# FACTORING TRINOMIALS IN THE FORM OF ax 2 + bx + c

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1 Tallahassee Community College 55 FACTORING TRINOMIALS IN THE FORM OF ax 2 + bx + c This kind of trinomial differs from the previous kind we have factored because the coefficient of x is no longer "1". EXAMPLES of trinomials in the form of 6a 2 + 7a 24 2 ax + bx + c : a = 6 b = 7 c = 24 3p 2 16p + 5 a = 3 b = 16 c = 5 9x x + 4 a = 9 b = 12 c = 4 NOTE that in each case the coefficient of the first term is not the number one, nor can we make it "1" by factoring anything out. In order to factor these trinomials, we must consider the factors of the coefficient of the first term of the trinomial as well as the factors of the third term of the trinomial. We still use the signs to help us as much as possible. The following factoring patterns are possible depending upon the signs of the trinomials. ax 2 + bx + c = ( x + )( x + ) ax 2 bx + c = ( x )( x ) ax 2 + bx c = ( x + )( x ) or ( x )( x + ) ax 2 bx c = ( x + )( x ) or ( x )( x + ) NOTE that we have two alternative factoring patterns when the signs of the binomials are different. Because we have to consider two different groups of factors, it will make a difference which one is positive and which one is negative.

2 EXAMPLE: Factor 6a 2 + 7a The terms of the trinomial do not share a common factor. (Nothing can be factored out.) 2. The third term of the trinomial is negative so the binomials will have different signs (one + and one ). 3. The first term of the trinomial (6a 2 ) comes from the F of FOIL and the third term of the trinomial ( 24) comes from the L of FOIL. 4. We must consider all of the factors of 6 and 24. Possibilities for 6 Possibilities for 24 1, 6 1, 24 2, 3 1, 24 2, 12 2, 12 3, 8 3, 8 4, 6 4, 6 We must now go through a trial and error process of trying each of the pairs of factors of 24 with each pair of factors of 6. Let s will try 2a and 3a first: (2a + 1)(3a 24) or (2a 24)(3a + 1) (2a 1)(3a + 24) or (2a + 24)(3a 1) (2a + 2)(3a 12) or (2a 12)(3a + 2) (2a 2)(3a + 12) or (2a + 12)(3a 2) (2a + 2)(3a 12) or (2a 12)(3a + 2) (2a + 3)(3a 8) or (2a 8)(3a + 3) (2a 3)(3a + 8) or (2a + 8)(3a 3) (2a + 4)(3a 6) or (2a 6)(3a + 4) (2a 4)(3a + 6) or (2a + 6)(3a 4) NOTE that we must be trying each pair of factors in both positions. In the first pairs we had to try 24 with both 2a and 3a and the +1 with both 2a and 3a. The next step is to check the O and I of FOIL to determine which combination of factors (if any) will give us the correct middle term. It looks like we have a lot of work to do but we can use one simple fact to help us cut down on the work. If the terms of the trinomial we are factoring have no common factors there will never be any common factors in the correct pair of binomials. 6a 2 + 7a 24 has no common factors. This means that we can immediately rule out many of the possible combinations. 2

3 trial factors trial factors (2a + 1)(3a 24) or (2a 24)(3a + 1) (2a 1)(3a + 24) or (2a + 24)(3a + 24) (2a + 2)(3a 12) or (2a 12)(3a + 3) (2a + 3)(3a 8) or (2a + 8)(3a 3) (2a 3)(3a + 8) or (2a + 8)(3a 3) (2a + 4)(3a 6) or (2a 6)(3a + 4) (2a 4)(3a + 6) or (2a + 6)(3a 4) By looking for common factors in the binomials, we can quickly rule out all but 2 possibilities: (2a + 3)(3a 8) and (2a 3)(3a + 8) Use the O and I of FOIL to check for the middle term. (2a + 3)(3a 8) (2a 3)(3a + 8) 16a + 9a = 7a 16a 9a = +7a The correct sum. The correct factorization of 6a 2 + 7a 24 is (2a 3)(3a + 8). If none of these combinations had given us the correct middle term, we would have repeated the process using 1a and 6a in place of 2a and 3a. NOTE that each possible combination gave us the correct first and third terms (6a 2 and 24), but only one gave us the correct middle term. FACTOR 3p 2 16p The third term is positive, so the signs of the binomials will be the same. 2. The middle term is negative, so the signs will both be negative. 3. The trinomial contains no common factors. 4. We must consider factors of 3 and negative factors of , 3 1, 5 Because 3 and 5 are both prime numbers we do not have any choice of factors. (3p 1)(p 5) or (3p 5)(p 1) REMEMBER that 5 and 1 must both be tried with 3p and p. 3

