2. 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 7 12 26 Guess 34 (x + 3)(x + 4) Example Solution y 2  8y + 15 Product Sum 88 15 Guess 33 55 Factor: y 2  8y + 15 = (y  3)(y  5)

3. Example Solution x 2  5x  24 Factor: x 2  5x  24 Product Sum 55  24 Guess 4 66 3 88 2  12 = (x + 3)(x  8)

4. Solution t 2 + 4t  32 Conclusion, the process of factoring trinomials is guess and check. Example Product Sum   32 Guess 4 88 44  8 = (t + 8)(t  4) Examples Factor Completely

5. Prime Polynomials A polynomial that cannot be factored is considered prime. Example: x 2  x + 7 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. Equations Containing Trinomials We can now use our new factoring skills to solve some polynomial equations.

6. Example Solve: x 2 + 5x + 6 = 0. Solution Algebraic x 2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 Factorization x + 2 = 0 or x + 3 = 0 x =  2 or x =  3 The solutions are x =  2 or  3. Graphical y1 = x^2 + 5*x + 6 The solutions are x =  2 or  3.

7. Example Solution—Algebraic The principle of zero products tells us we must have 0 on one side of the equation. Begin by multiplying the left side. (x – 6)(x + 1) = 8 x 2 – 5x  6 = 8 Multiplying x 2 – 5x  14 = 0 Adding 2 to both sides to get 0 on one side (x – 7)(x + 2) = 0 Factoring x – 7 = 0 or x + 2 = 0 x = 7 or x =  2 The solutions are x =  2 or x = 7. Solve: (x – 6)(x + 1) = 8.

8. Graphical Solution There is no need to simplify! The solution is x = 7 or x =  2.

9. Zeros and Factoring We can use the principle of zero products “in reverse” to factor a polynomial and write a function with given zeros. Example Write a polynomial function f(x) whose zeros are x =  2, x = 0 or x = 3. Solution Each zero of the polynomial function f is a linear factor of the polynomial. We write a linear factor for each given zero. If x =  2 is a zero of f (x) then x + 2 is a factor; If x = 0 is a zero of f (x) then x is a factor; If x = 3 is a zero of f (x) then x – 3 is a factor. The function is: f (x) = (x + 2)(x)(x – 3); multiplying gives us: f (x) = x 3 – x 2 – 6x.

10. x + 2 is a factor x is a factor x  3 is a factor