Step 1: Identify and a, b, c. Step 2: Multiply ac. Step 3: Identify pairs of factors of and find the pair of factors whose sum equals If there is ac, b. no pair of factors whose sum equals the original quadratic expression cannot be factored. b, Step 4: Split the middle linear term into two terms whose coefficients are the factors of ac. Step 5: Factor by grouping. For example, given the quadratic expression 2x2 + 11x + 12, = 2, = 11, and = 12. a b c Therefore, = 24, and factors of 24 include (1 x 24), (2 x 12), (3 x 8), and (4 x 6). (3 x 8) is the ac unique factor pair whose sum equals 11 (b = 11). Using these factors, we can split the middle term into two terms, and then factor by grouping. Factoring by grouping requires that we find a common factor in the first two terms, and find a common factor in the last two terms, and factor these terms out. Once a common binomial factor appears, we factor it out as well. 2x2 + 11x + 12 = 2x2 + 3x 8x + 12 = +3) + 4(2x+3) = (2x+3)(x + 4) + x(2x A more concrete approach is have students use area models or algebra tiles to examine the relationships between the quadratic expression and the binomial factors. This essentially involves undoing the multiplication of polynomials with area models demonstrated in Investigation 4. For example, to factor , start by assigning and 12 to two regions as shown. The next step is to find two numbers whose product is 12 and whose sum is 8. 6 and 2 fit the bill. So split the middle term 8x into 6x and 2x and assign these quantities to the two remaining rectangles. The final step is to find the common factor for each row and column as shown. The result is that the trinomial can be factored as (x + 6)(x + 2). Virtual algebra tiles may be found at In addition, students can check their factorizations using the Wolfram Alpha (). Differentiated Instruction (For Learners Needing More Help)
Step 1: Identify a, b, and c. Step 2: Multiply ac. Step 3: Identify pairs of factors of ac, and find the pair of factors whose sum equals b. If there is no pair of factors whose sum equals b, the original quadratic expression cannot be factored. Step 4: Split the middle linear term into two terms whose coefficients are the factors of ac. Step 5: Factor by grouping. For example, given the quadratic expression 2x2 + 11x + 12, a = 2, b = 11, and c = 12. Therefore, ac = 24, and factors of 24 include (1 x 24), (2 x 12), (3 x 8), and (4 x 6). (3 x 8) is the unique factor pair whose sum equals 11 (b = 11). Using these factors, we can split the middle term into two terms, and then factor by grouping. Factoring by grouping requires that we find a common factor in the first two terms, and find a common factor in the last two terms, and factor these terms out. Once a common binomial factor appears, we factor it out as well. 2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x(2x +3) + 4(2x+3) = (2x+3)(x + 4) A more concrete approach is have students use area models or algebra tiles to examine the relationships between the quadratic expression and the binomial factors. This essentially involves undoing the multiplication of polynomials with area models demonstrated in Investigation 4. For example, to factor , start by assigning and 12 to two regions as shown. The next step is to find two numbers whose product is 12 and whose sum is 8. 6 and 2 fit the bill. So split the middle term 8x into 6x and 2x and assign these quantities to the two remaining rectangles. The final step is to find the common factor for each row and column as shown. The result is that the trinomial can be factored as (x + 6)(x + 2). Virtual algebra tiles may be found at In addition, students can check their factorizations using the Wolfram Alpha (). Differentiated Instruction (For Learners Needing More Help)
alg_u8_i5_overview.docx