4 (3p 1)(p 5) or (3p 5)(p 1) 15p 1p = 16p 3p 5p = 8p Using the O and I of FOIL to check the middle term, we see that the correct factorization is (3p 1)(p 5) FACTOR: 9x x The third term of the trinomial is positive, so the signs of the binomials will be the same. 2. The middle term is positive, so the signs of the binomials will both be positive. 3. We must consider factors of 9 and , 9 1, 4 3, 3 2, 2 (3x + 2)(3x + 2) (3x + 1)(3x + 4) (x + 2)(9x + 2) (x + 1)(9x + 4) or (x + 4)(9x + 1) NOTE that in the first three pairs I had only one position to try for each factor. This is because the pairs were the same (3x and 3x or 2 and 2). Use the O and I of FOIL to check each possibility. (3x + 2)(3x + 2) (3x + 1)(3x + 4) Correct 6x + 6x = 12x 12x + 3x = 15x (x + 2)(9x + 2) (x + 1)(9x + 4) 2x + 18x = 20x 4x + 9x = 13x (x + 4)(9x + 1) x + 36 = 37x FACTOR COMPLETELY: 15b5 115b The terms of the trinomial have a common factor of The third term of the trinomial is positive, so the signs of the binomials will be the same. 4

5 3. The middle term of the trinomial is negative, so the signs of the binomials will both be negative. 4. We must factor out a 5. 5(3b 2 23b + 14) 5. We must consider factors of 3 and negative factors of , 3 1, 14 2, 7 (b 2)(3b 7) or (b 7)(3b 2) 7b 6b = 13b 2b 21b = 23b correct (b 1)(3b 14) or (b 14)(3b 1) 14b 3b = 17b b 42b = 43b Using the O and I of FOIL to check the middle term, we see that the correct factorization is 5(b 7)(3b 2). The GCF must be in the answer. FACTOR COMPLETELY: 24x 3 y+ 14x 2 y 20xy 1. The terms of the trinomial have a common factor of 2xy. 2. The third term of the trinomial is negative, so the signs of the binomials will be different. 3. We must factor out 2xy. 2xy(12x 2 + 7x 10) 4. We must consider factors of 12 and factors of , 12 1, 10 2, 6 1, 10 3, 4 2, 5 2, 5 By inspecting the pairs of factors carefully, we can see that we can rule out 2 and 6, because no matter which factors of 10 we used, we would have a common factor in one of the binomials. In every (2x 2)(6x + 5) or (2x + 5)(6x 2) case one (2x + 2)(6x 5) or (2x 5)(6x + 2) binomial (2x 1)(6x 10) or (2x + 10)(6x 1) has common (2x + 1)(6x 10) or (2x 10)(6x + 1) factors 5

6 Let's try 3 and 4. (3x 2)(4x + 5) or (3x + 5)(4x 2) (3x + 2)(4x 5) or (3x 5)(4x + 2) (3x 1)(4x + 10) or (3x + 10)(4x 1) (3x + 1)(4x 10) or (3x 10)(4x + 1) The only possibilities here are: (3x 2)(4x + 5) (3x + 2)(4x 5) (3x + 10)(4x 1) (3x 10)(4x + 1) The other pairs all have a common factor in one of the binomials. Checking the middle terms, we see that the correct factorization of 12x is (3x 2)(4x + 5), and the complete factorization of 24x 3 y + 14x 2 y 20xy is 2xy(3x 2)(4x + 5). The GCF must be included as part of the answer. If none of these possibilities had worked, we would go through the same procedure with 1 and 12. EXERCISES: Factor Completely: a. 18x 2 9x 5 f. 16x 2 16x 12 b. 2y 2 + 5y 12 g. 18x 2 27xy + 9y 2 c. 24y y + 12 h. 10x x 2 + 2x d. 18y 2 39y + 20 i. 12x 4 y 2 17x 3 y 3 + 6x 2 y 4 e. 8x 2 30x + 25 j. 26y y 24 KEY: a. (3x + 1)(6x 5) f. 4(2x + 1)(2x 3) b. (y + 4)(2y 3) g. 9(2x y)(x y) c. (3y + 4)(8y + 3) h. 2x(x + 1)(5x + 1) d. (3y 4)(6y 5) i. x5y5(3x 2y)(4x 3y) e. (2x 5)(4x 5) j. 2(y + 4)(13y 3) 6

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### When factoring, we look for greatest common factor of each term and reverse the distributive property and take out the GCF. Factoring: reversing the distributive property. The distributive property allows us to do the following: When factoring, we look for greatest common factor of each term and reverse the distributive property

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### EAP/GWL Rev. 1/2011 Page 1 of 5. Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. EAP/GWL Rev. 1/2011 Page 1 of 5 Factoring a polynomial is the process of writing it as the product of two or more polynomial factors. Example: Set the factors of a polynomial equation (as opposed to an 2.6.2 Cramer s Rule Determinants can be used to solve a linear system of equations using Cramer s Rule. Cramer s Rule for Two Equations in Two Variables Given the system This system has the unique solution FACTORING QUADRATIC EQUATIONS Summary 1. Difference of squares... 1 2. Mise en évidence simple... 2 3. compounded factorization... 3 4. Exercises... 7 The goal of this section is to summarize the methods