2.  FACTORING a polynomial means to break it apart into its prime factors.  For example:  x 2 – 4 = (x + 2)(x – 2)  x 2 + 6x + 5 = (x + 1)(x + 5)  3y 2 + 10y – 8 = (3y – 2)(y + 4)

4.  Let’s begin by grouping the first two terms and the last two terms:  (ax + ay) + (bx + by)  Next, factor out the GCF of each group:  a(x + y) + b(x + y)  Notice that (x + y) is a common factor of each term. This means: ◦ ax + ay + bx + by = (x + y)(a + b)

5.  First, group each set of terms:  (x 3 + 7x 2 ) + (2x + 14)  Then, take out GCFs:  x 2 (x + 7) + 2(x + 7)  Write final answer:  (x + 7)(x 2 + 2)

6.  First, group each set of terms:  (3x 2 + xy) – (12x + 4y)  Notice that because there was a minus here, we needed to change the sign in our second group!  Then, take out GCFs:  x(3x + y) – 4(3x + y)  Write final answer:  (3x + y)(x – 4)

7. 1. a 3 – 2a 2 + 5a – 10 2. 15y 3 + 24y 2 – 35y – 56

8. Answer: (a – 2)(a 2 + 5) Answer: (5y + 8)(3y 2 – 7) 1. a 3 – 2a 2 + 5a – 10 Group: (a 3 – 2a 2 ) + (5a – 10) GCFs: a 2 (a – 2) + 5(a – 2) 2. 15y 3 + 24y 2 – 35y – 56 Group:(15y 3 + 24y 2 ) – (35y + 56) **Notice the sign change! GCFs: 3y 2 (5y + 8) – 7(5y + 8)

10.  First, remember that your equation must be set equal to 0 before factoring.  3x 2 + 2x – 5 = 0  There isn’t a GCF to factor out. So, factor.  (3x + 5)(x – 1) = 0  Next, using the Zero Product Property, set each factor equal to 0 and solve for x.  3x + 5 = 0 or x – 1 = 0  x = -5/3 or x = 1  The solution set is: Note the graph shows us 2 zeros, so our solution makes sense!

12.  These three terms have a GCF that we will need to factor out first:  2x 3 (2x 2 – 3x – 2) = 0  Then, factor:  2x 3 (2x + 1)(x – 2) = 0  Set each factor equal to 0:  2x 3 = 0 or 2x + 1 = 0 or x – 2 = 0  Solve for x: Note the graph shows us 3 zeros, so our solution makes sense!

13.  First, we need to set our equation equal to 0:  3x 4 – 4x 2 + 1 = 0  Then, factor:  (3x 2 – 1)(x 2 – 1) = 0  Set each factor equal to 0:  3x 2 – 1 = 0 or x 2 – 1 = 0  Solve for x: Note the graph shows us 4 zeros, so our solution makes sense!

14.  First, factor:  (x 2 + 1)(x 2 – 1) = 0  (x 2 + 1)(x + 1)(x – 1) = 0  Set each factor equal to 0:  x 2 + 1 = 0 or x + 1 = 0 or x – 1 = 0  Solve for x: Note the graph shows us 2 zeros, because our calculator only shows REAL zeros.

16.  A company sells an item for x dollars. Their revenue y (in dollars) is given by the polynomial equation y = -10x 4 + 1000x 2. At what price will the company stop making money?

17.  Look at the graph of the function first:  We want to find the zeros, because then we will know when the company isn’t making any money (revenue of $0).  We can factor the polynomial equation to find these zeros!!!

18.  Let’s factor 0 = -10x 2 (x 2 – 100)  0 = -10x 2 (x + 10)(x – 10)  Now set each factor equal to 0 and solve for x:  -10x 2 = 0 or x + 10 = 0 or x – 10 = 0  -10 doesn’t make sense in the situation because the company wouldn’t charge -$10  0 doesn’t make sense in the situation because the company wouldn’t charge $0  10 does make sense! If the company charges $10, they won’t make any money…this would most likely be because people wouldn’t spend that much money for the item.