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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 β€˜simpler’polynomials. For example,5π‘₯π‘₯ 10 5(π‘₯π‘₯ 2), or π‘₯π‘₯ 2 9 (π‘₯π‘₯ 3)(π‘₯π‘₯ 3).In the above definition, β€˜simpler’ 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, πœ‹πœ‹π‘Ÿπ‘Ÿ 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 𝟏𝟏 or 𝟏𝟏 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 β€œlargest expression” 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 𝑦𝑦 3, 18π‘₯π‘₯ 5 𝑦𝑦, and 24π‘₯π‘₯ 4 𝑦𝑦 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 𝑦𝑦 3 , 𝑦𝑦, and 𝑦𝑦 2 ,which is 𝑦𝑦. Therefore,GCF(6π‘₯π‘₯ 2 𝑦𝑦 3 , 18π‘₯π‘₯ 5 𝑦𝑦, 24π‘₯π‘₯ 4 𝑦𝑦 2 ) 6π‘₯π‘₯ 2 𝑦𝑦This GCF can be used to factor the polynomial 6π‘₯π‘₯ 2 𝑦𝑦 3 18π‘₯π‘₯ 5 𝑦𝑦 24π‘₯π‘₯ 4 𝑦𝑦 2 by first seeingit as6π‘₯π‘₯ 2 𝑦𝑦 𝑦𝑦 2 6π‘₯π‘₯ 2 𝑦𝑦 3π‘₯π‘₯ 3 6π‘₯π‘₯ 2 𝑦𝑦 4π‘₯π‘₯ 2 𝑦𝑦,Greatest Common Factor and Factoring by Grouping

section F1 216and then, using the reverse distributing property, β€˜pulling’ the 6π‘₯π‘₯ 2 𝑦𝑦 out of the bracket toobtain6π‘₯π‘₯ 2 𝑦𝑦(𝑦𝑦 2 3π‘₯π‘₯ 3 4π‘₯π‘₯ 2 𝑦𝑦).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‘₯, 4𝑦𝑦) 2 or GCF( 2π‘₯π‘₯, 4𝑦𝑦) 2,depending on what expression we wish to leave after factoring the GCF out: 2π‘₯π‘₯ 4𝑦𝑦 ���𝑝𝑝𝑝𝐺𝐺𝐺𝐺𝐺𝐺( π‘₯π‘₯ 2𝑦𝑦) or 2π‘₯π‘₯ 4𝑦𝑦 οΏ½οΏ½ 2 (π‘₯π‘₯ 2𝑦𝑦) ��𝑛𝑛𝑛 ��𝑑𝑑𝑑𝑑𝑑𝑑Note 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)2π‘Žπ‘Žπ‘π‘ 2 , π‘Žπ‘Ž2 𝑏𝑏, 𝑏𝑏, π‘Žπ‘Žb.d.4πœ‹πœ‹(𝑦𝑦 π‘₯π‘₯), 8πœ‹πœ‹(π‘₯π‘₯ 𝑦𝑦)3π‘₯π‘₯ 1 𝑦𝑦 3 , π‘₯π‘₯ 2 𝑦𝑦 2 𝑧𝑧Since 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 ) πŸ‘πŸ‘πŸ‘πŸ‘(𝒙𝒙 𝟏𝟏)b.Since 𝑦𝑦 π‘₯π‘₯ is opposite to π‘₯π‘₯ 𝑦𝑦, then 𝑦𝑦 π‘₯π‘₯ can be written as (π‘₯π‘₯ 𝑦𝑦). So 4, πœ‹πœ‹, and(π‘₯π‘₯ 𝑦𝑦) is common for both expressions. Thus,GCF 4πœ‹πœ‹(𝑦𝑦 π‘₯π‘₯), 8πœ‹πœ‹(π‘₯π‘₯ 𝑦𝑦) πŸ’πŸ’πŸ’πŸ’(𝒙𝒙 π’šπ’š)Note: The Greatest Common Factor is unique up to the sign. Notice that in the aboveexample, we could write π‘₯π‘₯ 𝑦𝑦 as (𝑦𝑦 π‘₯π‘₯) and choose the GCF to be 4πœ‹πœ‹(𝑦𝑦 π‘₯π‘₯).c.The terms π‘Žπ‘Žπ‘π‘ 2 , π‘Žπ‘Ž2 𝑏𝑏, 𝑏𝑏, and π‘Žπ‘Ž have no common factor other than 1, soGCF(π‘Žπ‘Žπ‘π‘ 2 , π‘Žπ‘Ž2 𝑏𝑏, 𝑏𝑏, π‘Žπ‘Ž) 𝟏𝟏Factoring

