Factoring Polynomials Factor
2/13/2013Chapter 13§ 13.1FactoringPolynomialsThe Greatest CommonFactorChapter SectionsFactors13.1 – The Greatest Common Factor13.2 – Factoring Trinomials of the Formx2Factors (either numbers or polynomials)When an integer is written as a product ofintegers, each of the integers in the product is afactor of the original number.When a polynomial is written as a product ofpolynomials, each of the polynomials in theproduct is a factor of the original polynomial.Factoring – writing a polynomial as a product ofpolynomials. bx c13.3 – Factoring Trinomials of the Form ax2 bx c13.4 – Factoring Trinomials of the Form x2 bx cby Grouping13.5 – Factoring Perfect Square Trinomials andDifference of Two Squares13.6 – Solving Quadratic Equations by Factoring13.7 – Quadratic Equations and Problem SolvingMartin-Gay, Developmental Mathematics2Martin-Gay, Developmental Mathematics41
2/13/2013Greatest Common FactorGreatest Common FactorExampleGreatest common factor – largest quantity that is afactor of all the integers or polynomials involved.Find the GCF of each list of numbers.1) 6, 8 and 466 2·38 2·2·246 2 · 23So the GCF is 2.2) 144, 256 and 300144 2 · 2 · 2 · 3 · 3256 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2300 2 · 2 · 3 · 5 · 5So the GCF is 2 · 2 4.Finding the GCF of a List of Integers or Terms1) Prime factor the numbers.2) Identify common prime factors.3) Take the product of all common prime factors. If there are no common prime factors, GCF is 1.Martin-Gay, Developmental Mathematics5Greatest Common FactorMartin-Gay, Developmental Mathematics7Greatest Common FactorExampleExampleFind the GCF of each list of numbers.1) 12 and 812 2 · 2 · 38 2·2·2So the GCF is 2 · 2 4.2) 7 and 207 1·720 2 · 2 · 5There are no common prime factors so theGCF is 1.Martin-Gay, Developmental MathematicsFind the GCF of each list of terms.1) x3 and x7x3 x · x · xx7 x · x · x · x · x · x · xSo the GCF is x · x · x x32) 6x5 and 4x36x5 2 · 3 · x · x · x4x3 2 · 2 · x · x · xSo the GCF is 2 · x · x · x 2x36Martin-Gay, Developmental Mathematics82
2/13/2013Greatest Common FactorFactoring out the GCFExampleExampleFactor out the GCF in each of the followingpolynomials.Find the GCF of the following list of terms.a3b2, a2b5 and a4b7a3b2 a · a · a · b · ba2b5 a · a · b · b · b · b · ba4b7 a · a · a · a · b · b · b · b · b · b · bSo the GCF is a · a · b · b a2b21) 6x3 – 9x2 12x 3 · x · 2 · x2 – 3 · x · 3 · x 3 · x · 4 3x(2x2 – 3x 4)2) 14x3y 7x2y – 7xy 7 · x · y · 2 · x2 7 · x · y · x – 7 · x · y · 1 7xy(2x2 x – 1)Notice that the GCF of terms containing variables will use thesmallest exponent found amongst the individual terms for eachvariable.Martin-Gay, Developmental Mathematics9Factoring Polynomials11Factoring out the GCFExampleThe first step in factoring a polynomial is tofind the GCF of all its terms.Factor out the GCF in each of the followingpolynomials.Then we write the polynomial as a product byfactoring out the GCF from all the terms.1) 6(x 2) – y(x 2) 6 · (x 2) – y · (x 2) (x 2)(6 – y)2) xy(y 1) – (y 1) xy · (y 1) – 1 · (y 1) (y 1)(xy – 1)The remaining factors in each term will form apolynomial.Martin-Gay, Developmental MathematicsMartin-Gay, Developmental Mathematics10Martin-Gay, Developmental Mathematics123
