Factoring Trinomials of the form ax2 + bx + c Section 6.4 Factoring Trinomials of the form ax2 + bx + c

Using the Trial-and-Error Method Factor. 3x2 + 4x + 1 3x2 + 4x + 1 = (3x + m)(x + n) 3x2 + 4x + 1 3x2 + 4x + 1 = (3x + 1)(x + 1) In order for the coefficient of the x2-term to be 3, the coefficients of the x-terms in the factors must be 3 and 1. In order for the last term to be 1, the constants in the factors must both be 1.

Using the Trial-and-Error Method Factor. 9x2  13x + 4 Factorizations of 9 Factorizations of 4 3 and 3 1 and 4 1 and 9 2 and 2 Possible Factors Middle Term Correct? (3x – 2)(3x – 2) (3x – 1)(3x – 4) (x – 2)(9x – 2) (x – 1)(9x – 4) 9x2  13x + 4 = (x – 1)(9x – 4) –12x No –15x No –20x No –13x Yes

Using the Trial-and-Error Method Factor. 5x2 + 2x  7 Factors of 5 Factors of 7 1 and 5 1 and 7 Possible Factors Middle Term Correct? (x + 1)(5x  7) (x  1)(5x + 7) 5x2 + 2x  7 = (x – 1)(5x + 7) – 2x No 2x Yes

Grouping Method for Factoring Trinomials of the Form ax2 + bx + c. 1. Obtain the grouping number ac. 2. Find the two numbers whose product is the grouping number and whose sum is b. 3. Use those numbers to write bx as the sum of two terms. 4. Factor by grouping. 5. Multiply to check.

Example Factor. 2x2 + 9x + 4 1. The grouping number is 2(4) = 8. 2. The factors of 8 are 1(8) and (2)(4). We choose 1 and 8 because their product is 8 and sum is 9. 3. We write 9x as the sum x + 8x. 4. Factor by grouping. 2x2 + 9x + 4 = 2x2 + x + 8x + 4 = x(2x + 1) + 4(2x + 1) = (2x + 1)(x + 4)

Example Factor. 8x2 + 10x  3 1. The grouping number is (8)(3) = 24. 2. We want two numbers whose product is 24 and whose sum is 10. They are 12 and 2. 3. We write 10x as the sum 12x  2x. 4. Factor by grouping. 8x2 + 10x  3 = 8x2 + 12x  2x  3 = 4x(2x + 3)  1(2x + 3) = (2x + 3)(4x  1)

Example Factor. 9x2 + 3x  30 Remove the greatest common factor. 9x2 + 3x  30 = 3(3x2 + x – 10) = 3(3x – 5)(x + 2)