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 𝑦𝑦 3 and𝑦𝑦 2 is 𝑦𝑦 3 . Therefore,GCF(3π‘₯π‘₯ 1 𝑦𝑦 3 , π‘₯π‘₯ 2 𝑦𝑦 2 𝑧𝑧) 𝒙𝒙 𝟐𝟐 π’šπ’š πŸ‘πŸ‘Factoring out the Greatest Common FactorFactor each expression by taking the greatest common factor out. Simplify the factors, ifpossible.a.c.Solutiona.54π‘₯π‘₯ 2 𝑦𝑦 2 60π‘₯π‘₯𝑦𝑦 3b. π‘₯π‘₯(π‘₯π‘₯ 5) π‘₯π‘₯ 2 (5 π‘₯π‘₯) (π‘₯π‘₯ 5)2d.π‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Ž2 𝑏𝑏(π‘Žπ‘Ž 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 𝑦𝑦 2 , 60π‘₯π‘₯𝑦𝑦 3 ) 6π‘₯π‘₯𝑦𝑦 2 , we factor the 6π‘₯π‘₯𝑦𝑦 2 out by dividing each termof the polynomial 54π‘₯π‘₯ 2 𝑦𝑦 2 60π‘₯π‘₯𝑦𝑦 3 by 6π‘₯π‘₯𝑦𝑦 2 , as below.54π‘₯π‘₯ 2 𝑦𝑦 2 60π‘₯π‘₯𝑦𝑦 3 πŸ”πŸ”πŸ”πŸ”π’šπ’šπŸπŸ (πŸ—πŸ—πŸ—πŸ— 𝟏𝟏𝟏𝟏𝟏𝟏)54π‘₯π‘₯ 2 𝑦𝑦 2 9π‘₯π‘₯6π‘₯π‘₯𝑦𝑦 260π‘₯π‘₯𝑦𝑦 3 10𝑦𝑦6π‘₯π‘₯𝑦𝑦 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, π‘Žπ‘Žπ‘Žπ‘Ž and π‘Žπ‘Ž2 𝑏𝑏(π‘Žπ‘Ž 1). The greatestcommon factor for these two terms is π‘Žπ‘Žπ‘Žπ‘Ž, so we haveπ‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Ž2 𝑏𝑏(π‘Žπ‘Ž 1) π‘Žπ‘Žπ‘Žπ‘Ž 𝟏𝟏 π‘Žπ‘Ž(π‘Žπ‘Ž 1) π‘Žπ‘Žπ‘Žπ‘Ž(1 π‘Žπ‘Ž2 π‘Žπ‘Ž)2 π‘Žπ‘Žπ‘Žπ‘Ž( π‘Žπ‘Ž π‘Žπ‘Ž 1) 𝒂𝒂𝒂𝒂 π’‚π’‚πŸπŸ 𝒂𝒂 𝟏𝟏 remember to leave 1for the first termsimplify and arrangein decreasing powerstake the β€œ β€œ outGreatest Common Factor and Factoring by Grouping

section F1 218Note: Both factorizations, π‘Žπ‘Žπ‘Žπ‘Ž( π‘Žπ‘Ž2 π‘Žπ‘Ž 1) and π‘Žπ‘Žπ‘Žπ‘Ž(π‘Žπ‘Ž2 π‘Žπ‘Ž 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 (𝒙𝒙 πŸ“πŸ“) π’™π’™πŸπŸ 𝟐𝟐𝟐𝟐 πŸ“πŸ“ 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 𝒙𝒙 πŸ‘πŸ‘ π’™π’™πŸπŸ 𝟐𝟐𝟐𝟐 𝟏𝟏 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‘₯ 𝑦𝑦. 