2/13/2013FactoringFactoring TrinomialsRemember that factoring out the GCF from the terms ofa polynomial should always be the first step in factoringa polynomial.Recall by using the FOIL method thatFILx2(x 2)(x 4) 4x 2x 8 x2 6x 8This will usually be followed by additional steps in theprocess.To factor x2 bx c into (x one #)(x another #),note that b is the sum of the two numbers and c is theproduct of the two numbers.ExampleFactor 90 15y2 – 18x – 3xy2.90 15y2 – 18x – 3xy2 3(30 5y2 – 6x – xy2) 3(5 · 6 5 · y2 – 6 · x – x · y2) 3(5(6 y2) – x (6 y2)) 3(6 y2)(5 – x)Martin-Gay, Developmental MathematicsOSo we’ll be looking for 2 numbers whose product isc and whose sum is b.Note: there are fewer choices for the product, sothat’s why we start there first.13Martin-Gay, Developmental Mathematics15Factoring PolynomialsExample§ 13.2Factoring Trinomials of theForm x2 bx cFactor the polynomial x2 13x 30.Since our two numbers must have a product of 30 and asum of 13, the two numbers must both be positive.Positive factors of 30Sum of Factors1, 30312, 15173, 1013Note, there are other factors, but once we find a pairthat works, we do not have to continue searching.So x2 13x 30 (x 3)(x 10).Martin-Gay, Developmental Mathematics164
2/13/2013Factoring PolynomialsPrime PolynomialsExampleExampleFactor the polynomial x2 – 11x 24.Since our two numbers must have a product of 24 and asum of -11, the two numbers must both be negative.Negative factors of 24Sum of Factors– 1, – 24– 25– 2, – 12– 14– 3, – 8– 11Factor the polynomial x2 – 6x 10.Since our two numbers must have a product of 10 and asum of – 6, the two numbers will have to both be negative.Negative factors of 10Sum of Factors– 1, – 10– 11– 2, – 5–7Since there is not a factor pair whose sum is – 6,x2 – 6x 10 is not factorable and we call it a primepolynomial.So x2 – 11x 24 (x – 3)(x – 8).Martin-Gay, Developmental Mathematics17Factoring PolynomialsMartin-Gay, Developmental Mathematics19Check Your Result!ExampleFactor the polynomial x2 – 2x – 35.Since our two numbers must have a product of – 35 and asum of – 2, the two numbers will have to have different signs.Factors of – 35Sum of Factors– 1, 35341, – 35– 34– 5, 725, – 7–2You should always check your factoringresults by multiplying the factored polynomialto verify that it is equal to the originalpolynomial.Many times you can detect computationalerrors or errors in the signs of your numbersby checking your results.So x2 – 2x – 35 (x 5)(x – 7).Martin-Gay, Developmental Mathematics18Martin-Gay, Developmental Mathematics205
2/13/2013Factoring PolynomialsExample§ 13.3Factor the polynomial 25x2 20x 4.Possible factors of 25x2 are {x, 25x} or {5x, 5x}.Possible factors of 4 are {1, 4} or {2, 2}.We need to methodically try each pair of factors until we finda combination that works, or exhaust all of our possible pairsof factors.Factoring Trinomials ofthe Form ax2 bx cKeep in mind that, because some of our pairs are not identicalfactors, we may have to exchange some pairs of factors andmake 2 attempts before we can definitely decide a particularpair of factors will not work.Continued.23Martin-Gay, Developmental MathematicsFactoring TrinomialsFactoring PolynomialsExample ContinuedReturning to the FOIL method,FOI L2(3x 2)(x 4) 3x 12x 2x 8 3x2 14x 82To factor ax bx c into (#1·x #2)(#3·x #4), notethat a is the product of the two first coefficients, c isthe product of the two last coefficients and b is thesum of the products of the outside coefficients andinside coefficients.Note that b is the sum of 2 products, not just 2numbers, as in the last section.We will be looking for a combination that gives the sum of theproducts of the outside terms and the inside terms equal to 20x.Factors Factors ResultingProduct ofProduct ofSum ofof 25x2 of 4Binomials Outside Terms Inside Terms Products{x, 25x} {1, 4} (x 1)(25x 4)4x25x29x(x 4)(25x 1)x100x101x{x, 25x} {2, 2} (x 2)(25x 2)2x50x52x{5x, 5x} {2, 2} (5x 2)(5x 2)10x10x20xContinued.Martin-Gay, Developmental Mathematics22Martin-Gay, Developmental Mathematics246
