Factoring Polynomials Writing a polynomial as a product of polynomials of lower degree is called factoring. Factoring is an important procedure that is often used to simplify fractional expressions and to solve equations. The first step in any factorization of a polynomial is to use the distributive property to factor out the greatest common factor (GCF) of the terms in the polynomial. To find the GCF we will rewrite each term as a product of its factors. For example, if we have the term 27 a3b4 , we factor the coefficient 27 as 3 * 3 * 3 or 33, and can write the term as 33a3b4. For any polynomial, we will factor each term, then we compare the factors for each term looking for the GCF. Let’s try a few examples. Example 1: 10x3 + 6x Factor each term: 2*5*x3 + 2*3x After factoring each term, we see that both terms have 2 and x as factors. We will divide each term by 2x, placing 2x as a factor of the expression. 2x(5x2 + 3) Example 2: (m + 5)(x + 3) + (m + 5)(x – 10) In this example we have the sum of two terms each of which is composed of the product of two binomials. Examine each term to see if the same binomial appears in both. We see that m + 5 is a factor of each term. We will factor this term out of both terms. (m + 5) [(x + 3) + (x – 10)] I use the square brackets to group after removing the GCF m + 5. Now we examine what we have in the square brackets to see if it can be simplified and it can. (m + 5)(2x – 7) Example 3: 10x2y + 6xy –14xy2 We will factor each term: 2*5*x*x*y + 2*3*x*y –2*7*x*y*y Each term has 2xy as a factor, so we factor that out.

2xy(5x + 3 – 7y) FACTORING TRINOMIALS A trinomial is a polynomial with three terms. Some trinomials of the form x2 + bx + c, where b and c are constants, can be factored by a trial process. This method of factoring uses the FOIL method in reverse. For example, consider the following products: (x + 3)(x + 5) = x2 + 5x + 3x + (3)(5) = x2 + 8x + 15 (x - 2)(x - 7) = x2 - 7x - 2x + (-2)(-7) = x2 - 9x + 14 (x + 4)(x - 9) = x2 - 9x + 4x + (4)(-9) = x2 - 5x – 36 Notice that the coefficient of x is the sum of the constant terms of the binomials. Also, the constant term of the trinomial is the product of the constant terms of the binomials. We will use the following rules for factoring a trinomial of the form x2 + bx + c: 1. The constant term c of the trinomial is the product of the constant terms of the binomials. 2. The coefficient b in the trinomial is the sum of the constant terms in the binomials. 3. If the constant term c of the trinomial is positive, the constant terms of the binomials have the same sign as the coefficient of b. 4. If the constant term c of the trinomial is negative, the constant terms of the binomials have opposite signs. Let’s factor x2 + 7x –18 using these rules. By rule 1 we know that –18 is the product of the constant terms of the binomials. By rule 2 we know the b term is the sum of the constant terms of the binomials. We will create a table of the factors of –18, looking for two whose sum is 7. Factors of –18 1 * (-18) (-1) * 18 2 * (-9) (-2) * 9

Sum of the factors -17 17 -7 7

The –2 and 9 are the numbers whose product is –18 and whose sum is 7. We’ve found our constants for the binomials that are the factors of our trinomial. x2 + 7x –18 = (x –2)(x + 9) Use these rules to factor the following in Exercise 1-3: 19, 21, and 25.

This trial method can sometimes be used to factor trinomials of the form ax2 + bx + c, which do not have a leading coefficient of 1. Points to remember to factor x2 + bx + c, a > 0 1. If the constant term of the trinomial is positive, the constant terms of the binomials have the same sign as the coefficient of b in the trinomial. 2. If the constant term of the trinomial is negative, the constant terms of the binomials have opposite signs. 3. If the terms of the trinomial do not have a common factor, then neither binomial will have a common factor. Let’s apply these rules to the trinomial 6x2 - 11x + 4 By rule one, since the constant term of the trinomial is positive and the coefficient of the x term is negative, the constant terms of the binomials will both be negative. This time we find the factors of the first term as well as factors of the constant term. Factors of 6x2 x, 6x 2x, 3x

Factors of 4 (Both negative) -1, -4 -2, -2

Use these factors to write trial factors. Use the FOIL method to see whether any of the trial factors produce the correct middle term. If the terms of a trinomial do not have a common factor, then a binomial factor cannot have a common factor (rule 3). Such trial factors need not be checked. Trial factors (x –1)(6x – 4) (x – 4)(6x – 1) (x – 2)(6x – 2) (2x – 1)(3x – 4)

Middle term Common factor -1x – 24x = -25x Common factor -8x – 3x = -11x

This last is the correct middle term. So 6x2 - 11x + 4 = (2x – 1)(3x – 4). Try this these problems in Exercise 1-3: 5, 9, 11, and 17. Sometimes it is impossible to factor a polynomial into the product of two polynomials having integer coefficients. Such polynomials are said to be non-factorable over the integers. For example, x2 + 3x + 7 is non-factorable over the integers because there are no integers whose product is 7 and whose sum is 3. If you have difficulty factoring a trinomial, you may wish to use the following factorization theorem. It will indicate whether the trinomial is factorable over the integers.

