1. Factoring Quadratic Expressions Lesson 4-4 Part 2
Algebra 2
Factoring Quadratic Expressions
Lesson 4-4 Part 2

2. Goals Goal Rubric To find binomial factors of quadratic expressions
To factor special quadratic expressions.
Level 1 – Know the goals.
Level 2 – Fully understand the goals.
Level 3 – Use the goals to solve simple problems.
Level 4 – Use the goals to solve more advanced problems.
Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary
Perfect Square Trinomial
Difference of Two Squares

4. Big Idea: Solving Equations and Inequalities
Essential Question
Big Idea: Solving Equations and Inequalities
Why is it useful to recognize special quadratic expressions?

5. Factoring ax2 + bx + c
In the previous lesson you factored trinomials of the form x2 + bx + c. Now you will factor trinomials of the form ax + bx + c, where a ≠ 0.

6. THE FUN WAY TO FACTOR TRINOMIALS
X-BOX Method
THE FUN WAY TO FACTOR TRINOMIALS

7. X-Box Method
This is a guaranteed method for factoring trinomials of the form ax2 + bx +c — no guessing necessary!
We will learn how to factor quadratic equations using the x-box method
Background knowledge needed:
Basic x-factor problems
General form of a trinomial
The factoring box

8. m n mn m n m+n The X-Factor Given m & n on the sides of the x factor
m•n is the top of the x factor m n
m+n
m+n is the bottom of the x factor

9. The X-Factor mn n m
m+n

10. What’s on Top & Bottom? 5 3
Given m & n on the sides of the x factor ?
m•n is the top of the x factor 5 3 ?
m+n is the bottom of the x factor

11. 5 3 15 5 3 8 The X-Factor Given m & n on the sides of the x factor
m•n is the top of the x factor 5 3 8
m+n is the bottom of the x factor

12. The X-Factor -3 9 ? -3 9 ?

13. The X-Factor -3 9
-27 -3 9 6

14. The X-Factor 12
Given m•n 7
Given m+n 12 ? ?
Find m & n 7

15. The X-Factor 12
Given m•n 7
Given m+n 12 3 4
Find m & n 7

16. The X-Factor 8
Given m•n 9
Given m+n 8 ? ?
Find m & n 9

17. The X-Factor 8
Given m•n 9
Given m+n 8 8 1
Find m & n 9

18. The X-Factor
-84
Given m•n -5
Given m+n
-84
Find m & n -5

19. To solve a difficult X-Factor START AT 1 AND CHECK THE SUM
-84
-84
Given m•n
= -83
-41 = -39
-28 = -25
-21 = -17
-14 = -8
= -5 -5
Given m+n
-84
Find m & n
-12 7 -5

20. Why Know the “X” Factor?

21. Why Know the “X” Factor? The “X” factor is a Simple way to
FACTOR TRINOMIALS

22. General Form of a Trinomial
c is COEFFICIENT
of last term
b is COEFFICIENT
of x
a is COEFFICIENT
of x2

23. Identify a, b & c
b is 3
c is 9
a is 5

24. Factor the X-Box Method
ax2 + bx + c
Combine the x-factor and a 2⨯2 box to create the x-box method.
Factor Col. 1
Factor Col. 2
Product
ac=mn
First and Last Coefficients
1st Term
GCF Row 1
Factor n n m
Middle
Last term
Factor m
Factor Row 2
b=m+n
Sum

25. X-Box Procedure ax2 + bx + c 1st Term Factor n Factor m Last term
Create an x-factor with the product ac on the top, the middle term b on the bottom and the factors m & n on the sides.
Create a 2x2 box.
In the top left, put the 1st term. In the bottom right corner, put the last term.
Put the two factors m and n times the variable x in the open boxes.
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial.
1st Term
Factor n
Factor m
Last term

26. Factoring ax2+bx+c
Factor 5x2+11x+2
5•2= 10
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 1 10 11 x 2 5x x
10x
5x2 1 2
(x+2)(5x+1)