2/13/2013Factoring PolynomialsFactoring PolynomialsExample ContinuedExample ContinuedWe will be looking for a combination that gives the sum ofthe products of the outside terms and the inside terms equalto 41x.Check the resulting factorization using the FOIL method.FOIL(5x 2)(5x 2) 5x(5x) 5x(2) 2(5x) 2(2)Factors Factors ResultingProduct ofProduct ofSum ofof 21x2 of 10 Binomials Outside Terms Inside Terms Products 25x2 10x 10x 4 25x2 20x 4So our final answer when asked to factor 25x2 20x 4will be (5x 2)(5x 2) or (5x 2)2.25Martin-Gay, Developmental MathematicsFactoring Polynomials{x, 21x}{1, 10}(x – 1)(21x – 10)–10x 21x– 31x(x – 10)(21x – 1)–x 210x– 211x{x, 21x} {2, 5} (x – 2)(21x – 5)–5x 42x– 47x(x – 5)(21x – 2)–2x 105x– 107xContinued.27Martin-Gay, Developmental MathematicsFactoring PolynomialsExampleExample ContinuedFactor the polynomial 21x2 – 41x 10.Factors Factors ResultingProduct ofProduct ofSum ofof 21x2 of 10 Binomials Outside Terms Inside Terms ProductsPossible factors of 21x2 are {x, 21x} or {3x, 7x}.Since the middle term is negative, possible factors of 10must both be negative: {-1, -10} or {-2, -5}.We need to methodically try each pair of factors untilwe find a combination that works, or exhaust all of ourpossible pairs of factors.{3x, 7x}{1, 10}(3x – 1)(7x – 10) 30x 7x 37x(3x – 10)(7x – 1) 3x 70x 73x{3x, 7x} {2, 5} (3x – 2)(7x – 5) 15x 14x 29x(3x – 5)(7x – 2) 6x 35x 41xContinued.Martin-Gay, Developmental MathematicsContinued.26Martin-Gay, Developmental Mathematics287
2/13/2013Factoring PolynomialsFactoring PolynomialsExample ContinuedExample ContinuedCheck the resulting factorization using the FOIL method.FOIWe will be looking for a combination that gives the sum of theproducts of the outside terms and the inside terms equal to 7x.LFactorsof 6(3x – 5)(7x – 2) 3x(7x) 3x(-2) - 5(7x) - 5(-2) 21x2 – 6x – 35x 10 21x2ResultingBinomialsProduct ofProduct ofSum ofOutside Terms Inside Terms Products{ 1, 6} (3x – 1)(x – 6)– 41x 10(3x – 6)(x – 1){ 2, 3} (3x – 2)(x – 3)So our final answer when asked to factor 21x2 – 41x 10will be (3x – 5)(7x – 2).Martin-Gay, Developmental Mathematics(3x – 3)(x – 2) 18x x 19xCommon factor so no need to test. 9x 2x 11xCommon factor so no need to test.Continued.29Factoring PolynomialsMartin-Gay, Developmental Mathematics31Factoring PolynomialsExampleExample ContinuedFactor the polynomial 3x2 – 7x 6.Now we have a problem, because we haveexhausted all possible choices for the factors,but have not found a pair where the sum of theproducts of the outside terms and the insideterms is –7.So 3x2 – 7x 6 is a prime polynomial and willnot factor.The only possible factors for 3 are 1 and 3, so we know that, iffactorable, the polynomial will have to look like (3x)(x)in factored form, so that the product of the first two terms in thebinomials will be 3x2.Since the middle term is negative, possible factors of 6 must bothbe negative: { 1, 6} or { 2, 3}.We need to methodically try each pair of factors until we find acombination that works, or exhaust all of our possible pairs offactors.Continued.Martin-Gay, Developmental Mathematics30Martin-Gay, Developmental Mathematics328