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 𝑦𝑦. Observe what happens when we factor each group.2π‘₯π‘₯ π‘₯π‘₯ π‘₯π‘₯π‘₯π‘₯ 𝑦𝑦 π‘₯π‘₯(π‘₯π‘₯ 1) 𝑦𝑦(π‘₯π‘₯ 1)Factoring (π‘₯π‘₯ 1)(π‘₯π‘₯ 𝑦𝑦)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€œ ” or β€œ β€œ 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€œ ”, the expression would look like a product π‘₯π‘₯(π‘₯π‘₯ 1) 𝑦𝑦(π‘₯π‘₯ 1), which is notthe case. Also, since the expression π‘₯π‘₯(π‘₯π‘₯ 1) 𝑦𝑦(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 𝑦𝑦 8 2π‘₯π‘₯ 2 8𝑦𝑦b.d.5π‘₯π‘₯ 5𝑦𝑦 π‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Žπ‘Žπ‘Žπ‘₯π‘₯ 2 π‘₯π‘₯ 𝑦𝑦 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)𝟐𝟐 (𝒙𝒙 πŸ‘πŸ‘) πŸπŸπ’™π’™ 𝟏𝟏 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 – π‘Žπ‘Ž , rather than π‘Žπ‘Ž, out of the last twoterms. So,5π‘₯π‘₯ 5𝑦𝑦 π‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Žπ‘Žπ‘Ž 5(π‘₯π‘₯ 𝑦𝑦) π‘Žπ‘Ž(π‘₯π‘₯ 𝑦𝑦)reverse signs whenβ€˜pulling’ a β€œ β€œ out (𝒙𝒙 πŸ‘πŸ‘) πŸπŸπ’™π’™πŸπŸ 𝟏𝟏 c.Notice that 2 is the GCF of all terms, so we factor it out first.2π‘₯π‘₯ 2 𝑦𝑦 8 2π‘₯π‘₯ 2 8𝑦𝑦 2(π‘₯π‘₯ 2 𝑦𝑦 4 π‘₯π‘₯ 2 4𝑦𝑦)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‘₯𝑦𝑦 4𝑦𝑦 π‘₯π‘₯ ) 4 2 2[𝑦𝑦(π‘₯π‘₯ 4) (π‘₯π‘₯ 2 𝟐𝟐 π’™π’™πŸπŸ πŸ’πŸ’ (π’šπ’š 𝟏𝟏)d. 4)]reverse signs whenβ€˜pulling’ a β€œ ” outnow, there is no need for the squarebracket as multiplication is associativeThe polynomial π‘₯π‘₯ 2 π‘₯π‘₯ 𝑦𝑦 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‘₯ 𝑦𝑦 1 π‘₯π‘₯(π‘₯π‘₯ 1) (𝑦𝑦 1),which cannot be factored, as the expressions π‘₯π‘₯ 1 and 𝑦𝑦 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 π‘Žπ‘Žπ‘Žπ‘Ž 3π‘Žπ‘Ž 5 for π‘Žπ‘Ž.First, we move the terms containing the variable π‘Žπ‘Ž to one side of the equation,and then factor π‘Žπ‘Ž outπ‘Žπ‘Žπ‘Žπ‘Ž 3π‘Žπ‘Ž 5π‘Žπ‘Žπ‘Žπ‘Ž 3π‘Žπ‘Ž 5,π‘Žπ‘Ž(𝑏𝑏 3) 5.So, after dividing by 𝑏𝑏 3, we obtain𝒂𝒂 πŸ“πŸ“.𝒃𝒃 πŸ‘πŸ‘F.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‘₯ 8𝑦𝑦 is prime.123414The factorization π‘₯π‘₯ 𝑦𝑦 (2π‘₯π‘₯ 3𝑦𝑦) 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(π‘Žπ‘Ž 5), 12(5 π‘Žπ‘Ž)10. 21π‘Žπ‘Ž3 𝑏𝑏6 , 35π‘Žπ‘Ž7 𝑏𝑏5 , 28π‘Žπ‘Ž5 𝑏𝑏 812. π‘₯π‘₯(π‘₯π‘₯ 3)2 , π‘₯π‘₯ 2 (π‘₯π‘₯ 3)(π‘₯π‘₯ 2)14. (π‘₯π‘₯ 2𝑦𝑦)(π‘₯π‘₯ 1), (2𝑦𝑦 π‘₯π‘₯)(π‘₯π‘₯ 1)Greatest Common Factor and Factoring by Grouping

