Week 12 Day 2
Kristy Chimera
Casie Croff
Nichole Oswald

Title: Factoring Trinomials grades 9-12 Math A review

Materials:
	Whiteboard/chalkboard
	Whiteboard markers/chalk
	Worksheets and calculators

Lesson Overview:
	Students will learn how to factor a trinomial. Students will also learn how to check their solutions using both the F.O.I.L method, and the calculator.

Lesson Objectives:
	Students will be able to analyze graphs.
	Students will be able to explain why the solution works graphically.
	Students will be able to generate the factors for a trinomial.

NYS Standards:
	Key Idea 2: Number and Numeration
		2C. Apply the properties of real numbers to various subsets of numbers.
	Key Idea 3: Operations
		3A. Use addition, subtraction, multiplication, division, and exponentiation 
			with real numbers and algebraic expressions.
		3D. Use field properties to justify mathematical procedures.

Anticipatory Set: (4 minutes)
	Put the following on the board. Have the students do the problem in their notes on 
	a sheet of paper. Then have one volunteer come up to the board and complete and 	explain the problem. Use F.O.I.L to evaluate (x + 2)(x + 3)
	Answer: (x + 2)(x + 3) = (x2 + 3x + 2x + (2*3))
			 = x2 + 3x + 2x + 6
			 = x2 + (3 + 2)x + 6
			 = x2 + 5x + 6
	Now ask what the prefix “tri” means? Answers could include: tricycle, triangle, 	triathlon. Tell the students that today we’re going to factor trinomials!

Developmental Activity:
	1. Hand out worksheet titles: x2 + bx + c
	2. Go through the notes section of the worksheet. Fill in the blanks as you get to 
		them. Try to have the students come up with the answers. (2 minutes)
	3. Go through cases 1 and 2 with the students. Go slowly and answer any 				questions that come up. You may need to spend a little time on finding the 		factors of a number, they may not remember this. (10 minutes)
	4. Make sure the students understand how to use the calculator to check their 			answer when you get to the check in case 2. 

Guided Practice:
	1. Have the students work together (groups of 2 or 3) to work through cases 3 and 
		4. Walk around the room and help any students who are stuck. They may 			have a hard time coming up with all of the factors. Try to give some hints 			without giving away the answers. (15 minutes)
	2. Come back together as a class and go over what the answers are for cases 3 and 
		4. Answer any questions that come up along the way. (4 minutes)

Independent Practice: (5 minutes)
	Hand out the exit ticket. The students should be able to figure out how to do this. 			There is a hint on there for them as well. Collect the exit ticket before the 			students leave the room.

Name:_____________________________________________	Date: _____________

x2 + bx + c

Notes:
	Notice, in the previous example we had x2 + (3 + 2)x + (2*3) where m = 3, and 
n = 2. So bx = m + n, and c = n*m. This is true for all trinomials. The coefficient (number in front of the variable) of the middle term is the ______ (addition) of m and n and the last term is the __________ (multiplication) of m and n. We will use this pattern to factor quadratic trinomials of the form x2 + bx + c. Let’s look at 4 cases.

Case 1:  b and c are both positive.

Factor x2 + 6x + 8		b = _________	c = _________
We need two numbers whose sum is 6 and whose product is 8. Let’s look at the factors of 8: what can we multiply to get 8? Let’s look at a table:

Factors of 8	Sum of the factors	
      1, 8	              9	
      2, 4	              6	

Since we want our sum to be 6 and our product to be 8, we will use 4 and 2. 
So x2 + 6x + 8 = (x + m)(x + n) which is the pattern
		= (x + 2) (x + 4)  so m = 2 and n = 4

Let’s check using F.O.I.L
	(x + 2)(x + 4)
F       O     I	L
x2 + 4x + 2x + 8 = x2 + 6x + 8 which is what we started with.

Case 2: b is negative and c is positive

Factor x2 - 10x + 16		b = _________	c = _________
Now we need two numbers whose sum is -10 and whose product is 16, so we need to find the factors of 16. We need to look at negative numbers so when we add them, we will get a negative 10.

Factors of 16	Sum of the factors	
    -1, -16	          -17	
    -2, -8	          -10	
    -4, -4	           -8	

It looks like -2 and -8 will work. Let’s see what we get.
x2 - 10x + 16 = (x + m)(x + n)
	          = (x - 2)(x - 8)    so m = -2 and n = -8

Check: (x - 2)(x - 8) = x2 -8x -2x +16
                                 = x2 -10x + 16

Alternate way to check: Go to y = on your calculator. In Y1, type x2 - 10x + 16. In 
Y2 type (x - 2)(x - 8). Hit graph. Why does only one graph appear? _________________
_______________________________________________________________________

Case 3:  b is positive and c is negative

Factor x2 + x - 12		b = _________	c = _________
This time we need a sum of 1 and a product of -12. Let’s look at the factors of -12.

Factors of -12	Sum of the factors	

      	           	

		

The correct factors are ____ and _____. 
So x2 + x -12 = (x + m)(x + n)
                      = (x      )(x      )   	so m = _____ and n = ______

Check: (x    )(x    ) = 
		       = 

Case 4:  b is negative and c is negative

Factor x2 - 7x - 18			b = _________	c = _________
Now we need a sum of -7 and a product of -18 so let’s look at the factors of -18.

