Factoring Trinomials

Lesson #13 Factoring (Trinomials)

A. Factoring Trinomials: A trinomial is a polynomial with exactly three terms. These polynomials have a very special form since they are the typical polynomials that come out of the FOIL method for multiplying two binomials. The Key Number Method of factoring applies to any trinomials ax² + bx + c, where a, b, and c are integers and x represents any letter variable or string of variables.

	Key Number Method	Example 1: Factor 6x² + x - 15.
Step 1:	Multiply the coefficients a× c The result is the key number.	The key number is (6)(-15) = -90
Step 2:	Find two factors of the key number whose sum is b (the middle coefficient).	We need factors of 90 that add up to 1 (the middle term is 1x). Since 90 = 1× 90 or 2× 45 or 3× 30 or 5× 18 or 6× 15 or 9× 10, it looks like we need +10 and -9 since they multiply to -90 and add to +1.
Step 3:	Rewrite the trinomial as a four term polynomial, splitting the middle term into two terms, where the coefficients are the factors found in step 2.	We rewrite 6x² + x - 15 as 6x² + 10x - 9x - 15
Step 4:	Finish factoring by grouping.	6x² + 10x - 9x - 15 = (6x² + 10x) + (-9x - 15) = 2x(3x + 5) - 3(3x + 5) = (3x + 5)(2x - 3)
Step 5:	Check by multiplying (use FOIL).	(3x + 5)(2x - 3) = 6x² - 9x + 10x - 15 = 6x² + x - 15

The Key Number Method attempts to find the coefficients that come from the FOIL method of multiplying two binomials. You will need to be careful to choose the correct signs (sometimes more that one factor pair looks good at first). One hint is to start with the larger number in the chosen factor pair and make it the same sign as the middle term. If this number is still smaller than the middle term, then the other factor needs the same sign too. If the larger factor is bigger than the middle term, then the second factor needs the opposite sign to bring the difference back down to the middle term. If no factor pair works to build the middle term, then the polynomial is unfactorable (prime).

Example 2: Factor 4x² - 16x + 15.

The Key Number is 4× 15 = 60. The possible factorings of 60 are 1× 60, 2× 30, 3× 20, 4× 15, 5× 12, and 6× 10. We choose 6× 10 since 6 + 10 is also the middle coefficient 16.

So 4x² - 16x + 15 = 4x² - 10x - 6x + 15

(use both negative 10x and 6x so that the total middle term is -16x). Now factor by grouping:

4x² - 10x - 6x + 15 = 2x(2x - 5) - 3(2x - 5) = (2x - 5)(2x - 3).

B. General Factoring Procedure: Approach every factoring problem with the following order of steps.

1) Look for any common factor, and pull the GCF out front.

2) Count the number of terms in the polynomial.

4 terms: Use the grouping method.

3 terms: Use the Key Number Method to change the polynomial into 4 terms by splitting the middle term.

2 terms: Use a formula or special form to factor (Mat 105 material).

Repeat step 2 for any factor in parentheses that still contains exponents. If a factor with exponents is prime (will not factor), then just leave the prime factor in the answer. The best possible factoring is one that breaks the polynomial down into linear factors (no exponents on any variable).

Example 3: Factor 14x³ - 12x² - 2x.

1) Since 2 divides each coefficient, and x appears in every term, the GCF is 2x. Factor out the GCF to get 14x³ - 12x² - 2x = 2x(7x² - 6x - 1).

2) Inside the parentheses is still a trinomial (3 terms). The Key Number is 7 (7× 1). The only factoring of 7 is 7× 1, and 7 - 1 = 6 matches the middle term.

So write 7x² - 6x - 1 = 7x² - 7x + 1x - 1 and factor by grouping.

= 7x(x - 1) + 1(x - 1)

= (x - 1)(7x + 1)

Now put the two steps together to get the full factoring:

14x³ - 12x² - 2x = 2x(7x² - 6x - 1) = 2x(x - 1)(7x + 1)

Note that the GCF from step 1 remains out in front of the final factoring.

C. Solving Equations by Factoring: As we saw in Lesson #11, we use factoring to solve equations that involve exponents in one or more variable term. Use these steps to approach any "solve" problem that uses exponents in the equation:

1) Move terms (add or subtract) to get 0 alone on one side;

2) Factor the remaining polynomial side;

3) Set each factor = 0, and solve the simpler equations for the unknown variable.

Example 4. Solve y² + 14 = 9y.

Start by subtracting 9y from both sides to make 0 on the right hand side. The equation becomes y² - 9y + 14 = 0 (Note that we order the terms from highest exponent on down to view the polynomial in its usual form). This polynomial is a trinomial where the Key Number is 1× 14 = 14. Since 14 can also be 2× 7 and 2 + 7 = 9, we write the equation as y² - 7y - 2y + 14 = 0. Finish factoring by grouping.

y(y - 7) - 2(y - 7) = 0

(y - 7)(y - 2) = 0

y - 7 = 0 or y - 2 = 0

So y = 7 or y = 2.

Exercises

A. Factor the following trinomials:

1. 2x² + 13x - 24	2. y² - 4y - 12	3. w² - 14w + 40
4. 10x² + 17x + 3	5. 3x² - 2x - 5	6. 3z² + 25z + 8
7. 9x² - 44x - 60	8. 5s² + 12s - 9	9. 8x² + 2x - 21
10. 12c² - 5c - 2	11. 4x² - 16x + 15	12. u² - 9u + 14
13. x² + 5x + 4	14. x² - 2x -3	15. x² - 7x + 12
16. y² + 8y - 20	17. 2x² - 9x - 5	18. 6a² + a + 1
19. 10x² + 23x - 14	20. 4b² - b -14	21. x² - 2x - 48
22. a² - 15a + 54	23. x² - 11x - 12	24. 8x² + 29x - 12
25. 10x² - 29x + 12	26. 4y² + 12y + 9	27. b² - b - 6
28. 2a² + 17a + 30	29. x² - 6x + 9	30. a² + 4a - 21

B. Factor completely (look for a GCF first):

31. 12a³ + a² - 20a	32. 12x² - 26x + 10	33. 3a³b² - 12a²b² - 15ab²
34. 2x³ + 10x² - 48x	35. 3x³ - 9x² - 12x	36. 36rt² - 42rt - 30r

C. Solve each equation:

37. x² - x - 6 = 0	38. x² + 4x + 3 = 0	39. x² - 5x - 6 = 0
40. y² - 8y + 12 = 0	41. w² + 10w - 24 = 0	42. x² = 7x + 18
43. 2x² + 7x = 4	44. 5w² = 9w + 18	45. 1 + 3x = 4x²

Check the answers to Lesson 13 exercises

Go on to Lesson 14

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