2/13/2013Factoring PolynomialsFactoring PolynomialsExample ContinuedFactorsResultingProduct ofProduct ofSum ofof -30BinomialsOutside Terms Inside Terms Products{-1, 30} (3x – 1)(x 30)90x-x89xExampleFactor the polynomial 6x2y2 – 2xy2 – 60y2.Remember that the larger the coefficient, the greater theprobability of having multiple pairs of factors to check.So it is important that you attempt to factor out anycommon factors first.(3x 30)(x – 1){1, -30}Common factor so no need to test.(3x 1)(x – 30)-90x(3x – 30)(x 1){-2, 15}6x2y2 – 2xy2 – 60y2 2y2(3x2 – x – 30){2, -15}(3x – 2)(x 15)45x-2x(3x 2)(x – 15)-45x2x-43xCommon factor so no need to test.Continued.35Martin-Gay, Developmental MathematicsFactoring PolynomialsFactoring PolynomialsExample ContinuedExample ContinuedSince the product of the last two terms of the binomialswill have to be –30, we know that they must bedifferent signs.Factorsof –30{–3, 10}Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15},{2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}.ResultingBinomials(3x – 3)(x 10)(3x 10)(x – 3){3, –10}(3x 3)(x – 10)(3x – 10)(x 3)We will be looking for a combination that gives the sumof the products of the outside terms and the inside termsequal to –x.Product ofProduct ofSum ofOutside Terms Inside Terms ProductsCommon factor so no need to test.–9x10xxCommon factor so no need to test.9x–10xContinued.Martin-Gay, Developmental Mathematics43xCommon factor so no need to test.(3x – 15)(x 2)33Martin-Gay, Developmental Mathematics-89xCommon factor so no need to test.(3x 15)(x – 2)The only possible factors for 3 are 1 and 3, so we knowthat, if we can factor the polynomial further, it will have tolook like 2y2(3x)(x) in factored form.Continued.x–xContinued.34Martin-Gay, Developmental Mathematics369
2/13/2013Factoring PolynomialsFactoring by GroupingFactoring polynomials often involves additionaltechniques after initially factoring out the GCF.One technique is factoring by grouping.Example ContinuedCheck the resulting factorization using the FOIL method.FOILExample(3x – 10)(x 3) 3x(x) 3x(3) – 10(x) – 10(3)Factor xy y 2x 2 by grouping.Notice that, although 1 is the GCF for all fourterms of the polynomial, the first 2 terms have aGCF of y and the last 2 terms have a GCF of 2.xy y 2x 2 x · y 1 · y 2 · x 2 · 1 y(x 1) 2(x 1) (x 1)(y 2) 3x2 9x – 10x – 30 3x2 – x – 30So our final answer when asked to factor the polynomial6x2y2 – 2xy2 – 60y2 will be 2y2(3x – 10)(x 3).Martin-Gay, Developmental Mathematics37Martin-Gay, Developmental Mathematics39Factoring by Grouping§ 13.4Factoring Trinomials ofthe Form x2 bx cby GroupingFactoring a Four-Term Polynomial by Grouping1) Arrange the terms so that the first two terms have acommon factor and the last two terms have a commonfactor.2) For each pair of terms, use the distributive property tofactor out the pair’s greatest common factor.3) If there is now a common binomial factor, factor it out.4) If there is no common binomial factor in step 3, beginagain, rearranging the terms differently. If no rearrangement leads to a common binomialfactor, the polynomial cannot be factored.Martin-Gay, Developmental Mathematics4010
