Exponents, Polynomials and Functions 3 Copyright © Cengage Learning. All rights reserved.

3.4 Factoring Polynomials Copyright © Cengage Learning. All rights reserved.

Objectives Factor out the GCF. Factor by grouping. Factor quadratics using the AC method. Factor using trial and error.

Factoring Out the GCF

Factoring Out the GCF We will multiply the two polynomials using the distributive property. a(x + y) = ax + ay When working with polynomials, we sometimes want to “undo” the distributive property by changing a polynomial into a product of two or more factors. The process of changing a polynomial from terms that are added together to factors that are multiplied together is called factoring.

Factoring Out the GCF Factoring is one method that is used to solve some polynomial equations, and it has several other uses in algebra. The basis of factoring is to take a polynomial and rewrite it into simpler parts that are multiplied together. Standard form Factored form x2 + 8x + 15 (x + 3)(x + 5)

Factoring Out the GCF When the factored form is multiplied out, the expression will return to the standard form. (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15 The most basic step in any factoring process is to factor out the common elements from the terms of an expression. When we factor out a common element, it does not go away, but it becomes a factor that is in front of the remaining expression.

Factoring Out the GCF The common elements of an expression are called a common factor. The terms of the expression 12x + 10 both are divisible by 2. Therefore, a 2 can be factored out of the expression and be put in front as a factor. 12x + 10 = 2(6x + 5) If we multiply the 2 back into the expression 6x + 5, we would be right back to the original expression,12x + 10.

Factoring Out the GCF The terms in some expressions will have constants, variables, or both variables and constants in common. The greatest common factor (GCF) is the largest common factor shared in common with all the terms of the polynomial. When factoring, we will most often factor out the GCF.

Example 1 – Factoring out the greatest common factor Factor out the greatest common factor. a. 5x2 + 20x b. 15w3z – 9w2z2 c. 2x2 + 6x – 4 d. 5x(x – 8) + 2(x – 8) e. 3z(z – 5) – (z – 5)

Example 1 – Solution a. Both terms in the expression 5x2 + 20x have at least one x and also are divisible by 5, so we can factor out 5x from both terms. 5x(x + 4) The first term of this expression had x2, but one is factored out, leaving a single x. b. In the expression 15w3z – 9w2z2, both of the terms have a w2, have a z, and are divisible by 3. Factoring 3w2z out, we have 3w2z(5w – 3z)

Example 1 – Solution cont’d c. All of the terms in the polynomial 2x2 + 6x – 4 are divisible by 2. Because the last term does not have an x in it, x is not a common factor, 2(x2 + 3x – 2) d. In the expression 5x(x – 8) + 2(x – 8), both terms have the expression (x – 8). (x – 8)(5x + 2)

Example 1 – Solution cont’d Both terms in the expression 3z(z – 5) – (z – 5) have the expression (z – 5). Remember that the expression (z – 5) has a “hidden” coefficient of –1. When we factor out the (z – 5), a – 1 will remain in its place. (z – 5)(3z – 1)

Factoring by Grouping

Factoring by Grouping Factoring out what is in common will always be the first step in any factoring process. In many cases, there will not be anything in common, and we will move on to other steps. If a polynomial has four terms, we try a factoring technique called factoring by grouping. We will group the first two terms together and factor out the GCF of those two terms. Also we group the last two terms and factor out the GCF.

Factoring by Grouping We finish by factoring out the GCF from the remaining expressions.

Example 2 – Factoring by grouping Factor by grouping. a. 2x2 – 8x + 7x – 28 b. 6a2 + 8ab + 15a + 20b

Example 2(a) – Solution There are four terms in the expression 2x2 – 8x + 7x – 28, so we will factor by grouping. There are no common factors for all four terms. Group the first two terms and the last two terms, then factor out the GCFs. 2x2 – 8x + 7x – 28 = (2x2 – 8x) + (7x – 28) = 2x(x – 4) + 7(x – 4) = (x – 4)(2x + 7) Group the first two and last two terms Factor out the GCFs 2x and 7. Factor out the GCF (x – 4)

Example 2(a) – Solution cont’d To check the factorization, we can multiply out the factored form using the distributive property. (x – 4)(2x + 7) = 2x2 – 8x + 7x – 28 We got the original expression back, so the factored form we found is correct.