27. Factor 3x2 – x – 4 3 -4 -1 3x -4 x 3x -4x 3x2 1 -4 3•-4= -12
Factoring ax2+bx+c
Factor 3x2 – x – 4
3•-4=
-12
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 3 -4 -1 3x -4 x 3x
-4x
3x2 1 -4
(x+1)(3x – 4)

28. Factor 12x2 +5x – 2 8 -3 5 4x -1 3x 12x2 8x -3x 2 -2 12•-2= -24
Factoring ax2+bx+c
Factor 12x2 +5x – 2
12•-2=
-24
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 8 -3 5 4x -1 3x
12x2 8x
-3x 2 -2
(3x+2)(4x – 1)

29. Factor 15x2 +7x – 2 10 -3 7 5x -1 3x 10x -3x 15x2 2 -2 15•-2= -30
Factoring ax2+bx+c
Factor 15x2 +7x – 2
15•-2=
-30
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 10 -3 7 5x -1 3x
10x
-3x
15x2 2 -2
(3x+2)(5x – 1)

30. Factoring ax2+bx+c
Factor 10x2 +9x +2
10•2= 20
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 5 4 9 5x 2 2x 5x 4x
10x2 1 2
(2x+1)(5x +2)

31. Your Turn:
Factor: 2x² - 13x + 6

32. Factor 2x2 – 13x +6 -12 -1 -13 2x -1 x -12x -x 2x2 -6 6 2•6= 12
Factoring ax2+bx+c
Factor 2x2 – 13x +6
2•6= 12
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial.
-12 -1
-13 2x -1 x
-12x -x
2x2 -6 6
(2x – 1)(x – 6)

33. Your Turn:
Factor: 4x² + 11x – 3

34. Factor 4x²+11x – 3 12 -1 11 4x -1 x 12x -x 4x2 3 -3 4•-3= -12
Factoring ax2+bx+c
Factor 4x²+11x – 3
4•-3=
-12
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 12 -1 11 4x -1 x
12x -x
4x2 3 -3
(4x – 1)(x + 3)

35. Your Turn:
Factor: 3x²+8x+4

36. Factor 3x²+8x+4 6 2 8 3x 2 x 6x 2x 3x2 2 4 4•3= 12 Factoring ax2+bx+c
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 6 2 8 3x 2 x 6x 2x
3x2 2 4
(3x + 2)(x + 2)

37. Your Turn:
Factor: -5x²+6x-1

38. Factor -5x²+6x-1 5 1 6 -5x 1 x 5x x -5x2 -1 -1 -5•-1= 5
Factoring ax2+bx+c
Factor -5x²+6x-1
-5•-1= 5
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. 5 1 6
-5x 1 x 5x x
-5x2 -1 -1
(-5x + 1)(x – 1)

39. Factor Completely * Always Factor out the GCF First *
To factor a polynomial completely, first factor at the GCF of the polynomial’s terms.
Then factor the remaining polynomial until it is written as the product of polynomials that cannot be factored further.
* Always Factor out the GCF First *

40. Example: Factor Completely
Factor Completely: 18x2 – 33x + 12 First factor out the GCF 3 3(6x2 – 11x + 4) Then factor the remaining polynomial (6x2 – 11x + 4)

41. Factor 6x2 – 11x+4 -8 -3 -11 2x -1 3x -8x -3x 6x2 -4 4 6•4= 24
Factoring ax2+bx+c
Factor 6x2 – 11x+4
6•4= 24
Set up & Solve X-factor:
Put ac on top
Put b on bottom
Determine side factors
Set up & Solve “BOX”:
Put First Term in First Box
Put Last Term in Last Box
Put Side Factors Times x in Box
Determine the GCF of the upper boxes and the remaining top & side factors.
The sum of the factor’s for the top and the side are the factors of the polynomial. -8 -3
-11 2x -1 3x
-8x
-3x
6x2 -4 4
(3x – 4)(2x – 1)

42. Example: Factor Completely
Factor Completely: 18x2 – 33x + 12
First factor out the GCF 3
3(6x2 – 11x + 4)
Then factor the remaining polynomial (6x2 – 11x + 4)
(3x – 4)(2x – 1)
Completely Factored: 3(3x – 4)(2x – 1)

43. Your Turn: Completely factor 8x2 – 36x – 20 4(2x + 1)(x – 5)

44. Perfect Square Trinomial

45. Perfect Square Trinomials
Factor the polynomial x x + 4.
The result is (5x + 2)2, an example of a binomial squared.
Any trinomial that factors into a single binomial squared is called a perfect square trinomial.

