1. Copyright © Cengage Learning. All rights reserved. Factoring Polynomials and Solving Equations by Factoring 5

2. Copyright © Cengage Learning. All rights reserved. Section 5.3 Factoring Trinomials with a Leading Coefficient of 1

3. 3 Objectives Factor a trinomial of the form ax 2 + bx + c, a = 1, by trial and error. Factor a trinomial containing a negative greatest common factor. Identify a prime trinomial. Factor a polynomial completely. 1 1 2 2 3 3 4 4

4. 4 Objectives Factor a trinomial of the form ax 2 + bx + c, a = 1, by grouping (ac method). Factor a perfect-square trinomial. 5 5 6 6

5. 5 Factoring Trinomials with a Leading Coefficient of 1 We now discuss how to factor trinomials of the form x 2 + bx + c, where the coefficient of x 2 is 1 and there are no common factors.

6. 6 Factor a trinomial of the form ax 2 + bx + c, a = 1 by trial and error 1.

7. 7 Factor a trinomial of the form ax 2 + bx + c, a = 1 by trial and error The product of two binomials is often a trinomial. For example, (x + 3)(x + 3) = x 2 + 6x + 9 and (x – 3y)(x + 4y) = x 2 + xy – 12y 2 For this reason, we should not be surprised that many trinomials factor into the product of two binomials. To develop a method for factoring trinomials, we multiply (x + a) and (x + b).

8. 8 (x + a)(x + b) = x 2 + bx + ax + ab = x 2 + ax + bx + ab = x 2 + (a + b)x + ab From the result, we can see that the first term is the product of x and x. the coefficient of the middle term is the sum of a and b, and the last term is the product of a and b. We can use these facts to factor trinomials with leading coefficients of 1. Use the distributive property. Apply the commutative property of addition. Factor the GCF, x, out of ax + bx. Factor a trinomial of the form ax 2 + bx + c, a = 1 by trial and error

9. 9 Example 1 Factor: x 2 + 5x + 6 Solution: To factor this trinomial, we will write it as the product of two binomials. Since the first term of the trinomial is x 2, the first term of each binomial factor must be x because x  x = x 2. To fill in the following blanks, we must find two integers whose product is +6 and whose sum is +5. x 2 + 5x + 6 = (x ) (x )

10. 10 Example 1 – Solution The positive factorizations of 6 and the sums of the factors are shown in the following table. The last row contains the integers +2 and +3, whose product is +6 and whose sum is +5. So we can fill in the blanks with +2 and +3. x 2 + 5x + 6 = (x + 2)(x + 3) cont’d

11. 11 Example – Solution To check the result, we verify that (x + 2) times (x + 3) is x 2 + 5x + 6. (x + 2)(x + 3) = x 2 + 3x + 2x + 2  3 = x 2 + 5x + 6 cont’d

12. 12 Comment In Example 1, the factors can be written in either order due to the commutative property of multiplication. An equivalent factorization is x 2 + 5x + 6 = (x + 3)(x + 2). Factor a trinomial of the form ax 2 + bx + c, a = 1 by trial and error

13. 13 Factor a trinomial containing a negative greatest common factor 2.

14. 14 Factor a trinomial containing a negative greatest common factor When the coefficient of the first term is –1, we begin by factoring out –1.

15. 15 Example 6 Factor: –x 2 + 2x + 15. Solution: We factor out –1 and then factor the trinomial. –x 2 + 2x + 15 = –(x 2 – 2x – 15) = –(x – 5)(x + 3) Check: –(x – 5)(x + 3) = –(x 2 + 3x – 5x – 15) = –(x 2 – 2x – 15) = –x 2 + 2x + 15 Factor out –1. Factor x 2 – 2x – 15.

16. 16 Identify a prime trinomial 3.

17. 17 Identify a prime trinomial If a trinomial cannot be factored using only rational coefficients, it is called a prime polynomial over the set of rational numbers.

18. 18 Example 7 Factor: x 2 + 2x + 3 Solution: To factor the trinomial, we must find two integers whose product is +3 and whose sum is 2. The possible factorizations of 3 and the sums of the factors are shown in the table.

19. 19 Example 7 – Solution Since two integers whose product is +3 and whose sum is +2 do not exist, x 2 + 2x + 3 cannot be factored. It is a prime trinomial. cont’d

20. 20 Factor a polynomial completely 4.