2/13/2013Factoring by GroupingExample§ 13.5Factor each of the following polynomials by grouping.1) x3 4x x2 4 x · x2 x · 4 1 · x2 1 · 4 x(x2 4) 1(x2 4) (x2 4)(x 1)2) 2x3 – x2 – 10x 5 x2 · 2x – x2 · 1 – 5 · 2x – 5 · (– 1) x2(2x – 1) – 5(2x – 1) (2x – 1)(x2 – 5)Martin-Gay, Developmental MathematicsFactoring Perfect SquareTrinomials and theDifference of Two Squares41Factoring by GroupingPerfect Square TrinomialsExampleRecall that in our very first example in Section4.3 we attempted to factor the polynomial25x2 20x 4.Factor 2x – 9y 18 – xy by grouping.Neither pair has a common factor (other than 1).So, rearrange the order of the factors.2x 18 – 9y – xy 2 · x 2 · 9 – 9 · y – x · y 2(x 9) – y(9 x) 2(x 9) – y(x 9) (make sure the factors are identical)(x 9)(2 – y)Martin-Gay, Developmental MathematicsThe result was (5x 2)2, an example of abinomial squared.Any trinomial that factors into a singlebinomial squared is called a perfect squaretrinomial.42Martin-Gay, Developmental Mathematics4411
2/13/2013Perfect Square TrinomialsDifference of Two SquaresIn the last chapter we learned a shortcut for squaring abinomial(a b)2 a2 2ab b2(a – b)2 a2 – 2ab b2Another shortcut for factoring a trinomial is when wewant to factor the difference of two squares.a2 – b2 (a b)(a – b)A binomial is the difference of two square ifSo if the first and last terms of our polynomial to befactored are can be written as expressions squared, andthe middle term of our polynomial is twice the productof those two expressions, then we can use these twoprevious equations to easily factor the polynomial.a2 2ab b2 (a b)2a2 – 2ab b2 (a – b)2Martin-Gay, Developmental Mathematics1.both terms are squares and2.the signs of the terms are different.9x2 – 25y2– c4 d445Perfect Square TrinomialsMartin-Gay, Developmental Mathematics47Difference of Two SquaresExampleExampleFactor the polynomial 16x2 – 8xy y2.Since the first term, 16x2, can be written as (4x)2, andthe last term, y2 is obviously a square, we check themiddle term.8xy 2(4x)(y) (twice the product of the expressionsthat are squared to get the first and last terms of thepolynomial)Therefore 16x2 – 8xy y2 (4x – y)2.Note: You can use FOIL method to verify that thefactorization for the polynomial is accurate.Martin-Gay, Developmental MathematicsFactor the polynomial x2 – 9.The first term is a square and the last term, 9, can bewritten as 32. The signs of each term are different, sowe have the difference of two squaresTherefore x2 – 9 (x – 3)(x 3).Note: You can use FOIL method to verify that thefactorization for the polynomial is accurate.46Martin-Gay, Developmental Mathematics4812
2/13/2013Solving Quadratic EquationsSteps for Solving a Quadratic Equation byFactoring§ 13.61)2)3)4)5)Solving QuadraticEquations by FactoringWrite the equation in standard form.Factor the quadratic completely.Set each factor containing a variable equal to 0.Solve the resulting equations.Check each solution in the original equation.Martin-Gay, Developmental MathematicsZero Factor TheoremSolving Quadratic EquationsExampleQuadratic EquationsSolve x2 – 5x 24. First write the quadratic equation in standard form.x2 – 5x – 24 0 Now we factor the quadratic using techniques fromthe previous sections.x2 – 5x – 24 (x – 8)(x 3) 0 We set each factor equal to 0.x – 8 0 or x 3 0, which will simplify tox 8 or x – 3Continued. Can be written in the form ax2 bx c 0. a, b and c are real numbers and a 0. This is referred to as standard form.Zero Factor Theorem If a and b are real numbers and ab 0, then a 0or b 0. This theorem is very useful in solving quadraticequations.Martin-Gay, Developmental Mathematics5150Martin-Gay, Developmental Mathematics5213
