1. Unit 4: Factoring and Complex Numbers
Warm-Up: Complete the half-sheet on finding GCF.
How can you use the calculator to find factors of a number?
Use the functions “y=“ or “gcd”

2. When factoring: always find GCF first if it exists
Examples:
8x – 4
14x3 – 7x2
6x2 + 12x – 21
4n4 + 6n3 – 8n2
9 – 27x2

3. X-box Factoring

4. X- Box ax2 + bx + c Trinomial (Quadratic Equation)
Product of a & c
ax2 + bx + c
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. b

5. X- Box x2 + 9x + 20 Trinomial (Quadratic Equation)
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. 5 4 9

6. X- Box 2x2 -x - 21 Trinomial (Quadratic Equation)
-42
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. 6 -7 -1

7. X-box Factoring
This is a guaranteed method for factoring quadratic equations—no guessing necessary!
We will learn how to factor quadratic equations using the x-box method

8. LET’S TRY IT!
Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.
Objective: I can use the x-box method to factor non-prime trinomials.

9. First and Last Coefficients
Factor the x-box way
y = ax2 + bx + c
GCF
GCF
Product
ac=mn
First and Last Coefficients
1st Term
Factor n*
GCF n m
Middle
Last term
Factor m*
b=m+n
Sum
GCF
Factors “n” and “m” come from the “X” so include “x” in the box with them.

10. Factor the x-box way x2 -5x -10 2x Example: Factor x2 -3x -10
(1)(-10)=
-10 x -5 x2
-5x x
GCF -5 2 -3 2x
-10 +2
GCF
GCF
GCF
x2 -3x -10 = (x-5)(x+2)

11. Factor the x-box way Example: Factor 3x2 -13x -10 x -5 -30 3x 3x2 -15x+2
3x2 -13x -10 = (x-5)(3x+2)

12. Examples Factor using the x-box method. 1. x2 + 4x – 12 x +6 x2 6x x
a) b) x +6
-12 4
x2 6x
-2x -12 x 6 -2 -2
Solution: x2 + 4x – 12 = (x + 6)(x - 2)

13. Examples continued 2. x2 - 9x + 20 x -4 x x2 -4x -5x 20
a) b) x -4 20 -9 x
x2 -4x
-5x 20 -5
Solution: x2 - 9x + 20 = (x - 4)(x - 5)

14. Examples continued 3. 2x2 - 5x - 7 2x -7 x 2x2 -7x 2x -7
a) b) 2x -7
-14 -5 x
2x2 -7x
2x -7 +1
Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)

15. Examples continued 3. 15x2 + 7x - 2 3x +2 5x 15x2 10x -3x -2
a) b) 3x +2
-30 7 5x
15x2 10x
-3x -2 -1
Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)

16. Extra Practice
x2 +4x -32
4x2 +4x -3
3x2 + 11x – 20

17. Notes
What do you do if there are no factors that add to equal “b” and multiply to give “ac”?