Ch 10: Polynomials F) Factoring ax2 + bx + c Objective: To factor polynomials when a ≠ 1

Definitions Polynomial in Standard Form: ax2 + bx + c = 0 A polynomial written in descending order based on the exponents. Polynomial in Factored Form: (a1x + c1)(a2x + c2) = 0 A polynomial written as the product of two (or more) binomials.

Table Method a1 c1 a2 c2 b = a1 c2 + c1 a2 Arrange the polynomial in Standard Form everything on one side and zero on the other Create a 5 column table listing all of the factors of a in columns 1 and 3 List all of the factors of c in columns 2 & 4 Multiply the values in columns 1 & 4 and Multiply the values in columns 2 & 3. Place the sum of the products in column 5 Look for the value of b in column 5 and place the values from columns 1 & 2 and 3 & 4 into the Factored Form. (a1x + )(a2x + ) = 0 c1 c2 a1 c1 a2 c2 b = a1 c2 + c1 a2

Table Method ( ) ( ) Example 1 2x2 + 5x + 3 = 0 12 1 3 = 3 1 −1−3 Reverse Order 12 1 3 = 3 1 −1−3 = −3−1 a1 c2 c1 a2 a1 c1 a2 c2 b = + ( ) ( ) 1 3 + 1 1 2 3 5 = 12 1 −1 2 −3 1−3 + −12 1 3 2 1 1 1 + 32 1−1 + −32 1 −3 2 −1 ( x + )( x + ) = 0 a1 c1 a2 c2 (1x + 1) (2x + 3) = 0 (x + 1)(2x + 3) = 0

Table Method ( ) ( ) Example 2 2x2 − 9x + 7 = 0 12 1 7 = 7 1 −1−7 Reverse Order 12 1 7 = 7 1 −1−7 = −7−1 a1 c2 c1 a2 a1 c1 a2 c2 b = + 1 7 + 1 1 2 7 9 = 12 ( ) ( ) 1 −1 2 −7 -9 = 1−7 + −12 1 7 2 1 1 1 + 72 1−1 + −72 1 −7 2 −1 ( x + )( x + ) = 0 a1 c1 a2 c2 (1x +−1) (2x +−7) = 0 (x − 1)(2x − 7) = 0

Table Method ( ) ( ) −1 8 = 8−1 Example 3 1−8 = −8 1 17 −2 4 Reverse Order −1 8 = 8−1 Example 3 1−8 = −8 1 17 −2 4 = 4−2 7x2 − 55x − 8 = 0 2−4 = −4 2 a1 c2 c1 a2 a1 c1 a2 c2 b = + + 1 −1 7 8 1= 1 8 −17 1 1 7 −8 −1= 1−8 + 17 1 −2 7 4 −10 = 1 4 + −27 10 = 1−4 + 27 1 2 7 −4 1 8 −1 55 = 1−1 + 87 7 ( ) ( ) −55 = 1 1 + −87 1 −8 7 1 + 1 4 7 −2 1−2 47 1 −4 7 2 1 2 + −47 ( x + )( x + ) = 0 a1 c1 a2 c2 (1x +−8) (7x + 1) = 0 (x − 8)(7x + 1) = 0

Table Method ( ) ( ) 14 Example 4 −1 3 = 3−1 22 1−3 = −3 1 Reverse Order 14 Example 4 −1 3 = 3−1 22 1−3 = −3 1 4x2 + 4x – 3 = 0 a1 c2 c1 a2 a1 c1 a2 c2 b = + + 1 −1 4 3 −1= 1 3 −14 1 1 4 −3 1= 1−3 + 14 1 4 11= 1−1 + 34 3 −1 −11= 1 1 + −34 1 −3 4 1 ( ) ( ) −1 3 4 = 2 3 + −12 2 2 2−3 + 12 2 1 2 −3 + 2 3 2 −1 2−1 32 2 −3 2 1 2 1 + −32 ( x + )( x + ) = 0 a1 c1 a2 c2 (2x +−1) (2x + 3) = 0 (2x − 1)(2x + 3) = 0

Algebra Tiles x + 1 2x + 3 + + = (2x + 3) (x + 1) = 0 Arrange the polynomial in Standard Form everything on one side and zero on the other Lay out the tiles that represent the polynomial 2x2 + 5x + 3 = 0 3) Rearrange the tiles so they form a rectangle 4) The tiles across the bottom and down the right side represent the binomials in Factored Form. x + 1 2x + 3 + + = (2x + 3) (x + 1) = 0 2x2 5x 3

1) 2) 3) 4) 2x2 + 7x + 3 = 0 2x2 − 19x + 24 = 0 (2x + 1)(x + 3) = 0 Classwork 1) 2) 2x2 + 7x + 3 = 0 2x2 − 19x + 24 = 0 (2x + 1)(x + 3) = 0 (2x − 3)(x − 8) = 0 3) 4) 5x2 − 29x + 20 = 0 2x2 − 7x − 49 = 0 (5x − 4)(x − 5) = 0 (2x + 7)(x − 7) = 0

5) 6) 7) 8) 4x2 + 4x + 1 = 0 4x2 − 25 = 0 + 0x (2x + 1)(2x + 1) = 0