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Factoring ax 2 +bx +c ax 2 +bx +c a is the coefficient of the squared term or the term with a degree of 2. b is the coefficient of the middle term or the.

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Presentation on theme: "Factoring ax 2 +bx +c ax 2 +bx +c a is the coefficient of the squared term or the term with a degree of 2. b is the coefficient of the middle term or the."— Presentation transcript:

1 Factoring ax 2 +bx +c ax 2 +bx +c a is the coefficient of the squared term or the term with a degree of 2. b is the coefficient of the middle term or the term with a degree of 1. c is the constant term or the term with the degree of 0. Example: 2x 2 – 3x + 5 a = 2 b = - 3 c = 5

2 Factoring ax 2 +bx +c 2x 2 + x - 1 (2x )(x ) (2x 1)(x 1) Put in your signs last. The last term is negative. The middle sign is positive. We must have a difference of POSITIVE 1x when the outside and inside multiplications are combined. One sign must be positive and one sign must be negative to produce a negative product. (2x 1)(x 1) The outside multiplication gets the “+” sign since it is 2x It is larger than the inside multiplication of 1x that gets the negative sign to produce a positive sum when the inside and outside multiplications are combined. (2x - 1)(x + 1) Check 2x 2 + x - 1

3 Factoring ax 2 +bx +c 3x 2 +13x - 10 (3x )(x ) 1,10 or 2,5 (3x 5)(x 2) (3x 2)(x 5) Put in your signs last. 3x 2 +13x - 10 (3x - 2)(x + 5) Check 3x 2 +13x-10

4 Factoring ax 2 +bx +c 6x x +12 (3x )(2x ) 1,12 or 2,6 or 4,3 Since the middle term is odd we must multiply two odd numbers to get and odd product. If we add and odd factor (outside) to an even factor (inside), the result is an odd sum in the middle. Therefore we will eliminate the factor pair of 2,6 since both are even numbers will give an even result from both outside and inside multiplications. Try the factor pair: 4,3 (3x 4)(2x 3) Put in your signs last. 6x x + 12 (3x - 4)(2x - 3) Check 6x x + 12

5 Factoring ax 2 +bx +c 10x x – 21 1,10 or 2,5 (5x )(2x ) 1,21 or 7,3 (5x 1)(2x 21) (5x 21)(2x 1) 42-5 (5x 3)(2x 7 ) 35-6 (5x 7)(2x 3) 15-14

6 Factoring ax 2 +bx +c 10x x – 21 1,10 or 2,5 We tried 5x and 2x as the first terms in our binomials and they didn’t work so we go through the same process with 10x and 1x (10x )(1x ) 1,21 or 7,3 (10x 7)(1x 3) 30-7 (10x 3)(1x 7) 70-3 (10x 1)(1x 21 ) (10x 21)(1x 1) 21-10

7 Factoring ax 2 +bx +c Finally we got the correct combination to get a difference of 11 for our middle term. (10x 21)(1x 1) Outside 10x Inside 21x Difference is 11 Put the signs in last. Look at the original problem again. 10x x – 21 Since the middle term is negative 11x, the larger of the inside and outside multiplications must be negative so we put the negative in front of the 21 and the positive in front of the 1. (10x - 21)(1x + 1) Check by multiplying. 10x x – 21


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