46. A perfect square trinomial results after squaring a binomial
Example: (2x – 5)2
Multiply it out using FOIL
(2x – 5)2 = (2x – 5) (2x – 5)
= 4x2
– 10x
– 10x
+ 25 F O I L
= 4x2 – 20x
The first and last terms are perfect squares.
The middle term is double the product of the
square roots of the first and last terms.

47. Perfect Square Trinomials
(a + b)2 = a 2 + 2ab + b 2
(a – b)2 = a 2 – 2ab + b 2
So if the first and last terms of our polynomial to be factored can be written as expressions squared, and the middle term of our polynomial is twice the product of those two expressions, then we can use these two previous equations to easily factor the polynomial.
a 2 + 2ab + b 2 = (a + b)2
a 2 – 2ab + b 2 = (a – b)2

48. So how do we factor a Prefect Square Trinomial
When you have to factor a perfect square trinomial, the patterns make it easier
Product Doubled
Example: Factor 36x2 + 60x + 25
Perfect Square 6x
30x 5
Perfect Square
First you have to recognize that it’s a perfect square trinomial
Square Root
Product
Square Root
And so, the trinomial factors as:
Check:
(6x + 5)2

49. Example – First verify it is a Perfect Square Trinomial
= (3w + 4)2
= (4x – 9)2
= (5h – 16)(5h – 4)
= (a + 3)2
m2 – 4m + 4
9w2 + 24w + 16
16x2 – 72x + 81
25h2 – 100h + 64
a2 + 6a + 9 m 2m 2 3w
12w 4 4x
36x 9
Not a perfect square trinomial! 5h
40h 8 a 3a 3

50. Your Turn: = (m – 3)2 m2 – 6m + 9 = (2w + 7)2 4w2 + 28w + 49
= (9x – 1)2
= (3a + 2)2
m2 – 6m + 9
4w2 + 28w + 49
81x2 – 18x + 1
9a2 + 12a + 4 m 3m 3 2w
14w 7 9x 9x 1 3a 6a 2

51. Review: Perfect-Square Trinomial

52. Difference of Two Squares
D.O.T.S.

53. Conjugate Pairs (3x + 6) (3x - 6) (r - 5) (r + 5) (2b - 1) (2b + 1)
The following pairs of binomials are called conjugates. Notice that they all have the same terms, only the sign between them is different.
(3x + 6)
(3x - 6)
and
(r - 5)
(r + 5)
and
(2b - 1)
(2b + 1)
and
(x2 + 5)
(x2 - 5)
and

54. Multiplying Conjugates
When we multiply any conjugate pairs, the middle terms always cancel and we end up with a binomial.
(3x + 6)(3x - 6)
= 9x2 - 36
(r - 5)(r + 5)
= r2 - 25
(2b - 1)(2b + 1)
= 4b2 - 1

55. Only TWO terms (a binomial)
Difference of Two Squares
Binomials that look like this are called a Difference of Squares:
Only TWO terms (a binomial)
9x2 - 36
A MINUS between!
The first term is a Perfect Square!
The second term is a Perfect Square!

56. Difference of Two Squares
A binomial is the difference of two square if
both terms are squares and
the signs of the terms are different.
9x 2 – 25y 2
– c 4 + d 4

57. Factoring the Difference of Two Squares
A Difference of Squares!
A Conjugate Pair!

58. Difference of Two Squares
Example:
Factor the polynomial x 2 – 9.
The first term is a square and the last term, 9, can be written as 32. The signs of each term are different, so we have the difference of two squares
Therefore x 2 – 9 = (x – 3)(x + 3).
Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

59. Difference of Two Squares
Example: Factor x2 - 64
= (x + 8)(x - 8)
x2 = x • x
64 = 8 • 8
Example: Factor 9t2 - 25
= (3t + 5)(3t - 5)
9t2 = 3t • 3t
25 = 5 • 5

