π₯π₯π¦π¦ 3 ) 6π₯π₯π¦π¦ 2 , we factor the 6π₯π₯π¦π¦ 2 out by dividing each termof the polynomial 54π₯π₯ 2 π¦π¦ 2 60π₯π₯π¦π¦ 3 by 6π₯π₯π¦π¦ 2 , as below.54π₯π₯ 2 π¦π¦ 2 60π₯π₯π¦π¦ 3 ππππππππ (ππππ ππππππ)54π₯π₯ 2 π¦π¦ 2 9π₯π₯6π₯π₯π¦π¦ 260π₯π₯π¦π¦ 3 10π¦π¦6π₯π₯π¦π¦ 2Note: Since factoring is the reverse process of multiplication, it can be checked byfinding the product of the factors. If the product gives us the original polynomial, thefactorization is correct.b.First, notice that the polynomial has two terms, ππππ and ππ2 ππ(ππ 1). The greatestcommon factor for these two terms is ππππ, so we haveππππ ππ2 ππ(ππ 1) ππππ ππ ππ(ππ 1) ππππ(1 ππ2 ππ)2 ππππ( ππ ππ 1) ππππ ππππ ππ ππ remember to leave 1for the first termsimplify and arrangein decreasing powerstake the β β outGreatest Common Factor and Factoring by Grouping
section F1 218Note: Both factorizations, ππππ( ππ2 ππ 1) and ππππ(ππ2 ππ 1) are correct.However, we customarily leave the polynomial in the bracket with a positive leadingcoefficient.c.Observe that if we write the middle term π₯π₯ 2 (5 π₯π₯) as π₯π₯ 2 (π₯π₯ 5) by factoring thenegative out of the (5 π₯π₯), then (5 π₯π₯) is the common factor of all the terms of theequivalent polynomial π₯π₯(π₯π₯ 5) π₯π₯ 2 (π₯π₯ 5) (π₯π₯ 5)2 .Then notice that if we take (π₯π₯ 5) as the GCF, then the leading term of theremaining polynomial will be positive. So, we factor π₯π₯(π₯π₯ 5) π₯π₯ 2 (5 π₯π₯) (π₯π₯ 5)2 π₯π₯(π₯π₯ 5) π₯π₯ 2 (π₯π₯ 5) (π₯π₯ 5)2 (π₯π₯ 5) π₯π₯ π₯π₯ 2 (π₯π₯ 5) simplify and arrangein decreasing powers (ππ ππ) ππππ ππππ ππ d.The GCF(π₯π₯ 1 , 2π₯π₯ 2 , π₯π₯ 3 ) π₯π₯ 3, as 3 is the lowest exponent of the commonfactor π₯π₯. So, we factor out π₯π₯ 3 as below.π₯π₯ 1 2π₯π₯ 2 π₯π₯ 3 ππ ππ ππππ ππππ ππ the exponent 2 is found bysubtracting 3 from 1the exponent 1 is found bysubtracting 3 from 2To check if the factorization is correct, we multiplyadd exponentsπ₯π₯ 3 (π₯π₯ 2 2π₯π₯ 1) π₯π₯ 3 π₯π₯ 2 2π₯π₯ 3 π₯π₯ 1π₯π₯ 3 π₯π₯ 1 2π₯π₯ 2 π₯π₯ 3Since the product gives us the original polynomial, the factorization is correct.Factoring by GroupingWhen referring to acommon factor, wehave in mind acommon factor otherthan 1.Consider the polynomial π₯π₯ 2 π₯π₯ π₯π₯π₯π₯ π¦π¦. It consists of four terms that do not have anycommon factors. Yet, it can still be factored if we group the first two and the last two terms.The first group of two terms contains the common factor of π₯π₯ and the second group of twoterms contains the common factor of π¦π¦. Observe what happens when we factor each group.2π₯π₯ π₯π₯ π₯π₯π₯π₯ π¦π¦ π₯π₯(π₯π₯ 1) π¦π¦(π₯π₯ 1)Factoring (π₯π₯ 1)(π₯π₯ π¦π¦)now (π₯π₯ 1) is thecommon factor of theentire polynomial
section F1 219This method is called factoring by grouping, in particular, two-by-two grouping.Warning: After factoring each group, make sure to write the β β or β β between the terms.Failing to write these signs leads to the false impression that the polynomial is alreadyfactored. For example, if in the second line of the above calculations we would fail to writethe middle β β, the expression would look like a product π₯π₯(π₯π₯ 1) π¦π¦(π₯π₯ 1), which is notthe case. Also, since