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Factoring Polynomials

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Factoring Polynomials 2/13/2013 Chapter 13 § 13.1 Factoring The Greatest Common Polynomials Factor Chapter Sections Factors 13.1 – The Greatest Common Factor Factors (either numbers or polynomials) 13.2 – Factoring Trinomials of the Form x2 + bx + c When an integer is written as a product of 13.3 – Factoring Trinomials of the Form ax 2 + bx + c integers, each of the integers in the product is a factor of the original number. 13.4 – Factoring Trinomials of the Form x2 + bx + c When a polynomial is written as a product of by Grouping polynomials, each of the polynomials in the 13.5 – Factoring Perfect Square Trinomials and product is a factor of the original polynomial. Difference of Two Squares Factoring – writing a polynomial as a product of 13.6 – Solving Quadratic Equations by Factoring polynomials. 13.7 – Quadratic Equations and Problem Solving Martin-Gay, Developmental Mathematics 2 Martin-Gay, Developmental Mathematics 4 1 2/13/2013 Greatest Common Factor Greatest Common Factor Greatest common factor – largest quantity that is a Example factor of all the integers or polynomials involved. Find the GCF of each list of numbers. 1) 6, 8 and 46 6 = 2 · 3 Finding the GCF of a List of Integers or Terms 8 = 2 · 2 · 2 1) Prime factor the numbers. 46 = 2 · 23 2) Identify common prime factors. So the GCF is 2. 3) Take the product of all common prime factors. 2) 144, 256 and 300 144 = 2 · 2 · 2 · 3 · 3 • If there are no common prime factors, GCF is 1. 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4. Martin-Gay, Developmental Mathematics 5 Martin-Gay, Developmental Mathematics 7 Greatest Common Factor Greatest Common Factor Example Example Find the GCF of each list of numbers. Find the GCF of each list of terms. 1) 12 and 8 1) x3 and x7 12 = 2 · 2 · 3 x3 = x · x · x 8 = 2 · 2 · 2 x7 = x · x · x · x · x · x · x So the GCF is 2 · 2 = 4. So the GCF is x · x · x = x3 2) 7 and 20 2) 6x5 and 4 x3 7 = 1 · 7 6x5 = 2 · 3 · x · x · x 20 = 2 · 2 · 5 3 There are no common prime factors so the 4x = 2 · 2 · x · x · x GCF is 1. So the GCF is 2 · x · x · x = 2 x3 Martin-Gay, Developmental Mathematics 6 Martin-Gay, Developmental Mathematics 8 2 2/13/2013 Greatest Common Factor Factoring out the GCF Example Example Find the GCF of the following list of terms. Factor out the GCF in each of the following polynomials. a3b2, a2b5 and a4b7 a3b2 = a · a · a · b · b 1) 6 x3 – 9x2 + 12 x = a2b5 = a · a · b · b · b · b · b 3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 = a4b7 = a · a · a · a · b · b · b · b · b · b · b 3x(2 x2 – 3x + 4) 2 2 So the GCF is a · a · b · b = a b 2) 14 x3y + 7 x2y – 7xy = Notice that the GCF of terms containing variables will use the 7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 = smallest exponent found amongst the individual terms for each 7xy (2 x2 + x – 1) variable. Martin-Gay, Developmental Mathematics 9 Martin-Gay, Developmental Mathematics 11 Factoring Polynomials Factoring out the GCF The first step in factoring a polynomial is to Example find the GCF of all its terms. Factor out the GCF in each of the following Then we write the polynomial as a product by polynomials. factoring out the GCF from all the terms. 1) 6( x + 2) – y(x + 2) = The remaining factors in each term will form a 6 · (x + 2) – y · (x + 2) = polynomial. (x + 2) (6 – y) 2) xy (y + 1) – (y + 1) = xy · (y + 1) – 1 · (y + 1) = (y + 1) (xy – 1) Martin-Gay, Developmental Mathematics 10 Martin-Gay, Developmental Mathematics 12 3 2/13/2013 Factoring Factoring Trinomials Remember that factoring out the GCF from the terms of Recall by using the FOIL method that a polynomial should always be the first step in factoring F O I L a polynomial. (x + 2)( x + 4) = x2 + 4x + 2x + 8 This will usually be followed by additional steps in the = x2 + 6x + 8 process. To factor x2 + bx + c into ( x + one #)( x + another #), Example note that b is the sum of the two numbers and c is the Factor 90 + 15 y2 – 18 x – 3xy 2. product of the two numbers. 