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Factoring ax 2 +bx +c

Factoring ax 2 +bx +c. a x 2 + b x + 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: 2 x 2 – 3 x + 5 a = 2 b = - 3

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Factoring ax 2 +bx +c

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  1. Factoring ax2+bx +c • ax2+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: • 2x2 –3x + 5 • a = 2 • b = - 3 • c = 5

  2. Factoring ax2+bx +c • 2x2 + 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 ofPOSITIVE 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 • 2x2 +x - 1

  3. Factoring ax2+bx +c • 3x2+13x - 10 • (3x )(x ) • 1,10 or 2,5 • (3x 5)(x 2) • (3x 2)(x 5) • Put in your signs last. • 3x2+13x - 10 • (3x - 2)(x + 5) • Check • 3x2+13x-10

  4. Factoring ax2+bx +c • 6x2 - 17x +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 • (3x4)(2x3) Put in your signs last. • 6x2 - 17x +12 • (3x - 4)(2x - 3) • Check • 6x2 - 17x+12

  5. Factoring ax2+bx +c 10x2 - 11x – 21 1,10 or 2,5 • (5x )(2x ) • 1,21 or 7,3 • (5x 1)(2x 21) • 105-2 • (5x 21)(2x 1) • 42-5 • (5x 3)(2x 7 ) • 35-6 • (5x 7)(2x 3) • 15-14

  6. Factoring ax2+bx +c 10x2 - 11x – 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 ) • 210-1 • (10x 21)(1x 1) • 21-10

  7. Factoring ax2+bx +c • Finally we got the correct combination to get a difference of 11 for our middle term. (10x21)(1x1) Outside 10x Inside 21x Difference is 11 • Put the signs in last. Look at the original problem again. 10x2 - 11x – 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. 10x2 - 11x – 21

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