the expression π₯π₯(π₯π₯ 1) π¦π¦(π₯π₯ 1) is a sum, not a product, we shouldnot stop at this step. We need to factor out the common bracket (π₯π₯ 1) to leave it as aproduct.A two-by-two grouping leads to a factorization only if the binomials, after factoring outthe common factors in each group, are the same. Sometimes a rearrangement of terms isnecessary to achieve this goal.For example, the attempt to factor π₯π₯ 3 15 5π₯π₯ 2 3π₯π₯ by grouping the first and the lasttwo terms,2π₯π₯ 3 15 5π₯π₯ 3π₯π₯ (π₯π₯ 3 15) π₯π₯(5π₯π₯ 3)does not lead us to a common binomial that could be factored out.However, rearranging terms allows us to factor the original polynomial in the followingways:π₯π₯ 3 15 5π₯π₯ 2 3π₯π₯ π₯π₯ 3 5π₯π₯ 2 3π₯π₯ 15 π₯π₯ 2 (π₯π₯ 5) 3(π₯π₯ 5) (π₯π₯ 5)(π₯π₯ 2 3)orπ₯π₯ 3 15 5π₯π₯ 2 3π₯π₯2 π₯π₯ 3 3π₯π₯ 5π₯π₯ 15 π₯π₯(π₯π₯ 2 3) 5(π₯π₯ 2 3) (π₯π₯ 2 3)(π₯π₯ 5)Factoring by grouping applies to polynomials with more than three terms. However, not allsuch polynomials can be factored by grouping. For example, if we attempt to factor π₯π₯ 3 π₯π₯ 2 2π₯π₯ 2 by grouping, we obtain3 π₯π₯ 2 2π₯π₯ 2π₯π₯ π₯π₯ 2 (π₯π₯ 1) 2(π₯π₯ 1).Unfortunately, the expressions π₯π₯ 1 and π₯π₯ 1 are not the same, so there is no commonfactor to factor out. One can also check that no other rearrangments of terms allows us forfactoring out a common binomial. So, this polynomial cannot be factored by grouping.Factoring by GroupingFactor each polynomial by grouping, if possible. Remember to check for the GCF first.Greatest Common Factor and Factoring by Grouping
section F1 220a.c.Solutiona.2π₯π₯ 3 6π₯π₯ 2 π₯π₯ 32π₯π₯ 2 π¦π¦ 8 2π₯π₯ 2 8π¦π¦b.d.5π₯π₯ 5π¦π¦ ππππ πππππ₯π₯ 2 π₯π₯ π¦π¦ 1Since there is no common factor for all four terms, we will attempt the two-by-twogrouping method.32π₯π₯ 6π₯π₯ 2 π₯π₯ 3 2π₯π₯ 2 (π₯π₯ 3) 1(π₯π₯ 3)ππ (ππ ππ) ππππ ππ b.write the 1 forthe second termAs before, there is no common factor for all four terms. The two-by-two groupingmethod works only if the remaining binomials after factoring each group are exactlythe same. We can achieve this goal by factoring β ππ , rather than ππ, out of the last twoterms. So,5π₯π₯ 5π¦π¦ ππππ ππππ 5(π₯π₯ π¦π¦) ππ(π₯π₯ π¦π¦)reverse signs whenβpullingβ a β β out (ππ ππ) ππππππ ππ c.Notice that 2 is the GCF of all terms, so we factor it out first.2π₯π₯ 2 π¦π¦ 8 2π₯π₯ 2 8π¦π¦ 2(π₯π₯ 2 π¦π¦ 4 π₯π₯ 2 4π¦π¦)Then, observe that grouping the first and last two terms of the remaining polynomialdoes not help, as the two groups do not have any common factors. However,exchanging for example the second with the fourth term will help, as shown below.the square bracket isessential here becauseof the factor of 222 2(π₯π₯π¦π¦ 4π¦π¦ π₯π₯ ) 4 2 2[π¦π¦(π₯π₯ 4) (π₯π₯ 2 ππ ππππ ππ (ππ ππ)d. 4)]reverse signs whenβpullingβ a β β outnow, there is no need for the squarebracket as multiplication is associativeThe polynomial π₯π₯ 2 π₯π₯ π¦π¦ 1 does not have any common factors for all four terms.Also, only the first two terms have a common factor. Unfortunately, when attemptingto factor using the two-by-two grouping method, we obtainπ₯π₯ 2 π₯π₯ π¦π¦ 1 π₯π₯(π₯π₯ 1) (π¦π¦ 1),which cannot be factored, as the expressions π₯π₯ 1 and π¦π¦ 1 are different.One can also check that no other arrangement of terms allows for factoring of thispolynomial by grouping. So, this polynomial cannot be factored by grouping.Factoring