Example 2(b) – Solution cont’d There are four terms in the expression 6a2 + 8ab + 15a + 20b, so we will factor by grouping. There are no common factors for all four terms. Group the first two terms and the last two terms, then factor out the GCFs. 6a2 + 8ab + 15a + 20b = (6a2 + 8ab + 15a + 20b) = 2a(3a + 4b + 5(3a + 4b) = (3a + 4b)(2a + 5) Group the first two and last two terms. Factor out the GCFs 2a and 5. Factor out the GCF (3a + 4b).

Example 2(b) – Solution cont’d To check the factorization, we can multiply out the factored form using the distributive property. (3a + 4b)(2a + 5) = 6a2 + 8ab + 15a + 20b We got the original expression back, so the factored form we found is correct.

Factoring Using the AC Method

Factoring Using the AC Method One of the most common polynomial functions that we will work with is called a quadratic. Quadratic functions are second-degree polynomials of the form f (x) = ax2 + bx + c where a, b, and c are real numbers and a ≠ 0.

Factoring Using the AC Method Many methods are used to factor quadratics, but we will concentrate on one called the AC method. In working with a quadratic in standard form, this method will provide basic steps that will guide us through the factoring process.

Example 3 – Factor using the AC method Factor the following. a. x2 + 8x + 15 b. 6x2 + x – 35 c. 4x2 + 7xy + 3y2

Example 3(a) – Solution The quadratic x2 + 8x + 15 is in standard form ax2 + bx + c, so we will use the AC method. Step 1: Factor out the greatest common factor. The terms in this quadratic have no common factors. Step 2: Multiply a and c together. (Do this step off to the side, in the margin.) a = 1 and c = 15 so ac = 1(15) = 15

Example 3(a) – Solution cont’d Step 3: Find factors of ac that sum to b. (Do this off to the side also.) List both the positive and negative factors of 15. ac = 1(15) = 15 b = 8 1 15 1 + 15 = 16 3 5 3 + 5 = 8 –1 –15 –1 + (–15) = –16 –3 –5 –3 + (–5) = –8 In this list, the factors 3 and 5 will add up to 8. Not equal to b Equal to b Not equal to b Not equal to b

Example 3(a) – Solution cont’d Step 4: Rewrite the middle (bx) term, using the factors from step 3. The factors 3 and 5, found in the previous step, will be used to rewrite the bx term in the expression. x2 + 3x + 5x + 15 Step 5: Group and factor out what is in common. If we group the first two terms together and the last two terms together.

Example 3(a) – Solution We can factor out some common factors. cont’d We can factor out some common factors. x2 + 3x + 5x + 15 = (x2 + 3x) + (5x + 15) = x(x + 3) + 5(x + 3) = (x + 3)(x + 5) Group first and last two terms together. Factor out x from the first group and 5 from the second group. Factor out the (x + 3).

Example 3(a) – Solution cont’d To check the factorization, we can multiply out the factored form using the distributive property. (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15 We got the original expression back, so the factored form we found is correct.

Example 3(b)– Solution cont’d The quadratic 6x2 + x – 35 is in standard form ax2 + bx + c, so we will use the AC method. Step 1: Factor out the greatest common factor. The terms in this quadratic have nothing in common. Step 2: Multiply a and c together. (Do this step off to the side, in the margin.) a = 6 and c = –35 so ac = 6(–35) = – 210.