section F1 22215. 3π‘₯π‘₯ 2 𝑦𝑦 3 , 6π‘₯π‘₯ 3 𝑦𝑦 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. 6π‘Žπ‘Ž3 36π‘Žπ‘Ž4 18π‘Žπ‘Ž223. π‘Žπ‘Ž(π‘₯π‘₯ 2) 𝑏𝑏(π‘₯π‘₯ 2)18. 8π‘˜π‘˜ 3 24π‘˜π‘˜21. 10π‘Ÿπ‘Ÿ 2 𝑠𝑠 2 15π‘Ÿπ‘Ÿ 4 𝑠𝑠 226. (𝑛𝑛 2)(𝑛𝑛 3) (𝑛𝑛 2)(𝑛𝑛 3)28. (4π‘₯π‘₯ 𝑦𝑦) 4π‘₯π‘₯(𝑦𝑦 4π‘₯π‘₯)27. 𝑦𝑦(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. 3𝑝𝑝 5 𝑝𝑝 3 2𝑝𝑝 222. 5π‘₯π‘₯ 2 𝑦𝑦 3 10π‘₯π‘₯ 3 𝑦𝑦 224. π‘Žπ‘Ž(𝑦𝑦 2 3) 2(𝑦𝑦 2 3)25. (π‘₯π‘₯ 2)(π‘₯π‘₯ 3) (π‘₯π‘₯ 2)(π‘₯π‘₯ 5)31. 3π‘₯π‘₯ 3 π‘₯π‘₯ 219. 6𝑝𝑝3 3𝑝𝑝2 9𝑝𝑝430. 2(𝑝𝑝 3) 4(𝑝𝑝 3)2 (𝑝𝑝 3)332. π‘˜π‘˜ 2 2π‘˜π‘˜ 433. π‘₯π‘₯ 4 2π‘₯π‘₯ 3 7π‘₯π‘₯ 2Factor by grouping, if possible.35. 3π‘₯π‘₯ 3 𝑦𝑦 π‘₯π‘₯ 2 𝑦𝑦 236. 5π‘₯π‘₯ 2 𝑦𝑦 3 2π‘₯π‘₯ 1 𝑦𝑦 237. 20 5π‘₯π‘₯ 12𝑦𝑦 3π‘₯π‘₯π‘₯π‘₯38. 2π‘Žπ‘Ž3 π‘Žπ‘Ž2 14π‘Žπ‘Ž 739. π‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Žπ‘Žπ‘Ž 𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏43. 3𝑝𝑝2 9𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝 3π‘žπ‘ž 244. 3π‘₯π‘₯ 2 π‘₯π‘₯ 2 𝑦𝑦 𝑦𝑦𝑧𝑧 2 3𝑧𝑧 245. 2π‘₯π‘₯ 3 π‘₯π‘₯ 2 4π‘₯π‘₯ 240. 2π‘₯π‘₯π‘₯π‘₯ π‘₯π‘₯ 2 𝑦𝑦 6 3π‘₯π‘₯46. π‘₯π‘₯ 2 𝑦𝑦 2 π‘Žπ‘Žπ‘Žπ‘Ž π‘Žπ‘Žπ‘¦π‘¦ 2 𝑏𝑏π‘₯π‘₯ 249. π‘₯π‘₯π‘₯π‘₯ 6𝑦𝑦 3π‘₯π‘₯ 1841. 3π‘₯π‘₯ 2 4π‘₯π‘₯π‘₯π‘₯ 6π‘₯π‘₯π‘₯π‘₯ 8𝑦𝑦 242. π‘₯π‘₯ 3 π‘₯π‘₯π‘₯π‘₯ 𝑦𝑦 2 π‘₯π‘₯ 2 𝑦𝑦48. π‘₯π‘₯ 2 𝑦𝑦 π‘₯π‘₯π‘₯π‘₯ π‘₯π‘₯ 𝑦𝑦47. π‘₯π‘₯π‘₯π‘₯ π‘Žπ‘Žπ‘Žπ‘Ž 𝑏𝑏𝑏𝑏 π‘Žπ‘Žπ‘Žπ‘Ž50. π‘₯π‘₯ 𝑛𝑛 𝑦𝑦 3π‘₯π‘₯ 𝑛𝑛 𝑦𝑦 551. π‘Žπ‘Žπ‘›π‘› π‘₯π‘₯ 𝑛𝑛 2π‘Žπ‘Žπ‘›π‘› π‘₯π‘₯ 𝑛𝑛 2Factor completely. Remember to check for the GCF first.52. 5π‘₯π‘₯ 5π‘Žπ‘Žπ‘Žπ‘Ž 5π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž 5𝑏𝑏𝑏𝑏54. π‘₯π‘₯ 4 (π‘₯π‘₯ 1) π‘₯π‘₯ 3 (π‘₯π‘₯ 1) π‘₯π‘₯ 2 π‘₯π‘₯Discussion Point53. 6π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ 14𝑠𝑠 6π‘Ÿπ‘Ÿ 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 𝑦𝑦 5 8π‘₯π‘₯𝑦𝑦 3 is 2π‘₯π‘₯𝑦𝑦 3 (2π‘₯π‘₯𝑦𝑦 2 4). Is this a completefactorization?Use factoring the GCF strategy to solve each formula for the indicated variable.57. 𝐴𝐴 𝑷𝑷 π‘·π‘·π‘Ÿπ‘Ÿ, for 𝑷𝑷59. 2𝒕𝒕 𝑐𝑐 π‘˜π‘˜π’•π’•, for 𝒕𝒕Factoring121258. 𝑀𝑀 π’‘π’‘π‘žπ‘ž π’‘π’‘π‘Ÿπ‘Ÿ, for 𝒑𝒑60. π‘€π‘€π’šπ’š 3π’šπ’š π‘₯π‘₯, for π’šπ’š