section F1 221Factoring in Solving FormulasSolutionSolve ππππ 3ππ 5 for ππ.First, we move the terms containing the variable ππ to one side of the equation,and then factor ππ outππππ 3ππ 5ππππ 3ππ 5,ππ(ππ 3) 5.So, after dividing by ππ 3, we obtainππ ππ.ππ ππF.1 ExercisesVocabulary CheckComplete each blank with the most appropriate term or phrase from the given list:common factor, distributive, grouping, prime, product.1.To factor a polynomial means to write it as a of simpler polynomials.2.The greatest of two or more terms is the product of the least powers ofeach type of common factor out of all the terms.3.To factor out the GCF, we reverse the property of multiplication.4.A polynomial with four terms having no common factors can be still factored by its terms.5.A polynomial, other than a monomial, cannot be factored into two polynomials, both differentthat 1 or 1.Concept Check True or false.6.7.8.The polynomial 6π₯π₯ 8π¦π¦ is prime.123414The factorization π₯π₯ π¦π¦ (2π₯π₯ 3π¦π¦) is essential to be complete.The GCF of the terms of the polynomial 3(π₯π₯ 2) π₯π₯(2 π₯π₯) is (π₯π₯ 2)(2 π₯π₯).Concept Check Find the GCF with a positive coefficient for the given expressions.9.8π₯π₯π₯π₯, 10π₯π₯π₯π₯, 14π₯π₯π₯π₯11. 4π₯π₯(π₯π₯ 1), 3π₯π₯ 2 (π₯π₯ 1)13. 9(ππ 5), 12(5 ππ)10. 21ππ3 ππ6 , 35ππ7 ππ5 , 28ππ5 ππ 812. π₯π₯(π₯π₯ 3)2 , π₯π₯ 2 (π₯π₯ 3)(π₯π₯ 2)14. (π₯π₯ 2π¦π¦)(π₯π₯ 1), (2π¦π¦ π₯π₯)(π₯π₯ 1)Greatest Common Factor and Factoring by Grouping
section F1 22215. 3π₯π₯ 2 π¦π¦ 3 , 6π₯π₯ 3 π¦π¦ 516. π₯π₯ 2 (π₯π₯ 2) 2 , π₯π₯ 4 (π₯π₯ 2) 1Factor out the greatest common factor. Leave the remaining polynomial with a positive leading coeficient.Simplify the factors, if possible.17. 9π₯π₯ 2 81π₯π₯20. 6ππ3 36ππ4 18ππ223. ππ(π₯π₯ 2) ππ(π₯π₯ 2)18. 8ππ 3 24ππ21. 10ππ 2 π π 2 15ππ 4 π π 226. (ππ 2)(ππ 3) (ππ 2)(ππ 3)28. (4π₯π₯ π¦π¦) 4π₯π₯(π¦π¦ 4π₯π₯)27. π¦π¦(π₯π₯ 1) 5(1 π₯π₯)29. 4(3 π₯π₯)2 (3 π₯π₯)3 3(3 π₯π₯)Factor out the least power of each variable.34. 3ππ 5 ππ 3 2ππ 222. 5π₯π₯ 2 π¦π¦ 3 10π₯π₯ 3 π¦π¦ 224. ππ(π¦π¦ 2 3) 2(π¦π¦ 2 3)25. (π₯π₯ 2)(π₯π₯ 3) (π₯π₯ 2)(π₯π₯ 5)31. 3π₯π₯ 3 π₯π₯ 219. 6ππ3 3ππ2 9ππ430. 2(ππ 3) 4(ππ 3)2 (ππ 3)332. ππ 2 2ππ 433. π₯π₯ 4 2π₯π₯ 3 7π₯π₯ 2Factor by grouping, if possible.35. 3π₯π₯ 3 π¦π¦ π₯π₯ 2 π¦π¦ 236. 5π₯π₯ 2 π¦π¦ 3 2π₯π₯ 1 π¦π¦ 237. 20 5π₯π₯ 12π¦π¦ 3π₯π₯π₯π₯38. 2ππ3 ππ2 14ππ 739. ππππ ππππ ππππ ππππ43. 3ππ2 9ππππ ππππ 3ππ 244. 3π₯π₯ 2 π₯π₯ 2 π¦π¦ π¦π¦π§π§ 2 3π§π§ 245. 2π₯π₯ 3 π₯π₯ 2 4π₯π₯ 240. 2π₯π₯π₯π₯ π₯π₯ 2 π¦π¦ 6 3π₯π₯46. π₯π₯ 2 π¦π¦ 2 ππππ πππ¦π¦ 2 πππ₯π₯ 249. π₯π₯π₯π₯ 6π¦π¦ 3π₯π₯ 1841. 3π₯π₯ 2 4π₯π₯π₯π₯ 6π₯π₯π₯π₯ 8π¦π¦ 242. π₯π₯ 3 π₯π₯π₯π₯ π¦π¦ 2 π₯π₯ 2 π¦π¦48. π₯π₯ 2 π¦π¦ π₯π₯π₯π₯ π₯π₯ π¦π¦47. π₯π₯π₯π₯ ππππ ππππ ππππ50. π₯π₯ ππ π¦π¦ 3π₯π₯ ππ π¦π¦ 551. ππππ π₯π₯ ππ 2ππππ π₯π₯ ππ 2Factor completely. Remember to check for the GCF first.52. 5π₯π₯ 5ππππ 5ππππππ 5ππππ54. π₯π₯ 4 (π₯π₯ 1) π₯π₯ 3 (π₯π₯ 1) π₯π₯ 2 π₯π₯Discussion Point53. 6ππππ 14π π 6ππ 1455. π₯π₯ 3 (π₯π₯ 2)2 2π₯π₯ 2 (π₯π₯ 2) (π₯π₯ 2)(π₯π₯ 2)56. One of possible factorizations of the polynomial 4π₯π₯ 2 π¦π¦ 5 8π₯π₯π¦π¦ 3 is 2π₯π₯π¦π¦ 3 (2π₯π₯π¦π¦ 2 4). Is this a completefactorization?Use factoring the GCF strategy to solve each formula for the indicated variable.57. π΄π΄ π·π· π·π·ππ, for π·π·59. 2ππ ππ ππππ, for ππFactoring121258. ππ ππππ ππππ, for ππ60. π€π€ππ 3ππ π₯π₯, for ππ