FACTORIZATION THEOREM: The trinomial ax2 + bx + c, with integer coefficients a, b and c, can be factored as the product of two binomials with integer coefficients if and only if b2 – 4ac is a perfect square. This should save you quite a bit of effort, not to mention frustration, when factoring trinomials. Self-review exercise: Determine if the following trinomials are factorable over the integers. If it is factorable, factor it. Answers will be posted in the mailing list on Friday of week one. 1. 2. 3. 4. 5.

4x2 + 8x –7 6x2 - 5x – 4 8x2 + 26x + 15 4x2 - 5x + 6 6x2 - 14x + 5

SPECIAL FACTORING FORMULAS Special Form Difference of two squares Perfect-square trinomials Sum of cubes Difference of cubes

Formula(s) x2 – y2 = (x – y)(x + y) x2 + 2xy + y2 = (x + y)2 x2 - 2xy + y2 = (x - y)2 x3 + y3 = (x + y)( x2 – xy + y2) x3 - y3 = (x - y)( x2 + xy + y2)

Let’s apply these factoring formulas. Factor: 49x2 –144. Recognize the difference of squares form since the square root of 49x2 is 7x and the square root of 144 is 12. The factors of 49x2 –144 are (7x – 12)(7x + 12). A perfect-square trinomial is a trinomial that is the square of a binomial. For example, x2 + 6x +9 is a perfect-square trinomial because (x + 3)2 = x2 + 6x + 9. Every perfect-square trinomial can be factored by the trial method, but it is generally faster to factor perfect-square trinomials by using the factoring formula. Factor: 16m2 –40mn +25n2. First since the last term is positive and the middle term is negative, the binomials second terms will be negative. We recognize that the first term and last term are perfect squares since 16m2 = (4m)2 and 25n2 = (5n)2. Is twice the product of the square roots of those terms equal to the middle term? Is 2(4m * 5n) = 40mn? Yes, it is. Our factors are: 16m2 –40mn +25n2 = (4m – 5n)(4m – 5n) = (4m – 5n)2

Factor: 8a3 + b3 Because our variables are cubed, we will check to see if the rules for cubes will apply. Since the operator is a +, we will use the sum of cubes rule. The cube root of 8a3 is 2a. The cube root of b3 is b. 8a3 + b3 = (2a + b)( 4a2 - 2ab + b2) Factor: a3 – 64 Because our variables are cubed, we will check to see if the rules for cubes will apply. Since the operator is a -, we will use the difference of cubes rule. The cube root of a3 is a. The cube root of 64 is 4. a3 – 64 = (a – 4)(a2 + 4a + 16) FACTOR BY GROUPING Some polynomials can be factored by grouping. Pairs of terms that have common factors are first grouped together. The process makes repeated use of the distributive property as shown in the following factorization of 6y3 –21y2 –4y + 14. 6y3 –21y2 –4y + 14 = (6y3 –21y2) – (4y – 14) Group the first two terms and the last two terms because they have common factors. Since we placed the minus sign outside the parenthesis, we changed each sign in the parenthesis. = 3y2(2y – 7) – 2(2y – 7) Factor out the GCF from each group. = (2y – 7)[3y2 – 2] Factor out the GCF (2y – 7). When you factor by grouping, some experimentation may be necessary to find a grouping that is of the form of one of the special factoring formulas. Factor by grouping: a2 + 10ab + 25b2 – c2 If we group the first two and the last two terms, we have a common factor of a in the first group and no common factor for the second group. The second group is the difference of two squares. That grouping won’t provide us with a common factor for the two groupings. If we group the first three terms we have a perfect-square trinomial. a2 + 10ab + 25b2 – c2 = (a2 + 10ab + 25b2)– c2 = (a + 5b)2 – c2 = m2 – c2 = (m – c)(m + c) = (a + 5b –c)(a + 5b + c)

Factor the first group Now we have the difference of two squares. Let m = a + 5b Apply the difference of squares rule Substitute a + 5b for m.

GENERAL FACTORING Here is a general factoring strategy for polynomials: 1. Factor out the GCF of all terms. 2. Try to factor a binomial as a. the difference of two squares b. the sum or difference of two cubes 3. Try to factor a trinomial a. as a perfect-square trinomial. b. using the trial method. 4. Try to factor a polynomial with more than three terms by grouping. 5. After each factorization, examine the new factors to see whether they can be factored.