section F1 Analytic Skills61.4π‘₯π‘₯Write the area of each shaded region in factored οΏ½οΏ½οΏ½π‘Ÿπ‘Ÿπ‘ŸGreatest 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 𝑏𝑏𝑏𝑏 𝑐𝑐, then quadratic trinomialsof the form π‘Žπ‘Žπ‘₯π‘₯ 2 𝑏𝑏𝑏𝑏 𝑐𝑐, where π‘Žπ‘Ž 1, and finally trinomials reducibleto quadratic by means of substitution.Factorization of Quadratic Trinomials π’™π’™πŸπŸ 𝒃𝒃𝒃𝒃 𝒄𝒄Factorization of a quadratic trinomial π‘₯π‘₯ 2 𝑏𝑏𝑏𝑏 𝑐𝑐 is the reverse process of the FOILmethod of multiplying two linear binomials. Observe that(π‘₯π‘₯ 𝑝𝑝)(π‘₯π‘₯ π‘žπ‘ž) π‘₯π‘₯ 2 π‘žπ‘žπ‘₯π‘₯ 𝑝𝑝π‘₯π‘₯ 𝑝𝑝𝑝𝑝 π‘₯π‘₯ 2 (𝑝𝑝 π‘žπ‘ž)π‘₯π‘₯ 𝑝𝑝𝑝𝑝So, to reverse this multiplication, we look for two numbers 𝑝𝑝 and π‘žπ‘ž, such that the product𝑝𝑝𝑝𝑝 equals to the free term 𝑐𝑐 and the sum 𝑝𝑝 π‘žπ‘ž equals to the middle coefficient 𝑏𝑏 of thetrinomial.π‘₯π‘₯ 2 𝑏𝑏 π‘₯π‘₯ 𝑐𝑐 (π‘₯π‘₯ 𝑝𝑝)(π‘₯π‘₯ π‘žπ‘ž)(𝒑𝒑 𝒒𝒒)𝒑𝒑𝒑𝒑For 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 𝒄𝒄 in a trinomial π‘₯π‘₯ 2 𝑏𝑏π‘₯π‘₯ 𝑐𝑐 tells us that theintegers 𝑝𝑝 and π‘žπ‘ž in the factorization (π‘₯π‘₯ 𝑝𝑝)(π‘₯π‘₯ π‘žπ‘ž) are both of the same sign and theirsum is the middle coefficient 𝑏𝑏. In addition, if 𝑏𝑏 is positive, both 𝑝𝑝 and π‘žπ‘ž are positive, andif 𝑏𝑏 is negative, both 𝑝𝑝 and π‘žπ‘ž are negative.The negative constant 𝒄𝒄 in a trinomial π‘₯π‘₯ 2 𝑏𝑏π‘₯π‘₯ 𝑐𝑐 tells us that the integers 𝑝𝑝 and π‘žπ‘ž in thefactorization (π‘₯π‘₯ 𝑝𝑝)(π‘₯π‘₯ π‘žπ‘ž) are of different signs and a difference of their absolutevalues is the middle coefficient 𝑏𝑏. In addition, the integer whose absolute value is largertakes the sign of the middle coefficient 𝑏𝑏.These observations are summarized in the following Table of Signs.Assume that 𝑝𝑝 π‘žπ‘ž .sum 𝒃𝒃product 𝒄𝒄 𝒑𝒑 𝒒𝒒 comments𝑏𝑏 is the sum of 𝑝𝑝 and π‘žπ‘žπ‘π‘ is the sum of 𝑝𝑝 and π‘žπ‘žπ‘π‘ is the difference 𝑝𝑝 π‘žπ‘ž 𝑏𝑏 is the difference π‘žπ‘ž 𝑝𝑝 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‘₯ 40𝑦𝑦 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 tabledoesn’t include signs of theintegers. The signs aredetermined according tothe Table of Signs.product 𝟐𝟐𝟐𝟐(pairs of factors of 24)𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 πŸπŸπŸπŸπŸ‘πŸ‘ πŸ–πŸ–πŸ’πŸ’ πŸ”πŸ”sum 𝟏𝟏𝟏𝟏(sum of ��𝑩!