Example 3(b) – Solution cont’d Step 3: Find factors of ac that sum to b. (Do this off to the side also.) List the factors of –210. Once we have listed the factors of –120, then we will switch the signs of the factors to come up with the entire list of factors for –120. The product must be negative, so one factor must be positive and the other must be negative. Be sure to switch the signs of the factors to find all the factorizations of –120.

Example 3(b) – Solution cont’d In this list, the factors –14 and 15 will add up to 1. Step 4: Rewrite the middle (bx) term, using the factors from step 3. The factors –14 and 15 will replace the b value in the expression. 6x2 –14x + 15x – 35

Example 3(b) – Solution cont’d Step 5: Group and factor out what is in common. If we group the first two terms together and the last two terms together, we can factor out some common factors. 6x2 – 14x + 15x – 35 = (6x2 – 14x) + (15x – 35) = 2x(3x – 7) + 5(3x – 7) = (3x – 7)(2x + 5) Group the first and last two terms together. Factor out2x from the first group and 5 from the second group. Factor out the (3x – 7).

Example 3(b) – Solution cont’d We will check this factorization by multiplying the factored form out using the distributive property. (3x – 7)(2x + 5) = 6x2 + 15x – 14x – 35 = 6x2 + x – 35 This is the same expression that we started with, so the factored form is correct.

Example 3(c) – Solution cont’d The polynomial 4x2 + 7xy + 3y2 has two variables, but each variable has a squared term and a first-degree term. We will factor it using the AC method focusing on the x variables, and the y variables will follow along the process. Step 1: Factor out the greatest common factor. The terms in this polynomial have nothing in common.

Example 3(c) – Solution cont’d Step 2: Multiply a and c together. (Do this step off to the side, in the margin.) a = 4 and c = 3 so ac = 4(3) = 12 Step 3: Find factors of ac that sum to b. (Do this off to the side also.) List the factors of 12. 12 12 1 12 –1 –12 2 6 –2 –6 3 4 –3 –4 In this list, the factors 3 and 4 will add up to 7.

Example 3(c) – Solution cont’d Step 4: Rewrite the middle (bx) term, using the factors from step 3. The factors 3 and 4 will replace the b value in the expression. 4x2 + 4xy + 3xy + 3y2 Step 5: Group and factor out what is in common. If we group the first two terms together and the last two terms together, we can factor out some common factors. 4x2 + 4xy + 3xy + 3y2 = (4x2 + 4xy) + (3xy + 3y2) Group the first two terms and the last two terms. Factor out 4x from the first group and 3y from the second group. Factor out the (x + y)

Example 3(c) – Solution = 4x(x + y) + 3y(x + y ) = (x + y)(4x + 3y) cont’d = 4x(x + y) + 3y(x + y ) = (x + y)(4x + 3y) We will check this factorization by multiplying the factored form out using the distributive property. (x + y)(4x + 3y) = 4x2 + 3xy + 4xy + 3y2 = 4x2 + 7xy + 3y2 This is the same expression that we started with, so the factored form is correct.

Factoring Using Trial and Error

Factoring Using Trial and Error Another method used to factor quadratics and other polynomials is trial and error. This method uses our knowledge of multiplying and adding numbers to help us guess the factors of the quadratic. The first step of this method will also be to take out any common factors. After factoring out anything common, we will use factors of a and c to try to find the factored form of the expression.

Factoring Using Trial and Error If a = 1 in the standard form, we will have to consider only the factors of c. a = 1 x2 + 5x + 6 (x + factor of c)(x + factor of c) (x + 2)(x + 3) If a does not equal 1, we will have to consider both the factors of a and c. a ≠ 1 10x2 + 33x + 20 (factor of a  x + 1 factor of c)(factor of a  x + factor of c) (2x + 5)(5x + 4)

Factoring Using Trial and Error The signs of the factors are another aspect of the factored form that we will have to be cautious about. If both of the terms in the original polynomial expression have plus signs, we will have two plus signs in the factored form. If the polynomial has one or more negative signs, we will have at least one negative sign in the factored form. x2 + 9x + 20 x2 + 2x – 8 x2 – 3x + 2 x2 – 4x – 5 (x + 5)(x + 4) (x + 4)(x – 2) (x – 2)(x – 1) (x – 5)(x + 1) It is best not to attempt to memorize every situation. Learn to try something, check it, and try again if it does not work.