60. Difference of Two Squares
Example:
Factor x 2 – 16.
Since this polynomial can be written as x 2 – 42,
x 2 – 16 = (x – 4)(x + 4).
Factor 9x2 – 4.
Since this polynomial can be written as (3x)2 – 22,
9x 2 – 4 = (3x – 2)(3x + 2).
Factor 16x 2 – 9y 2.
Since this polynomial can be written as (4x)2 – (3y)2,
16x 2 – 9y 2 = (4x – 3y)(4x + 3y).

61. A Sum of Squares?
A Sum of Squares, like x2 + 64, can NOT be factored!
It is a PRIME polynomial.

62. Difference of Two Squares
Example:
Factor x 8 – y 6.
Since this polynomial can be written as
(x 4)2 – (y 3)2,
x 8 – y 6 = (x 4 – y 3)(x 4 + y 3).
Factor x2 + 4.
Oops, this is the sum of squares, not the difference of squares, so it can’t be factored. This polynomial is a prime polynomial.

63. Difference of Two Squares
Example:
Factor 36x 2 – 64.
Remember that you should always factor out any common factors, if they exist, before you start any other technique.
Factor out the GCF.
36x 2 – 64 = 4(9x 2 – 16)
Since the polynomial can be written as (3x)2 – (4)2,
(9x 2 – 16) = (3x – 4)(3x + 4).
So our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4).

64. Recognizing D.O.T.S. Reading Math
Recognize a difference of two squares: the coefficients of variable terms are perfect squares, powers on variable terms are even, and constants are perfect squares.
Reading Math

65. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
3p 2 – 9q 4
3p 2 – 9q 4
3q 2  3q 2
3p2 is not a perfect square.
3p 2 – 9q 4 is not the difference of two squares because 3p 2 is not a perfect square.

66. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
100x 2 – 4y 2
100x 2 – 4y 2
2y 2y 
10x 10x
The polynomial is a difference of two squares.
(10x)2 – (2y)2
a = 10x, b = 2y
(10x + 2y)(10x – 2y)
Write the polynomial as (a + b)(a – b).
100x2 – 4y2 = (10x + 2y)(10x – 2y)

67. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
x 4 – 25y 6
x 4 – 25y 6
5y 3 5y 3 
x 2 x 2
The polynomial is a difference of two squares.
(x 2)2 – (5y 3)2
a = x2, b = 5y3
Write the polynomial as (a + b)(a – b).
(x 2 + 5y 3)(x 2 – 5y 3)
x 4 – 25y 6 = (x 2 + 5y 3)(x 2 – 5y 3)

68. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
1 – 4x 2
1 – 4x 2
2x 2x 
The polynomial is a difference of two squares.
(1) – (2x)2
a = 1, b = 2x
(1 + 2x)(1 – 2x)
Write the polynomial as (a + b)(a – b).
1 – 4x 2 = (1 + 2x)(1 – 2x)

69. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
p 8 – 49q 6
p 8 – 49q 6
7q 3 7q 3 ●
p 4 p 4
The polynomial is a difference of two squares.
(p 4)2 – (7q 3)2
a = p4, b = 7q3
(p 4 + 7q 3)(p 4 – 7q 3)
Write the polynomial as
(a + b)(a – b).
p 8 – 49q 6 = (p 4 + 7q 3)(p 4 – 7q 3)

70. Your Turn:
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
16x 2 – 4y 5
16x 2 – 4y 5
4x  4x
4y5 is not a perfect square.
16x 2 – 4y 5 is not the difference of two squares because 4y 5 is not a perfect square.

71. Big Idea: Solving Equations and Inequalities
Essential Question
Big Idea: Solving Equations and Inequalities
Why is it useful to recognize special quadratic expressions?
It is easier to factor a quadratic expression if you identify it as a perfect square trinomial or a difference of squares. Use the corresponding property to factor the quadratic expression.

72. Assignment
Section 4-4 Part 2, Pg 234 – 236; #1 – 5 all, 8 – 44 even.