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2. 2 6.3 More About Factoring Trinomials

3. 3 What You Will Learn  Factor trinomials of the form ax 2 + bx + c  Factoring trinomials completely  Factor trinomials by grouping

4. 4 Factoring Trinomials of the Form ax 2 + bx + c

5. 5 In this section, you will learn how to factor trinomials whose leading coefficients are not 1. To see how this works, consider the following. The goal is to find a combination of factors of a and c such that the outer and inner products add up to the middle term bx.

6. 6 Example 1 – Factoring a Trinomial Factor the trinomial 4x 2 – 4x – 3. Solution: First, observe that 4x 2 – 4x – 3 has no common monomial factor. For this trinomial, a = 4 and c = –3. You need to find a combination of the factors of 4 and –3 such that the outer and inner products add up to –4x.

7. 7 Example 1 – Factoring a Trinomial The possible combinations are as follows. Factors O + I –3x + 4x = x (x – 1)(4x + 3) 3x – 4x = –x (x + 3)(4x – 1) –x + 12x = 11x x does not equal –4x. –x does not equal –4x. 11x does not equal –4x. cont’d

8. 8 Example 1 – Factoring a Trinomial cont’d (x – 3)(4x + 1) x – 12x = –11x (2x + 1)(2x – 3) –6x + 2x = –4x (2x – 1)(2x + 3) 6x – 2x = 4x So, the correct factorization is 4x 2 – 4x – 3 = (2x + 1)(2x – 3). –11x does not equal –4x. –4x equals –4x. 4x does not equal –4x.

9. 9 Factoring Trinomials of the Form ax 2 + bx + c The following guidelines can help shorten the list of possible factorizations.

10. 10 Example 3 – Factoring a Trinomial Factor the trinomial 6x 2 + 5x – 4. Solution: First, observe that 6x 2 + 5x – 4 has no common monomial factor. For this trinomial, a = 6 and c = –4. You need to find a combination of the factors of 6 and –4 such that the outer and inner products add up to 5x. FactorsO + I (x + 1)(6x – 4) –4x + 6x = 2x (x – 1)(6x + 4) 4x – 6x = –2x

11. 11 Example 3 – Factoring a Trinomial (x + 4)(6x – 1) –x + 24x = 23x (x – 4)(6x + 1) x – 24x = –23x (x + 2)(6x – 2) –2x + 12x = 10x (x – 2)(6x + 2) 2x – 12x = –10x (2x + 1)(3x – 4) –8x + 3x = –5x (2x – 1)(3x + 4) 8x – 3x = 5x (2x + 4)(3x – 1) –2x + 12x = 10x (2x – 4)(3x + 1) 2x – 12x = –10x 23x does not equal 5x. 5x equals 5x. –23x does not equal 5x. 10x does not equal 5x. –10x does not equal 5x. –5x does not equal 5x. 10x does not equal 5x. –10x does not equal 5x. cont’d

12. 12 Example 3 – Factoring a Trinomial cont’d (2x + 2)(3x – 2) –4x + 6x = 2x (2x – 2)(3x + 2) 4x – 6x = –2x So, the correct factorization is 6x 2 + 5x – 4 = (2x – 1)(3x + 4). 2x does not equal 5x. –2x does not equal 5x.

13. 13 Factoring Trinomials of the Form ax 2 + bx + c Using these guidelines, you can shorten the list in Example 3 to the following. (x + 4)(6x – 1) = 6x 2 + 23x – 4 (2x + 1)(3x – 4) = 6x 2 – 5x – 4 (2x – 1)(3x + 4) = 6x 2 + 5x – 4 23x does not equal 5x. Opposite sign Correct factorization

14. 14 Factoring Completely

15. 15 Factoring Completely Remember that if a trinomial has a common monomial factor, the common monomial factor should be factored out first. The complete factorization will then show all monomial and binomial factors.

16. 16 Example 4 – Factoring Completely Factor 4x 3 – 30x 2 + 14x completely. Solution: Begin by factoring out the common monomial factor. 4x 3 – 30x 2 + 14x = 2x(2x 2 – 15x + 7) Now, for the new trinomial 2x 2 – 15x + 7, a = 2 and c = 7.

17. 17 Example 4 – Factoring Completely The possible factorizations of this trinomial are as follows. (2x – 7)(x – 1) = 2x 2 – 9x + 7 (2x – 1)(x – 7) = 2x 2 – 15x + 7 So, the complete factorization of the original trinomial is 4x 3 – 30x 2 + 14x = 2x(2x 2 – 15x + 7) = 2x(2x – 1)(x – 7). Correct factorization cont’d

18. 18 Factoring Completely In factoring a trinomial with a negative leading coefficient, first factor –1 out of the trinomial, as demonstrated in Example 5.

19. 19 Example 5 – A Negative Leading Coefficient Factor the trinomial –5x 2 + 7x + 6. Solution: This trinomial has a negative leading coefficient, so you should begin by factoring –1 out of the trinomial. –5x 2 + 7x + 6 = (–1)(5x 2 – 7x – 6) Now, for the new trinomial 5x 2 – 7x – 6, you have a = 5 and c = –6.