2/13/2013Solving Quadratic EquationsSolving Quadratic EquationsExample ContinuedExample Continued Check both possible answers in the original equation. Check both possible answers in the originalequation.82 – 5(8) 64 – 40 24 true(–3)2 – 5(–3) 9 – (–15) 24 true So our solutions for x are 8 or –3.4( 18 ) (8 ( 18 ) 9 ) 4 ( 18 ) (1 9 ) 4 ( 81 ) (1 0 ) 12 (1 0 ) 5true( )( ( ) ) ( )( )55554 8 9 4 ( 10 9 ) 4 ( 1) ( 5)( 1) 54444true So our solutions for x areMartin-Gay, Developmental Mathematics535or 4 .Martin-Gay, Developmental Mathematics55Finding x-interceptsSolving Quadratic EquationsExampleSolve 4x(8x 9) 5 First write the quadratic equation in standard form.32x2 36x 532x2 36x – 5 0 Now we factor the quadratic using techniques from theprevious sections.32x2 36x – 5 (8x – 1)(4x 5) 0 We set each factor equal to 0.8x – 1 0 or 4x 5 018x 1 or 4x – 5, which simplifies to x or 5 .84Continued.Martin-Gay, Developmental Mathematics1854Recall that in Chapter 3, we found the x-intercept oflinear equations by letting y 0 and solving for x.The same method works for x-intercepts in quadraticequations.Note: When the quadratic equation is written in standardform, the graph is a parabola opening up (when a 0) ordown (when a 0), where a is the coefficient of the x2term.The intercepts will be where the parabola crosses thex-axis.Martin-Gay, Developmental Mathematics5614
2/13/2013Finding x-interceptsStrategy for Problem SolvingExampleGeneral Strategy for Problem Solving1) Understand the problem Read and reread the problem Choose a variable to represent the unknown Construct a drawing, whenever possible Propose a solution and check2) Translate the problem into an equation3) Solve the equation4) Interpret the result Check proposed solution in problem State your conclusionFind the x-intercepts of the graph of y 4x2 11x 6.The equation is already written in standard form, sowe let y 0, then factor the quadratic in x.0 4x2 11x 6 (4x 3)(x 2)We set each factor equal to 0 and solve for x.4x 3 0 or x 2 04x –3 or x –2x –¾ or x –2So the x-intercepts are the points (–¾, 0) and (–2, 0).Martin-Gay, Developmental Mathematics5759Martin-Gay, Developmental MathematicsFinding an Unknown Number§ 13.7ExampleThe product of two consecutive positive integers is 132. Find thetwo integers.1.) UnderstandRead and reread the problem. If we letQuadratic Equationsand Problem Solvingx one of the unknown positive integers, thenx 1 the next consecutive positive integer.ContinuedMartin-Gay, Developmental Mathematics6015
2/13/2013Finding an Unknown NumberFinding an Unknown NumberExample continuedExample continued2.) Translate4.) InterpretCheck: Remember that x is suppose to represent a positiveinteger. So, although x -12 satisfies our equation, it cannot be asolution for the problem we were presented.The product ofis132If we let x 11, then x 1 12. The product of the two numbersis 11 · 12 132, our desired result.two consecutive positive integersState: The two positive integers are 11 and 12.x (x 1) 132Continued61Martin-Gay, Developmental MathematicsFinding an Unknown NumberPythagorean TheoremIn a right triangle, the sum of the squares of thelengths of the two legs is equal to the square of thelength of the hypotenuse.(leg a)2 (leg b)2 (hypotenuse)23.) Solvex(x 1) 132x2 x – 132 063The Pythagorean TheoremExample continuedx2 x 132Martin-Gay, Developmental Mathematics(Distributive property)(Write quadratic in standard form)(x 12)(x – 11) 0(Factor quadratic polynomial)x 12 0 or x – 11 0x –12 or x 11leg ahypotenuse(Set factors equal to 0)(Solve each factor for x)ContinuedMartin-Gay, Developmental Mathematics62leg bMartin-Gay, Developmental Mathematics6416