section F1 Analytic Skills61.4π₯π₯Write the area of each shaded region in factored οΏ½οΏ½οΏ½πππGreatest Common Factor and Factoring by Grouping
section F2 F.2224Factoring TrinomialsIn this section, we discuss factoring trinomials. We start with factoringquadratic trinomials of the form π₯π₯ 2 ππππ ππ, then quadratic trinomialsof the form πππ₯π₯ 2 ππππ ππ, where ππ 1, and finally trinomials reducibleto quadratic by means of substitution.Factorization of Quadratic Trinomials ππππ ππππ ππFactorization of a quadratic trinomial π₯π₯ 2 ππππ ππ is the reverse process of the FOILmethod of multiplying two linear binomials. Observe that(π₯π₯ ππ)(π₯π₯ ππ) π₯π₯ 2 πππ₯π₯ πππ₯π₯ ππππ π₯π₯ 2 (ππ ππ)π₯π₯ ππππSo, to reverse this multiplication, we look for two numbers ππ and ππ, such that the productππππ equals to the free term ππ and the sum ππ ππ equals to the middle coefficient ππ of thetrinomial.π₯π₯ 2 ππ π₯π₯ ππ (π₯π₯ ππ)(π₯π₯ ππ)(ππ ππ)ππππFor example, to factor π₯π₯ 2 5π₯π₯ 6, we think of two integers that multiply to 6 and add to5. Such integers are 2 and 3, so π₯π₯ 2 5π₯π₯ 6 (π₯π₯ 2)(π₯π₯ 3). Since multiplication iscommutative, the order of these factors is not important.This could also be illustrated geometrically, using algebra tiles.π₯π₯ 2π₯π₯1The area of a square with the side length π₯π₯ is equal to π₯π₯ 2 . The area of a rectangle with thedimensions π₯π₯ by 1 is equal to π₯π₯, and the area of a unit square is equal to 1. So, the trinomialπ₯π₯ 2 5π₯π₯ 6 can be represented asπ₯π₯ 2π₯π₯ π₯π₯ π₯π₯ π₯π₯ π₯π₯1 1 1 1 1 1To factor this trinomial, we would like to rearrange these tiles to fulfill a rectangle. π₯π₯ 2π₯π₯ 3 π₯π₯ 2π₯π₯π₯π₯π₯π₯ π₯π₯ π₯π₯1 1 11 1 1The area of such rectangle can be represented as the product of its length, (π₯π₯ 3), andwidth, (π₯π₯ 2) which becomes the factorization of the original trinomial.In the trinomial examined above, the signs of the middle and the last terms are both positive.To analyse how different signs of these terms influence the signs used in the factors, observethe next three examples.Factoring
section F2 225To factor π₯π₯ 2 5π₯π₯ 6, we look for two integers that multiply to 6 and add to 5. Suchintegers are 2 and 3, so π₯π₯ 2 5π₯π₯ 6 (π₯π₯ 2)(π₯π₯ 3).To factor π₯π₯ 2 π₯π₯ 6, we look for two integers that multiply to 6 and add to 1. Suchintegers are 2 and 3, so π₯π₯ 2 π₯π₯ 6 (π₯π₯ 2)(π₯π₯ 3).To factor π₯π₯ 2 π₯π₯ 6, we look for two integers that multiply to 6 and add to 1. Suchintegers are 2 and 3, so π₯π₯ 2 π₯π₯ 6 (π₯π₯ 2)(π₯π₯ 3).Observation: The positive constant ππ in a trinomial π₯π₯ 2 πππ₯π₯ ππ tells us that theintegers ππ and ππ in the factorization (π₯π₯ ππ)(π₯π₯ ππ) are both of the same sign and theirsum is the middle