Martin-Gay, Beginning Algebra, 5ed 22 Example Solution Think of FOIL in reverse. (x + )(x + ) We need 2 constant terms that have a product of 12 and a sum of 7. We list some pairs of numbers that multiply to 12. Product Sum Guess 34

Martin-Gay, Beginning Algebra, 5ed 33 Example continued Factor: x 2 + 7x + 12 Since 3  4 = 12 and = 7, the factorization of x 2 + 7x + 12 is (x + 3)(x + 4). To check we simply multiply the two binomials. Check: (x + 3)(x + 4) = x 2 + 4x + 3x + 12 = x 2 + 7x + 12

Martin-Gay, Beginning Algebra, 5ed 44

55 Example Factor: y 2  8y + 15 Solution Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 15 in which both factors are negative. Their sum must be  8. y 2  8y + 15 = (y  3)(y  5) Product Sum 88 15 Guess 33 55

Martin-Gay, Beginning Algebra, 5ed 66 Example Solution The constant term must be expressed as the product of a negative number and a positive number. Since the sum of the two numbers must be negative, the negative number must have the greater absolute value. x 2  5x  24 = (x + 3)(x  8) Factor: x 2  5x  24 Product Sum 55  24 Guess 4 66 3 88 2  12

Martin-Gay, Beginning Algebra, 5ed 77 Solution We need one positive and one negative factor. The sum must be 4, so the positive factor must have the larger absolute value. t 2 + 4t  32 = (t + 8)(t  4) Conclusion, the process of factoring trinomials is guess and check. Example Product Sum   32 Guess 4 88 44  8

Martin-Gay, Beginning Algebra, 5ed 88

99

10 Often factoring requires two or more steps. Remember, when told to factor, we should factor completely. This means the final factorization should contain only prime polynomials.

Martin-Gay, Beginning Algebra, 5ed 11

Martin-Gay, Beginning Algebra, 5ed 12 Examples Factor Completely Often factoring requires two or more steps. Remember, when told to factor, we should factor completely. This means the final factorization should contain only prime polynomials.