Section R.4 Factoring

R.4 Factoring Factor polynomials by removing a common factor. Factor polynomials by grouping. Factor trinomials of the type x2 + bx + c. Factor trinomials of the type ax2 + bx + c, a  1, using the FOIL method and the grouping method. Factor special products of polynomials.

Terms with Common Factors When factoring, we should always look first to factor out a factor that is common to all the terms. Example:

Factoring by Grouping In some polynomials, pairs of terms have a common binomial factor that can be removed in the process called factoring by grouping. Example: x3 + 3x2  5x  15 = (x3 + 3x2) + (5x  15) = x2(x + 3)  5(x + 3) = (x + 3)(x2 − 5)

Trinomials of the Type x2 + bx + c Factor: x2 + 5x + 6. Solution: 1. Look for a common factor. 2. Find the factors of 14, whose sum is 9. Pairs of Factors Sum 1, 6 7 2, 3 5 The numbers we need are 2 and 3. 3. The factorization is (x + 2)(x + 3).

Trinomials of the Type ax2 + bx + c, a  1 FOIL method 1. Factor out the largest common factor. 2. Find two First terms whose product is ax2. 3. Find two Last terms whose product is c. 4. Repeat steps (2) and (3) until a combination is found for which the sum of the Outside and Inside products is bx.

Example Factor: 3x2 − 10x − 8. Solution 1. There is no common factor (other than 1 or -1). 2. Factor the first term, 3x2. The only possibility (with positive integer coefficients) is 3x · x. The factorization, if it exists, must be of the form (3x + )(x + ). 3. Next, factor the constant term, −8. The possibilities are −8(1), 8(−1), −2(4), and 2(−4). The factors can be written in the opposite order as well: 1(−8), −1(8), 4(−2), and −4(2).

Example continued 4. Find a pair of factors for which the sum of the outer product and the inner product is the middle term, −10x. Each possibility should be checked by multiplying. Some trials show that the desired factorization is (3x + 2)(x – 4).

The Grouping Method (or ac-method) ax2 + bx + c The factor ax2 + bx + c, a ≠ 1, using the grouping method: 1. Factor out the largest common factor. 2. Multiply the leading coefficient a and the constant c. 3. Try to factor the product ac so that the sum of the factors is b. That is, find integers p and q such that pq = ac and p + q = b. 4. Split the middle term. That is, write it as a sum using the factors found in step (3). 5. Factor by grouping.

Example Factor: 12x3 + 10x2  8x. 1. Factor out the largest common factor, 2x. 2x(6x2 + 5x  4) 2. Multiply a and c: (6)(4) = 24. 3. Try to factor 24 so that the sum of the factors is the coefficient of the middle term, 5. (−3)(8) = 24 and -3 + 8 = 5 4. Split the middle term using the numbers found in (3). 5x = −3x + 8x. 5. Factor by grouping. 6x2 + 5x  4 = 6x2 – 3x + 8x – 4 = 3x(2x – 1) + 4(2x – 1) =(2x – 1)(3x + 4). Be sure to include the common factor to get the complete factorization. 12x3 + 10x3 – 8x = 2x(2x – 1)(3x + 4).

Special Factorizations Difference of Squares A2  B2 = (A + B)(A  B) Example x2  16 = x2 – 42 = (x + 4)(x  4)

Special Factorizations Squares of Binomials A2 + 2AB + B2 = (A + B)2; A2  2AB + B2 = (A  B)2 Example x2 + 8x + 16 = x2 + 2 · x · 4 + 42 = (x + 4)2

More Factorizations Sum or Difference of Cubes A3 + B3 = (A + B)(A2  AB + B2); A3  B3 = (A  B)(A2 + AB + B2). Example x3 + 27 = (x)3 + (3)3 = (x + 3)(x2  3x + 9)