2. Previously, we have learned how to factor and have explored various factoring techniques. First, we studied how to find and use the GCF. Next, we looked at the distributive property, and then we looked at factoring trinomials. Some examples of factoring are shown below. GCF & Distributive Property Factor into the product of two polynomials. Difference of two squares. Perfect Square Trinomials

3. Now that we know how to factor, we can apply this knowledge to the solution of equations. How would you solve the following equation? x2 x2 – 36 = 0 Step 1:1:Factor the polynomial. (x - 6)(x + 6) = 0

4. Step 2:2:Apply the zero product property which states that For all numbers a and b, if ab = 0, then a = 0, b = 0, 0, or both a and b equal 0. (x - 6)(x + 6) = 0 Therefore (x – 6) = 0 or (x + 6) = 0. x – 6 = 0 + 6 +6 x = 6 or x + 6 = 0 - 6 -6 x = -6 This equation has two solutions or zeros: x = 6 or x = -6.

5. 1.Get a value of zero on one side of the equation. 2.Factor the polynomial if possible. 3.Apply the zero product property by setting each factor equal to zero. 4.Solve for the variable.

6. Solve the following equation: x 2 – 10x + 5 = 29. x 2 – 10x + 5 = 29 -29 -29 x2 x2 – 10x –24 = 0 (x – 12) (x + 2) = 0The factors of –24 are: -1,24, 1,-24-2,122,-12 -3,83,-8-4,64,-6 Get zero on one side of the equation Factor the trinomial. Set each factor equal to 0. Solve for the variable. x – 12 = 0 + 12 +12 or x + 2 = 0 - 2 -2 x = 12 x = -2 x = 12 or x = -2

7. Solve the following equations. 1.x 2 – 25 = 0 2.x 2 + 7x – 8 = 0 3.x 2 – 12x + 36 = 0 4.c 2 – 8c = 0 5.5b 3 + 34b 2 = 7b

8. x 2 – 25 = 0 Factor the polynomial. Set each factor equal to zero. Solve for x.x. xx  5050 or +5+5 -5-5 x= 5 x=-5 Solution:x = 5 or x = -5

9. x 2 + 7x – 8 = 0 Factor the polynomial. Set each factor equal to zero. Solve for x.x. x  80 -8-8 x=-8 or x  10 +1+1 x= 1 Solution:x = 1 or x = -8

10. x 2 – 12x + 36 = 0 Factor the polynomial. Set each factor equal to zero. Solve for x.x. x  60 +6+6 x=6 Solution:x = 6There is only one solution for this equation since both of the factors are the same.

11. c 2 – 8c = 0 Factor the polynomial. Set each factor equal to zero. or Solve for c.c. c  80 +8+8 c=8 Solution:c = 0 or c = 8.

12. 5b 3 + 34b 2 = 7b Factor the polynomial. Get a value of zero on one side of the equation. 5b 3 + 34b 2 = 7b -7b -7b 5b 3 + 34b 2 –7b = 0 5b 3 + 34b 2 – 7b = b(5b 2 + 34b – 7) =b(5b – 1)(b + 7) Set each factor equal to zero. b = 0or 5b – 1 = 0 or b + 7 = 0 Solve for b.b. b = 0 or 5b – 1 = 0 +1 +1 5b = 1 5 5 b = 1/5 b + 7 = 0 - 7 -7 b = -7 Solution: b = 0 or b = 1/5 or b = -7.