20. 20 Example 5 – A Negative Leading Coefficient After testing the possible factorizations, you can conclude that (x – 2)(5x + 3) = 5x 2 – 7x – 6. So, a correct factorization is –5x 2 + 7x + 6 = (–1)(x – 2)(5x + 3) = (–x + 2)(5x + 3). Another correct factorization is (x – 2)(–5x – 3). Correct factorization Distributive Property cont’d

21. 21 Example 6 – Geometry: The Dimensions of a Sandbox The sandbox shown in the figure below has a height of x feet and a width of (x + 2) feet. The volume of the sandbox is (2x 3 + 7x 2 + 6x) cubic feet. a.Find the length of the sandbox. b.How many cubic hards of the sand are needed to fill the sandbox when x = 2?

22. 22 Example 6 – Geometry: The Dimensions of a Sandbox Solution a. Verbal Model: Labels:Volume = 2x 3 + 7x 2 + 6x (cubic feet) Width = x + 2 (feet) Height = x (feet) Equation: 2x 3 + 7x 2 + 6x= x(2x 2 + 7x + 6) = x(x + 2)(2x + 3) So, the length of the sandbox is (2x + 3) feet. cont’d

23. 23 Example 6 – Geometry: The Dimensions of a Sandbox b. When x = 2, the volume of the box is Volume= 2(2 3 ) + 7(2 2 ) + 6(2) = 2(8) + 7(4) + 12 = 56 ft 3. One cubic yard contains 27 cubic feet. So, you need 56/27 or about 2 cubic yards of sand to fill the sandbox. cont’d

24. 24 Factoring Trinomials by Grouping

25. 25 Factoring Trinomials by Grouping The examples in this section have shown how to use the guess, check, and revise strategy to factor trinomials. An alternative technique to use is factoring by grouping. Recall that the polynomial x 3 + 2x 2 + 3x + 6 was factored by first grouping terms and then applying the Distributive Property. x 3 + 2x 2 + 3x + 6 = (x 3 + 2x 2 ) + (3x + 6) = x 2 (x + 2) + 3(x + 2) = (x + 2)(x 2 + 3) Group terms. Factor out common monomial factor in each group. Distributive Property

26. 26 Factoring Trinomials by Grouping By rewriting the middle term of the trinomial as 2x 2 + x – 15 = 2x 2 + 6x – 5x – 15 you can group the first two terms and the last two terms and factor the trinomial as follows. 2x 2 + x – 15 = 2x 2 + 6x – 5x – 15 = (2x 2 + 6x) + (–5x – 15) = 2x(x + 3) – 5(x + 3) = (x + 3)(2x – 5) Group terms. Factor out common monomial factor in each group. Distributive Property Rewrite middle term.

27. 27 Factoring Trinomials by Grouping

28. 28 Example 7 – Factoring Trinomials by Grouping Use factoring by grouping to factor the trinomial 2x 2 + 5x – 3. Solution: In the trinomial 2x 2 + 5x – 3, a = 2 and c = –3, which implies that the product ac is –6. Now, because –6 factors as (6)(–1), and 6 – 1 = 5 = b, you can rewrite the middle term as 5x = 6x – x. This produces the following. 2x 2 + 5x – 3 = 2x 2 + 6x – x – 3 = (2x 2 + 6x) + (–x – 3) Group terms. Rewrite middle term.

29. 29 Example 7 – Factoring Trinomials by Grouping = 2x(x + 3) – (x + 3) = (x + 3)(2x – 1) So, the trinomial factors as 2x 2 + 5x – 3 = (x + 3)(2x – 1). Factor out common monomial factor in each group. Factor out common binomial factor. cont’d

30. 30 Example 9 – Geometry: The Dimensions of a Swimming Pool The swimming pool shown in the figure below has a depth of d feet and a length of (5d + 2) feet. The volume of the swimming pool is (15d 3 – 14d 2 – 8d) cubic feet. a.Find the width of the swimming pool. b.How many cubic feet of water are needed to fill the swimming pool when d = 6?

31. 31 Solution a. Verbal Model: Labels:Volume = 15d 3 – 14d 2 – 8d (cubic feet) Width = 5d + 2 (feet) Depth = d (feet) Equation:15d 3 – 14d 2 – 8d = d(15d 2 – 14d – 8) = d(15d 2 + 6d – 20d – 8) = d[3d(5d + 2) – 4(5d + 2)] = d(5d + 2)(3d – 4) So, the length of the sandbox is (3d – 4) feet. cont’d Example 9 – Geometry: The Dimensions of a Swimming Pool

32. 32 b. When d = 6, the dimensions of the swimming pool are 6 feet by 32 feet by 14 feet. So, the volume is 6(32)(14) = 2688 cubic feet. There are about 7.5 gallons in 1 cubic feet. So, the swimming pool holds about 20,160 gallons of water. cont’d Example 9 – Geometry: The Dimensions of a Swimming Pool