coefficient ππ. In addition, if ππ is positive, both ππ and ππ are positive, andif ππ is negative, both ππ and ππ are negative.The negative constant ππ in a trinomial π₯π₯ 2 πππ₯π₯ ππ tells us that the integers ππ and ππ in thefactorization (π₯π₯ ππ)(π₯π₯ ππ) are of different signs and a difference of their absolutevalues is the middle coefficient ππ. In addition, the integer whose absolute value is largertakes the sign of the middle coefficient ππ.These observations are summarized in the following Table of Signs.Assume that ππ ππ .sum ππproduct ππ ππ ππ commentsππ is the sum of ππ and ππππ is the sum of ππ and ππππ is the difference ππ ππ ππ is the difference ππ ππ Factoring Trinomials with the Leading Coefficient Equal to 1Factor each trinomial, if possible.a.c.Solutiona.π₯π₯ 2 10π₯π₯ 24π₯π₯ 2 39π₯π₯π₯π₯ 40π¦π¦ 2b.d.π₯π₯ 2 9π₯π₯ 36π₯π₯ 2 7π₯π₯ 9To factor the trinomial π₯π₯ 2 10π₯π₯ 24, we look for two integers with a product of 24and a sum of 10. The two integers are fairly easy to guess, 4 and 6. However, ifone wishes to follow a more methodical way of finding these numbers, one can list thepossible two-number factorizations of 24 and observe the sums of these numbers.For simplicity, the tabledoesnβt include signs of theintegers. The signs aredetermined according tothe Table of Signs.product ππππ(pairs of factors of 24)ππ ππππππ ππππππ ππππ ππsum ππππ(sum of οΏ½οΏ½π©!Factoring Trinomials
section F2 226Since the product is positive and the sum is negative, both integers must be negative.So, we take 4 and 6.Thus, π₯π₯ 2 10π₯π₯ 24 (ππ ππ)(ππ ππ). The reader is encouraged to check thisfactorization by multiplying the obtained binomials.b.To factor the trinomial π₯π₯ 2 9π₯π₯ 36, we look for two integers with a product of 36and a sum of 9. So, let us list the possible factorizations of 36 into two numbers andobserve the differences of these numbers.product ππππ(pairs of factors of 36)ππ ππππππ ππππππ ππππππ ππππ ππsum ππ(difference of factors)3516950This row contains thesolution, so there is noneed to list any of thesubsequent rows.Since the product is negative and the sum is positive, the integers are of different signsand the one with the larger absolute value assumes the sign of the sum, which ispositive. So, we take 12 and 3.Thus, π₯π₯ 2 9π₯π₯ 36 (ππ ππππ)(ππ ππ). Again, the reader is encouraged to checkthis factorization by mltiplying the obtained binomials.c.To factor the trinomial π₯π₯ 2 39π₯π₯π₯π₯ 40π¦π¦ 2 , we look for two binomials of the form(π₯π₯ ? π¦π¦)(π₯π₯ ? π¦π¦) where the question marks are two integers with a product of 40and a sum of 39. Since the two integers are of different signs and the absolute valuesof these integers differ by 39, the two integers must be 40 and 1.Therefore, π₯π₯ 2 39π₯π₯π₯π₯ 40π¦π¦ 2 (ππ ππππππ)(ππ ππ).Suggestion: Create a table of pairs of factors only if guessing the two integers with thegiven product and sum becomes too difficult.d.When attempting to factor the trinomial π₯π₯ 2 7π₯π₯ 9, we look for a pair of integersthat would multiply to 9 and add to 7. There are only two possible factorizations of 9:9 1 and 3 3. However, neither