Example 6 – Factor using trial and error Factor the following using trial and error. a. x2 + 7x + 6 b. x2 – 4x – 12 c. 2x2 + 9x + 4 d. 24x2 – 2x – 15

Example 6(a) – Solution cont’d Because a is 1 in the expression x2 + 7x + 6, we have to consider only the factors of c. The factors of 6 are 1 and 6 or 2 and 3. Because 1 and 6 add up to 7, the factorization is x2 + 7x + 6 = (x + 6)(x + 1) We check this by multiplying the factored form out. (x – 6)(x + 1) = x2 + x + 6x + 6 = x2 + 7x + 6

Example 6(b) – Solution cont’d Again, a is 1 in the expression x2 – 4x – 12, so we consider only the factors of c. The factors of 12 are 1 and 12, 2 and 6, or 3 and 4. Because we have negatives in the original expression, we will have one or more negatives in the factored form. The factors 2 and 6 can make –4 if the 2 is positive and the 6 is negative.

Example 6(b) – Solution This gives us the factored form. cont’d This gives us the factored form. x2 – 4x – 12 = (x – 6)(x + 2) We check this factored form by multiplying it out. (x – 6)(x + 2) = x2 + 2x – 6x – 12 = x2 – 4x – 12

Example 6(c) – Solution cont’d Because a is not 1 in the expression 2x2 + 9x + 4, we have to consider both the factors of a and c. The factors of 2 are 1 and 2, and the factors of 4 are 1 and 4 or 2 and 2. Using these, we find the factored form. 2x2 + 9x + 4 = (2x + 1)(x + 4) We check this factored form by multiplying it out. (2x + 1)(x + 4) = 2x2 + 8x + x + 4 = 2x2 + 9x + 4

Example 6(d) – Solution cont’d Again, a is not 1 in the expression 24x2 – 2x – 15, so we have to consider the factors of a and c. The factors of 24 are 1 and 24, 2 and 12, 3 and 8, and 4 and 6. The factors of 15 are 1 and 15 or 3 and 5. The negatives in the original expression mean that we will have one or more negatives in the factored form. 24x2 – 2x – 15 Try 1 Try 2 (2x + 5)(12x – 3) (3x + 5)(8x – 3)

Example 6(d) – Solution 24x2 – 6x + 60x – 15 24x2 – 9x + 40x – 15 cont’d 24x2 – 6x + 60x – 15 24x2 – 9x + 40x – 15 24x2 + 54x – 15 24x2 + 31x – 15 Wrong middle term. Wrong middle term. Try 3 Try 4 (4x + 1)(6x – 15) (4x + 3)(6x – 5) 24x2 – 60x + 6x – 15 24x2 – 20x + 18x – 15 24x2 – 54x – 15 24x2 – 2x – 15 Wrong middle term. This one works!

Prime Polynomials

Prime Polynomials Some polynomials will not be able to be factored by using rational numbers. When a polynomial is not factorable, it is called prime. When using the AC method, we know that a polynomial is not factorable if in step 3 there is no combination of factors for ac that add up to b. If this happens, we will stop and say that the polynomial is prime.

Prime Polynomials When factoring by trial and error, we must try every combination of factors of a and c to see whether any of the combinations work. If no combination of factors and positive and negative signs work, we will say the polynomial is not factorable over the rational numbers. 6x2 + 10x + 5 6(5) = 30 1 30 –1 – 30 2 15 –2 –15 3 10 –3 –10 5 6 –5 –6

Prime Polynomials None of the factors of ac add up to b, so this quadratic is not factorable over the rational numbers. This is a prime polynomial.