21. 21 Example Factor: –3ax 2 + 9a – 6ax Solution: We write the trinomial in descending powers of x and factor out the common factor of –3a. –3ax 2 + 9a – 6ax = –3ax 2 – 6ax + 9a = –3a(x 2 + 2x – 3) Then we factor the trinomial x 2 + 2x – 3. –3ax 2 + 9a – 6ax = –3a(x + 3)(x – 1)

22. 22 Example – Solution Check: –3a(x + 3)(x – 1) = –3a(x 2 + 2x – 3) = –3ax 2 – 6ax + 9a = –3ax 2 + 9a – 6ax cont’d

23. 23 Factor a trinomial of the form ax 2 + bx + c, a = 1 by grouping (ac method) 5.

24. 24 Factor a trinomial of the form ax 2 + bx + c, a = 1 by grouping (ac method) An alternate way of factoring trinomials of the form ax 2 + bx + c, a = 1 uses the technique of factoring by grouping, sometimes referred to as the ac method. For example, to factor x 2 + x – 12 by grouping, we proceed as follows: 1. Determine the values of a and c (a = 1 and c = –12) and find ac: (1)(–12) = –12 This number is called the key number.

25. 25 2. Find two factors of the key number –12 whose sum is b = 1. Two such factors are +4 and –3. +4(–3) = –12 and+4 + (–3) = 1 3. Use the factors +4 and –3 as the coefficients of two terms to be placed between x 2 and –12 to replace x. x 2 + x – 12 = x 2 + 4x – 3x – 12 x = +4x – 3x Factor a trinomial of the form ax 2 + bx + c, a = 1 by grouping (ac method)

26. 26 4. Factor the right side of the previous equation by grouping. x 2 + 4x – 3x – 12 = x(x + 4) – 3(x + 4) = (x + 4)(x – 3) Check this factorization by multiplication. Factor x out of (x 2 + 4x) and –3 out of (–3x – 12). Factor out (x + 4). Factor a trinomial of the form ax 2 + bx + c, a = 1 by grouping (ac method)

27. 27 Example Factor y 2 + 7y + 10 by grouping. Solution: We note that this equation is in the form y 2 + by + c, with a = 1, b = 7, and c = 10. First, we determine the key number ac: ac = 1(10) = 10 Then, we find two factors of 10 whose sum is b = 7. Two such factors are +2 and +5.

28. 28 Example 10 – Solution We use these factors as the coefficients of two terms whose sum is 7y. y 2 + 7y + 10 = y 2 + 2y + 5y + 10 Finally, we factor the right side of the previous equation by grouping. y 2 + 2y + 5y + 10 = y(y + 2) + 5(y + 2) = (y + 2)(y + 5) 7y = +2y + 5y Factor out y from (y 2 + 2y) and factor out 5 from (5y + 10). Factor out (y + 2). cont’d

29. 29 Factor a perfect-square trinomial 6.

30. 30 Factor a perfect-square trinomial We have discussed the following special-product relationships used to square binomials. 1. (x + y) 2 = x 2 + 2xy + y 2 2. (x – y) 2 = x 2 – 2xy + y 2 These relationships can be used in reverse order to factor special trinomials called perfect-square trinomials. Perfect-Square Trinomials 1. x 2 + 2xy + y 2 = (x + y) 2 2. x 2 – 2xy + y 2 = (x – y) 2

31. 31 Factor a perfect-square trinomial In words, Formula 1 states that if a trinomial is the square of one quantity, plus twice the product of the two quantities, plus the square of the second quantity, it factors into the square of the sum of the quantities. Formula 2 states that if a trinomial is the square of one quantity, minus twice the product of the two quantities, plus the square of the second quantity, it factors into the square of the difference of the quantities.

32. 32 Factor a perfect-square trinomial The trinomials on the left sides of the previous equations are perfect-square trinomials because they are the results of squaring a binomial. Although we can factor perfect-square trinomials by using the techniques discussed in this section, we usually can factor them by inspecting their terms.

33. 33 Factor a perfect-square trinomial For example, x 2 + 8x + 16 is a perfect-square trinomial, because The first term x 2 is the square of x. The last term 16 is the square of 4. The middle term 8x is twice the product of x and 4. Thus, x 2 + 8x + 16 = x 2 + 2(x)(4) + 4 2 = (x + 4) 2

34. 34 Example Factor: x 2 – 10x + 25 Solution: x 2 – 10x + 25 is a perfect-square trinomial, because The first term x 2 is the square of x. The last term 25 is the square of 5. The middle term –10x is the negative of twice the product of x and 5. Thus, x 2 – 10x + 25 = x 2 – 2(x)(5) + 5 2 = (x – 5) 2