of the sums, 9 1 or 3 3, are equal to 7. So, thereis no possible way of factoring π₯π₯ 2 7π₯π₯ 9 into two linear binomials with integralcoefficients. Therefore, if we admit only integral coefficients, this polynomial is notfactorable.Factorization of Quadratic Trinomials ππππππ ππππ ππ with ππ 0Before discussing factoring quadratic trinomials with a leading coefficient different than 1,let us observe the multiplication process of two linear binomials with integral coefficients.(πππ₯π₯ ππ)(πππ₯π₯ ππ) πππππ₯π₯ 2 πππππ₯π₯ πππππ₯π₯ ππππ ππ π₯π₯ 2 ππππFactoring ππ(ππππ ππππ)π₯π₯ ππππππ
section F2 227To reverse this process, notice that this time, we are looking for four integers ππ, ππ, ππ, andππ that satisfy the conditionsππππ ππ, ππππ ππ, ππππ ππππ ππ,where ππ, ππ, ππ are the coefficients of the quadratic trinomial that needs to be factored. Thisproduces a lot more possibilities to consider than in the guessing method used in the caseof the leading coefficient equal to 1. However, if at least one of the outside coefficients, ππor ππ, are prime, the guessing method still works reasonably well.For example, consider 2π₯π₯ 2 π₯π₯ 6. Since the coefficient ππ 2 ππππ is a prime number,there is only one factorization of ππ, which is 1 2. So, we can assume that ππ 2 and ππ 1. Therefore,2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ ππ )(π₯π₯ ππ )Since the constant term ππ 6 ππππ is negative, the binomial factors have different signsin the middle. Also, since ππππ is negative, we search for such ππ and ππ that the inside andoutside products differ by the middle term ππ π₯π₯, up to its sign. The only factorizations of6 are 1 6 and 2 3. So we try2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ 1)(π₯π₯ 6)π₯π₯Observe that these two trialscan be disregarded at onceas 2 is not a common factorof all the terms of thetrinomial, while it is acommon factor of the termsof one of the binomials.12π₯π₯2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ 6)(π₯π₯ 1)6π₯π₯2π₯π₯2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ 2)(π₯π₯ 3)2π₯π₯6π₯π₯2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ 3)(π₯π₯ 2)3π₯π₯4π₯π₯differs by 11π₯π₯ too muchdiffers by 4π₯π₯ still too muchdiffers by 4π₯π₯ still too muchdiffers by π₯π₯ perfect!Then, since the difference between the inner and outer products should be positive, thelarger product must be positive and the smaller product must be negative. So, we distributethe signs as below.2π₯π₯ 2 π₯π₯ 6 (2π₯π₯ 3)(π₯π₯ 2) 3π₯π₯4π₯π₯In the end, it is a good idea to multiply the product to check if it results in the originalpolynomial. We leave this task to the reader.What if the outside coefficients of the quadratic trinomial are both composite? Checkingall possible distributions of coefficients ππ, ππ, ππ, and ππ might be too cumbersome. Luckily,there is another method of factoring, called decomposition.Factoring Trinomials
