SLIDE 1
X- BOX
Product of a & c
b
Trinomial (Quadratic Equation)
ax2 + bx + c
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’.
ax 2 + bx + c Product of a & c Fill the 2 empty sides with 2 - - PowerPoint PPT Presentation
ax 2 + bx + c Product of a & c Fill the 2 empty sides with 2 - - PowerPoint PPT Presentation
D AY 139 X-BOX F ACTORING X- B OX Trinomial (Quadratic Equation) ax 2 + bx + c Product of a & c Fill the 2 empty sides with 2 numbers that are factors of ac and add to b give you b. X- B OX Trinomial (Quadratic
SLIDE 2
SLIDE 3
X- BOX
20
9
Trinomial (Quadratic Equation)
x2 + 9x + 20
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’.
4 5
SLIDE 4
X- BOX
- 42
- 1
Trinomial (Quadratic Equation)
2x2 -x - 21
Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’.
- 7
6
SLIDE 5
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
SLIDE 6
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.
SLIDE 7
FACTOR THE X-BOX WAY
Example: Factor x2 -3x -10
- 3
(1)(-10)=
- 10
- 5
2
- 10
- 5x
2x x2
x
- 5
x +2 x2 -3x -10 = (x-5)(x+2)
GCF GCF GCF GCF
SLIDE 8
FACTOR THE X-BOX WAY
Middle b=m+n Sum Product ac=mn m n
First and Last Coefficients
y = ax2 + bx + c
Last term 1st Term Factor n Factor m Base 1 Base 2 GCF Height
SLIDE 9
FACTOR THE X-BOX WAY
Example: Factor 3x2 -13x -10
- 13
- 30
- 15
2
- 10
- 15x
2x 3x2 x
- 5
3x +2 3x2 -13x -10 = (x-5)(3x+2)
SLIDE 10
EXAMPLES
Factor using the x-box method.
- 1. x2 + 4x – 12
a) b)
x
- 12
4 6
- 2
x2 6x
- 2x -12
x
- 2
+6 Solution: x2 + 4x – 12 = (x + 6)(x - 2)
SLIDE 11
EXAMPLES
CONTINUED
- 2. x2 - 9x + 20
a) b) 20
- 9
x2
- 4x
- 5x 20
x x
- 4
- 5
Solution: x2 - 9x + 20 = (x - 4)(x - 5)
- 4
- 5
SLIDE 12
EXAMPLES
CONTINUED
- 3. 2x2 - 5x - 7
a) b)
- 14
- 5
2x2 -7x 2x -7 x 2x
- 7
+1
Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)
- 7
2
SLIDE 13
EXAMPLES
CONTINUED
- 3. 15x2 + 7x - 2
a) b)
- 30
7
15x2 10x
- 3x -2
5x 3x +2
- 1
Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)
10
- 3
SLIDE 14
EXTRA PRACTICE
1.
x2 +4x -32
2.
4x2 +4x -3
3.
3x2 + 11x – 20
SLIDE 15