Factors of -18	Sum of the factors	

     	           	

The factors we should use are _____ and _____.

x2 - 7x - 18 = (x + m)(x + n)
                    = (x    )(x    )            so m = _____ and n = _____

Check: use the calculator to check. Enter x2 - 7x - 18 into Y1 and (x    )(x    ) into Y2.
Is this the solution? _______

Name: ______________________________________________	Date: _____________

Exit Ticket: Solve an Equation by Factoring

Solve x2 + 5x = 6  Then check your solutions.	

HINT: Get all terms on one side of the equation, factor, then set parts equal to 0.

Name: _________Answer Key___________________________	Date: _____________

Exit Ticket: Solve an Equation by Factoring

Solve x2 + 5x = 6  Then check your solutions.	

HINT: Get all terms on one side of the equation, factor, then set parts equal to 0.

x2 + 5x = 6						Check: Substitute 1 and -6 for x 
x2 + 5x - 6 = 0						into the original equation.
(x - 1)(x + 6) = 0				     x2 + 5x = 6		        x2 + 5x = 6
x - 1 = 0	or	x + 6 = 0		   12 + 5*1 = 6		(-6)2 + 5(-6) = 6
x       = 1		x        = 6		    1 + 5     = 6		        36 - 30 = 6
Solution set is {1, -6}				       6        = 6		             6     = 6

Name:_______Answer Key____________________________	Date: _____________

x2 + bx + c

Notes:
	Notice, in the previous example we had x2 + (3 + 2)x + (2*3) where m = 3, and 
n = 2. So bx = m + n, and c = n*m. This is true for all trinomials. The coefficient (number in front of the variable) of the middle term is the _Sum__ (addition) of m and n and the last term is the Product___ (multiplication) of m and n. We will use this pattern to factor quadratic trinomials of the form x2 + bx + c. Let’s look at 4 cases.

Case 1:  b and c are both positive.

Factor x2 + 6x + 8		b = ___6_____	c = ____8____
We need two numbers whose sum is 6 and whose product is 8. Let’s look at the factors of 8: what can we multiply to get 8? Let’s look at a table:

Factors of 8	Sum of the factors	
      1, 8	              9	
      2, 4	              6	

Since we want our sum to be 6 and our product to be 8, we will use 4 and 2. 
So x2 + 6x + 8 = (x + m)(x + n) which is the pattern
		= (x + 2) (x + 4)  so m = 2 and n = 4

Let’s check using F.O.I.L
	(x + 2)(x + 4)
F       O     I	L
x2 + 4x + 2x + 8 = x2 + 6x + 8 which is what we started with.

Case 2: b is negative and c is positive

Factor x2 - 10x + 16		b = _-10_____	c = ___16____
Now we need two numbers whose sum is -10 and whose product is 16, so we need to find the factors of 16. We need to look at negative numbers so when we add them, we will get a negative 10.

Factors of 16	Sum of the factors	
    -1, -16	          -17	
    -2, -8	          -10	
    -4, -4	           -8	

It looks like -2 and -8 will work. Let’s see what we get.
x2 - 10x + 16 = (x + m)(x + n)
	          = (x - 2)(x - 8)    so m = -2 and n = -8

Check: (x - 2)(x - 8) = x2 -8x -2x +16
                                 = x2 -10x + 16

Alternate way to check: Go to y = on your calculator. In Y1, type x2 - 10x + 16. In 
Y2 type (x - 2)(x - 8). Hit graph. Why does only one graph appear? Because the factors of the trinomial form the same line as the trinomial, so the graphs are on top of each other. 

Case 3:  b is positive and c is negative

Factor x2 + x - 12		b = ___1_____	c = ___-12___
This time we need a sum of 1 and a product of -12. Let’s look at the factors of -12.

Factors of -12	Sum of the factors	
      1, -12	          -11	
     -1, 12	           11	
      2, -6	           -4	
     -2, 6	            4	
      3, -4	           -1	
     -3, 4	            1	

The correct factors are _-3_ and __4__. 
So x2 + x -12 = (x + m)(x + n)
                      = (x - 3)(x + 4)   	so m = _-3__ and n = __4___

Check: (x - 3)(x + 4) = x2 + 4x - 3x - 12
		          = x2 + x - 12

Case 4:  b is negative and c is negative

Factor x2 - 7x - 18			b = _-7______	c = __-18____
Now we need a sum of -7 and a product of -18 so let’s look at the factors of -18.

Factors of -18	Sum of the factors	
      1, -18	          -17	
     -1, 18	           17	
       2, -9	           -7	
      -2, 9	            7	

The factors we should use are 2 and -9.

x2 - 7x - 18 = (x + m)(x + n)
                    = (x + 2)(x - 9)            so m = 2 and n = -9

Check: use the calculator to check. Enter x2 - 7x - 18 into Y1 and (x + 2)(x - 9) into Y2.
Is this the solution? _Yes___