section F2 228The decomposition method is based on the reverse FOIL process.Suppose the polynomial 6π₯π₯ 2 19π₯π₯ 15 factors into (πππ₯π₯ ππ)(πππ₯π₯ ππ). Observe that theFOIL multiplication of these two binomials results in the four term polynomial,πππππ₯π₯ 2 πππππ₯π₯ πππππ₯π₯ ππππ,which after combining the two middle terms gives us the original trinomial. So, reversingthese steps would lead us to the factored form of 6π₯π₯ 2 19π₯π₯ 15.To reverse the FOIL process, we would like to: This product is oftenreferred to as themaster product orthe ππππ-product. Express the middle term, 19π₯π₯, as a sum of two terms, πππππ₯π₯ and πππππ₯π₯, such that theproduct of their coefficients, ππππππππ, is equal to the product of the outsidecoefficients ππππ 6 15 90.Then, factor the four-term polynomial by grouping.Thus, we are looking for two integers with the product of 90 and the sum of 19. One cancheck that 9 and 10 satisfy these conditions. Therefore,6π₯π₯ 2 19π₯π₯ 15 6π₯π₯ 2 9π₯π₯ 10π₯π₯ 15 3π₯π₯(2π₯π₯ 3) 5(2π₯π₯ 3) (2π₯π₯ 3)(3π₯π₯ 5)Factoring Trinomials with the Leading Coefficient Different than 1Factor completely each trinomial.a.c.Solutiona.6π₯π₯ 3 14π₯π₯ 2 4π₯π₯18ππ2 19ππππ 12ππ 2b.d. 6π¦π¦ 2 10 19π¦π¦2(π₯π₯ 3)2 5(π₯π₯ 3) 12First, we factor out the GCF, which is 2π₯π₯. This gives us6π₯π₯ 3 14π₯π₯ 2 4π₯π₯ 2π₯π₯(3π₯π₯ 2 7π₯π₯ 2)The outside coefficients of the remaining trinomial are prime, so we can apply theguessing method to factor it further. The first terms of the possible binomial factorsmust be 3π₯π₯ and π₯π₯ while the last terms must be 2 and 1. Since both signs in the trinomialare positive, the signs used in the binomial factors must be both positive as well. So,we are ready to give it a try:2π₯π₯(3π₯π₯ 2 )(π₯π₯ 1 ) or2π₯π₯3π₯π₯Factoring2π₯π₯(3π₯π₯ 1 )(π₯π₯ 2 )π₯π₯6π₯π₯The first distribution of coefficients does not work as it would give us 2π₯π₯ 3π₯π₯ 5π₯π₯for the middle term. However, the second distribution works as π₯π₯ 6π₯π₯ 7π₯π₯, whichmatches the middle term of the trinomial. So,6π₯π₯ 3 14π₯π₯ 2 4π₯π₯ ππππ(ππππ ππ)(ππ ππ)
section F2 b.229Notice that the trinomial is not arranged in decreasing order of powers of π¦π¦. So, first,we rearrange the last two terms to achieve the decreasing order. Also, we factor outthe 1, so that the leading term of the remaining trinomial is positive. 6π¦π¦ 2 10 19π¦π¦ 6π¦π¦ 2 19π¦π¦ 10 (6π¦π¦ 2 19π¦π¦ 10)Then, since the outside coefficients are composite, we will use the decompositionmethod of factoring. The ππππ-product equals to 60 and the middle coefficient equals to 19. So, we are looking for two integers that multiply to 60 and add to 19. Theintegers that satisfy these conditions are 15 and 4. Hence, we factorthe square bracket isessential because of thenegative sign outside (6π¦π¦ 2 19π¦π¦ 10) (6π¦π¦ 2 15π¦π¦ 4π¦π¦ 10) [3π¦π¦(2π¦π¦ 5) 2(2π¦π¦ 5)]remember toreverse the sign! (ππππ ππ)(ππππ ππ)c.There is no common factor to take out of the polynomial 18ππ2 19ππππ 12ππ 2 . So,we will attempt to factor it into two binomials of the type (ππππ ππππ)(ππππ ππππ), usingthe decomposition method. The ππππ-product equals 12 18 2 2 2 3 3 3 andthe middle coefficient equals 19. To find the two integers that multiply to the ππππproduct and add to 19, it is convenient to group the factors of the product2 2 2 3 3 3in such a way that the products of each group differ by 19. It turns out that groupingall the 2βs and all the 3βs satisfy this condition, as 8 and 27 differ by 19. Thus, thedesired integers are 27 and 8, as the sum of them must be 19. So, we factor18ππ2 19ππππ 12ππ2 18ππ2 27ππππ 8ππππ 12ππ 2 9ππ(2ππ 3ππ) 4ππ(2ππ 3ππ) (ππππ ππππ)(ππππ ππππ)IN FORMd.To factor 2(π₯π₯ 3)2 5(π₯π₯ 3) 12,