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 (𝒙𝒙 πŸ’πŸ’)(𝒙𝒙 πŸ”πŸ”). 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 πŸ‘πŸ‘πŸ‘πŸ‘(pairs of factors of 36)𝟏𝟏 πŸ‘πŸ‘πŸ‘πŸ‘πŸπŸ πŸπŸπŸπŸπŸ‘πŸ‘ πŸπŸπŸπŸπŸ’πŸ’ πŸ—πŸ—πŸ”πŸ” πŸ”πŸ”sum πŸ—πŸ—(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 (𝒙𝒙 𝟏𝟏𝟏𝟏)(𝒙𝒙 πŸ‘πŸ‘). 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‘₯ 40𝑦𝑦 2 , we look for two binomials of the form(π‘₯π‘₯ ? 𝑦𝑦)(π‘₯π‘₯ ? 𝑦𝑦) 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‘₯ 40𝑦𝑦 2 (𝒙𝒙 πŸ’πŸ’πŸ’πŸ’πŸ’πŸ’)(𝒙𝒙 π’šπ’š).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 π’‚π’‚π’‚π’‚πŸπŸ 𝒃𝒃𝒃𝒃 𝒄𝒄 with 𝒂𝒂 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.(π’Žπ’Žπ‘₯π‘₯ 𝑝𝑝)(𝒏𝒏π‘₯π‘₯ π‘žπ‘ž) π‘šπ‘šπ‘šπ‘šπ‘₯π‘₯ 2 π‘šπ‘šπ‘žπ‘žπ‘₯π‘₯ 𝑛𝑛𝑝𝑝π‘₯π‘₯ 𝑝𝑝𝑝𝑝 𝒂𝒂 π‘₯π‘₯ 2 π’Žπ’Žπ’Žπ’ŽFactoring 𝒃𝒃(π’Žπ’Žπ’’π’’ 𝒏𝒏𝒑𝒑)π‘₯π‘₯ 𝒄𝒄𝒑𝒑𝒑𝒑

section F2 227To reverse this process, notice that this time, we are looking for four integers π‘šπ‘š, 𝑛𝑛, 𝑝𝑝, andπ‘žπ‘ž that satisfy the conditionsπ‘šπ‘šπ‘šπ‘š π‘Žπ‘Ž, 𝑝𝑝𝑝𝑝 𝑐𝑐, π‘šπ‘šπ‘žπ‘ž 𝑛𝑛𝑝𝑝 𝑏𝑏,where π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐 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, π‘Žπ‘Žor 𝑐𝑐, are prime, the guessing method still works reasonably well.For example, consider 2π‘₯π‘₯ 2 π‘₯π‘₯ 6. Since the coefficient π‘Žπ‘Ž 2 π‘šπ‘šπ‘šπ‘š is a prime number,there is only one factorization of π‘Žπ‘Ž, which is 1 2. So, we can assume that π‘šπ‘š 2 and 𝑛𝑛 1. Therefore,2π‘₯π‘₯ 2 π‘₯π‘₯ 6 (2π‘₯π‘₯ 𝑝𝑝 )(π‘₯π‘₯ π‘žπ‘ž )Since the constant term 𝑐𝑐 6 𝑝𝑝𝑝𝑝 is negative, the binomial factors have different signsin the middle. Also, since 𝑝𝑝𝑝𝑝 is negative, we search for such 𝑝𝑝 and π‘žπ‘ž that the inside andoutside products differ by the middle term 𝑏𝑏 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 