2.  A method for breaking up a quadratic equation in the form ax 2 + bx + c into factors (expressions which multiply to give you the original trinomial).  Use Algebra tiles to help you develop the patterns in factoring.

3.  x 2 + 5x + 4  x 2 + 7x + 12  x 2 + 6x + 8

4.  x 2 – 6x + 8  x 2 - 9x + 20  x 2 – 4x + 4  Notice how the answer changes if the b term is negative.

5.  x 2 + 4x – 12  x 2 + 3x – 18  x 2 – 7x – 8  x 2 – 3x – 10

6.  Factor each of the following expressions.  x 2 + 14x + 48  x 2 – 22x – 48  x 2 – 13x + 42  x 2 + 5x - 24

7.  Difference of squares  a 2 – b 2 = (a + b)(a – b)  Examples  x 2 – 49 = (x + 7)(x – 7)  4x 2 – 1 = (2x + 1)(2x – 1)  Practice  Factor: x 2 – 64  Factor: 9x 2 - 16

8.  Check to see if each term in the expression has a common factor.  Examples:  8x 3 – 50x What factors do they have in common?  2x(4x 2 – 25)  2x(2x + 5)(2x – 5)  Factor the following  5x 4 – 10x 3  2x 2 + 28x + 96

9.  1. Try to factor out a common monomial factor from all terms first.  2. Then look to see if the difference of squares format exists.  3. Use your techniques for factoring a quadratic trinomial if applicable.  Do-Now: Factor the following….  2x 2 – 18x + 40  3x 3 – 9x 2 – 30x

10.  What makes this expression more difficult to factor than what we have done before?  2x 2 + 7x + 6  You can use guess and check using FOIL, but it can become tedious for more difficult problems. Try these using guess and check.  3x 2 – 10x + 8  6x 2 + x – 15

11.  You may want to use this method on more complicated examples, such as……..  6x 2 + x – 12  First identify the values of a and c. ▪ a = 6, c = -12  Next, multiply a times c. ▪ (6)(-12) = -72  Then find two numbers that multiply to be (a)(c) and add to be the value of b. ▪ (9)(-8) = -72, 9 + -8 = 1

12.  Now rewrite the original expression with the x-term written as the sum of the two numbers you found.  6x 2 + x – 12 ▪ 6x 2 – 8x + 9x – 12  Finally, break the expression into two parts and factor twice.  (6x 2 – 8x) + (9x – 12)  2x(3x – 4) + 3(3x – 4)  (3x – 4)(2x + 3)

13.  2x 2 + 5x – 3  9x 2 + 12x + 4  12x 2 + 17x + 6  15x 2 + 8x – 16  20x 2 – 54x + 36

14.  It allows us to solve: ax 2 + bx + c = 0 using the “Zero Product Property”  If the product of two expressions is equal to zero, then………  one or both of the expressions must equal zero.  Examples. Solve the following equations.  x 2 – x – 42 = 0  x 2 = 7x

15.  Factor:  3x 2 + 17x + 10  8x 2 + 4x – 24  Solve the following equation.  4x 2 + 10x – 24 = 0

16.  When you solve y = ax 2 + bx + c when y = 0, you are finding the roots or zeros of the function. These are the points where the parabola crosses the x-axis.