Factoring . Factoring. Factoring is the reverse process of multiplication. Factoring polynomials in algebra has similar role as factoring numbers in arithmetic. Any number can be expressed as a product of prime numbers. For example, 2 3. 6 Similarly, any
Factoring . Factoring. Factoring is the reverse process of multiplication. Factoring polynomials in algebra has similar role as factoring numbers in arithmetic. Any number can be expressed as a product of prime numbers. For example, 2 3. 6 Similarly, any
241 Algebraic equations can be used to solve a large variety of problems involving geometric relationships. 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring Trinomials of the Form x2 bx c 5.4 Factoring Trinomials of the Form ax2 bx c 5.5 Factoring, Solving Equations, and Problem Solving
Factoring Polynomials Martin-Gay, Developmental Mathematics 2 13.1 β The Greatest Common Factor 13.2 β Factoring Trinomials of the Form x2 bx c 13.3 β Factoring Trinomials of the Form ax 2 bx c 13.4 β Factoring Trinomials of the Form x2 bx c by Grouping 13.5 β Factoring Perfect Square Trinomials and Difference of Two Squares
Grouping & Case II Factoring Factoring by Grouping: A new type of factoring is factoring by grouping. This type of factoring requires us to see structure in expressions. We usually factor by grouping when we have a polynomial that has four or more terms. Example Steps x3 2x2 3x 6 1. _ terms together that have
Factoring Trinomials Guided Notes . 2 Algebra 1 β Notes Packet Name: Pd: Factoring Trinomials Clear Targets: I can factor trinomials with and without a leading coefficient. Concept: When factoring polynomials, we are doing reverse multiplication or βun-distributing.β Remember: Factoring is the process of finding the factors that would .
Move to the next page to learn more about factoring and how it relates to polynomials. [page 2] Factoring Trinomials by Grouping There is a systematic approach to factoring trinomials with a leading coefficient greater than 1 called . If you need a refresher on factoring by grouping
Factoring Polynomials Factoring Quadratic Expressions Factoring Polynomials 1 Factoring Trinomials With a Common Factor Factoring Difference of Squares . Alignment of Accuplacer Math Topics and Developmental Math Topics with Khan Academy, the Common Core and Applied Tasks from The New England Board of Higher Education, April 2013
pile resistances or pile resistances calculated from profiles of test results into characteristic resistances. Pile load capacity β calculation methods 85 Case (c) is referred to as the alternative procedure in the Note to EN 1997-1 Β§7.6.2.3(8), even though it is the most common method in some countries. Characteristic pile resistance from profiles of ground test results Part 2 of EN 1997 .