90 + 15 y2 – 18 x – 3xy 2 = 3(30 + 5 y2 – 6x – xy 2) = So we’ll be looking for 2 numbers whose product is 3( 5 · 6 + 5 · y2 – 6 · x – x · y2) = c and whose sum is b. 3( 5(6 + y2) – x (6 + y2)) = Note: there are fewer choices for the product, so 3(6 + y2)(5 – x) that’s why we start there first. Martin-Gay, Developmental Mathematics 13 Martin-Gay, Developmental Mathematics 15 Factoring Polynomials Example § 13.2 Factor the polynomial x2 + 13 x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of 30 Sum of Factors 1, 30 31 2, 15 17 Factoring Trinomials of the 3, 10 13 2 Note, there are other factors, but once we find a pair Form x + bx + c that works, we do not have to continue searching. So x2 + 13 x + 30 = ( x + 3)( x + 10). Martin-Gay, Developmental Mathematics 16 4 2/13/2013 Factoring Polynomials Prime Polynomials Example Example Factor the polynomial x2 – 11 x + 24. Factor the polynomial x2 – 6x + 10. Since our two numbers must have a product of 24 and a Since our two numbers must have a product of 10 and a sum of -11, the two numbers must both be negative. sum of – 6, the two numbers will have to both be negative. Negative factors of 24 Sum of Factors Negative factors of 10 Sum of Factors – 1, – 24 – 25 – 1, – 10 – 11 – 2, – 12 – 14 – 2, – 5 – 7 – 3, – 8 – 11 Since there is not a factor pair whose sum is – 6, 2 2 x – 6x +10 is not factorable and we call it a prime So x – 11 x + 24 = ( x – 3)( x – 8). polynomial . Martin-Gay, Developmental Mathematics 17 Martin-Gay, Developmental Mathematics 19 Factoring Polynomials Check Your Result! Example You should always check your factoring Factor the polynomial x2 – 2x – 35. results by multiplying the factored polynomial Since our two numbers must have a product of – 35 and a to verify that it is equal to the original sum of – 2, the two numbers will have to have different signs. polynomial. Factors of – 35 Sum of Factors Many times you can detect computational – 1, 35 34 errors or errors in the signs of your numbers 1, – 35 – 34 by checking your results. – 5, 7 2 5, – 7 – 2 So x2 – 2x – 35 = ( x + 5)( x – 7). Martin-Gay, Developmental Mathematics 18 Martin-Gay, Developmental Mathematics 20 5 2/13/2013 Factoring Polynomials Example § 13.3 Factor the polynomial 25 x2 + 20 x + 4. Possible factors of 25 x2 are { x, 25 x} or {5 x, 5 x}. Possible factors of 4 are {1, 4} or {2, 2}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Factoring Trinomials of Keep in mind that, because some of our pairs are not identical 2 factors, we may have to exchange some pairs of factors and the Form ax + bx + c make 2 attempts before we can definitely decide a particular pair of factors will not work. Continued. Martin-Gay, Developmental Mathematics 23 Factoring Trinomials Factoring Polynomials Returning to the FOIL method, Example Continued F O I L We will be looking for a combination that gives the sum of the (3x + 2)(x + 4) = 3x2 + 12 x + 2x + 8 products of the outside terms and the inside terms equal to 20 x. = 3x2 + 14 x + 8 Factors Factors Resulting Product of Product of Sum of 2 of 25 x2 of 4 Binomials Outside Terms Inside Terms Products To factor ax + bx + c into (# 1·x + # 2)(# 3·x + # 4), note that a is the product of the two first coefficients, c is {x, 25 x} {1, 4} ( x + 1)( 25 x + 4) 4x 25 x 29 x the product of the two last coefficients and b is the sum of the products of the outside coefficients and (x + 4)( 25 x + 1) x 100 x 101 x inside coefficients. {x, 25 x} {2, 2} ( x + 2)( 25 x + 2) 2x 50 x 52 x Note that b is the sum of 2 products, not just 2 {5 x, 5 x} {2, 2} ( 5x + 2)( 5x + 2) 10 x 10 x 20 x numbers, as in the last section. Continued. Martin-Gay, Developmental Mathematics 22 Martin-Gay, Developmental Mathematics 24 6 2/13/2013 Factoring Polynomials Factoring Polynomials Example Continued Example Continued We will be looking for a combination that gives the sum of Check the resulting factorization using the FOIL method. the products of the outside terms and the inside terms equal F O I L to −41 x. (5 x + 2)(5 x + 2) = 5x(5 x) + 5 x(2) + 2(5 x) + 2(2) Factors Factors Resulting Product of Product of Sum of of 21 x2 of 10 Binomials Outside Terms Inside Terms Products = 25 x2 + 10 x + 10 x + 4 {x, 21 x}{1, 10}( x – 1)( 21 x – 10 ) –10 x −21 x – 31 x = 25 x2 + 20 x + 4 (x – 10 )( 21 x – 1) –x −210 x – 211 x So our final answer when asked to factor 25 x2 + 20 x + 4 {x, 21 x} {2, 5} ( x – 2)( 21 x – 5) –5x −42 x – 47 x will be (5 x + 2)(5 x + 2) or (5 x + 2) 2.
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