π‘šπ‘š, 𝑛𝑛, 𝑝𝑝, and π‘žπ‘ž 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 (π‘šπ‘šπ‘₯π‘₯ 𝑝𝑝)(𝑛𝑛π‘₯π‘₯ π‘žπ‘ž). Observe that theFOIL multiplication of these two binomials results in the four term polynomial,π‘šπ‘šπ‘šπ‘šπ‘₯π‘₯ 2 π‘šπ‘šπ‘žπ‘žπ‘₯π‘₯ 𝑛𝑛𝑝𝑝π‘₯π‘₯ 𝑝𝑝𝑝𝑝,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 𝒂𝒂𝒂𝒂-product. Express the middle term, 19π‘₯π‘₯, as a sum of two terms, π‘šπ‘šπ‘žπ‘žπ‘₯π‘₯ and 𝑛𝑛𝑝𝑝π‘₯π‘₯, such that theproduct of their coefficients, π‘šπ‘šπ‘šπ‘šπ‘π‘π‘π‘, is equal to the product of the outsidecoefficients π‘Žπ‘Žπ‘Žπ‘Ž 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‘₯18π‘Žπ‘Ž2 19π‘Žπ‘Žπ‘Žπ‘Ž 12𝑏𝑏 2b.d. 6𝑦𝑦 2 10 19𝑦𝑦2(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‘₯ 𝟐𝟐𝟐𝟐(πŸ‘πŸ‘πŸ‘πŸ‘ 𝟏𝟏)(𝒙𝒙 𝟐𝟐)

section F2 b.229Notice that the trinomial is not arranged in decreasing order of powers of 𝑦𝑦. 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. 6𝑦𝑦 2 10 19𝑦𝑦 6𝑦𝑦 2 19𝑦𝑦 10 (6𝑦𝑦 2 19𝑦𝑦 10)Then, since the outside coefficients are composite, we will use the decompositionmethod of factoring. The π‘Žπ‘Žπ‘Žπ‘Ž-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 (6𝑦𝑦 2 19𝑦𝑦 10) (6𝑦𝑦 2 15𝑦𝑦 4𝑦𝑦 10) [3𝑦𝑦(2𝑦𝑦 5) 2(2𝑦𝑦 5)]remember toreverse the sign! (𝟐𝟐𝟐𝟐 πŸ“πŸ“)(πŸ‘πŸ‘πŸ‘πŸ‘ 𝟐𝟐)c.There is no common factor to take out of the polynomial 18π‘Žπ‘Ž2 19π‘Žπ‘Žπ‘Žπ‘Ž 12𝑏𝑏 2 . So,we will attempt to factor it into two binomials of the type (π‘šπ‘šπ‘Žπ‘Ž 𝑝𝑝𝑏𝑏)(π‘›π‘›π‘Žπ‘Ž π‘žπ‘žπ‘π‘), usingthe decomposition method. The π‘Žπ‘Žπ‘Žπ‘Ž-product equals 12 18 2 2 2 3 3 3 andthe middle coefficient equals 19. To find the two integers that multiply to the π‘Žπ‘Žπ‘Žπ‘Žproduct 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 2’s and all the 3’s 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 factor18π‘Žπ‘Ž2 19π‘Žπ‘Žπ‘Žπ‘Ž 12𝑏𝑏2 18π‘Žπ‘Ž2 27π‘Žπ‘Žπ‘Žπ‘Ž 8π‘Žπ‘Žπ‘Žπ‘Ž 12𝑏𝑏 2 9π‘Žπ‘Ž(2π‘Žπ‘Ž 3𝑏𝑏) 4𝑏𝑏(2π‘Žπ‘Ž 3𝑏𝑏) (𝟐𝟐𝟐𝟐 πŸ‘πŸ‘πŸ‘πŸ‘)(πŸ—πŸ—πŸ—πŸ— πŸ’πŸ’πŸ’πŸ’)IN FORMd.To factor 2(π‘₯π‘₯ 3)2 5(π‘₯π‘₯ 3) 12,

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