2/13/2013The Pythagorean TheoremThe Pythagorean TheoremExample continuedExample4.) InterpretFind the length of the shorter leg of a right triangle if the longer legis 10 miles more than the shorter leg and the hypotenuse is 10 milesless than twice the shorter leg.1.) UnderstandRead and reread the problem. If we letx the length of the shorter leg, thenCheck: Remember that x is suppose to represent the length ofthe shorter side. So, although x 0 satisfies our equation, itcannot be a solution for the problem we were presented.2 x - 10If we let x 30, then x 10 40 and 2x – 10 50. Since 302 402 900 1600 2500 502, the Pythagorean Theoremchecks out.xx 10 the length of the longer leg andx 10State: The length of the shorter leg is 30 miles. (Remember thatis all we were asked for in this problem.)2x – 10 the length of the hypotenuse.ContinuedMartin-Gay, Developmental Mathematics65Martin-Gay, Developmental Mathematics67The Pythagorean TheoremExample continued2.) TranslateBy the Pythagorean Theorem,(leg a)2 (leg b)2 (hypotenuse)2x2 (x 10)2 (2x – 10)23.) Solvex2 (x 10)2 (2x – 10)222x x 20x 100 4x2 – 40x 100(multiply the binomials)2x2 20x 100 4x2 – 40x 100(simplify left side)0 2x2 – 60x0 2x(x – 30)x 0 or x 30(subtract 2x2 20x 100 from both sides)(factor right side)(set each factor 0 and solve)ContinuedMartin-Gay, Developmental Mathematics6617
Factoring Polynomials Martin-Gay, Developmental Mathematics 2 13.1 – The Greatest Common Factor 13.2 – Factoring Trinomials of the Form x2 bx c 13.3 – Factoring Trinomials of the Form ax 2 bx c 13.4 – Factoring Trinomials of the Form x2 bx c by Grouping 13.5 – Factoring Perfect Square Trinomials and Difference of Two Squares
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
Section P.5 Factoring Polynomials 49 1 Factor out the greatest common factor of a polynomial. 2 Factor by grouping. Factoring is the process of writing a polynomial as the product of two or more polynomials.The factors of are and In this section, we will be factoring over the set of integers,meaning that the coefficients in the factors are integers.
Section 5.4 Factoring Polynomials 231 5.4 Factoring Polynomials Factoring Polynomials Work with a partner. Match each polynomial equation with the graph of its related polynomial function. Use the x-intercepts of the graph to write each polynomial in factored form. Explain your reason
Factoring Polynomials by Grouping You have used the Distributive Property to factor out a greatest common monomial from a polynomial. Sometimes, you can factor out a common binomial. You may be able to use the Distributive Property to factor polynomials with four terms, as described below. Factoring by Grouping Factor each polynomial by .
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
method for factoring trinomials (polynomials with three terms). In order to completely discuss trinomials, I will first talk about the greatest common factor and factoring by grouping. Greatest Common Factor (GCF) – Every pair of numbers, or terms of a polynomial, has what is referred to as the greatest common factor. The GCF is the largest .
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
of tank wall, which would be required by each design method for this example tank. The API 650 method is a working stress method, so the coefficient shown in the figure includes a factor of 2.0 for the purposes of comparing it with the NZSEE ultimate limit state approach. For this example, the 1986 NZSEE method gave a significantly larger impulsive mode seismic coefficient and wall thickness .
