Math 20-1 Chapter 4 Quadratic Equations

4.2 Factoring Quadratic Equations

Teacher Notes

4.2A Factoring Quadratic Expressions

4.2A.2

Factor

Greatest Common

Factor

Simple Trinomial

Perfect Square

Trinomial

Difference of

Squares

Decomposition

Fractions

Complex Bases

24 2x x

2 5 6x x

2 10 25x x

2 25x 26 5 4x x

212 5

5x x

22 3 2 2x x

2 (2 1)x x

3 2x x

2

5 5

5

x x

x

5 5x x

2 1 3 4x x

2

2

110 25

51

55

x x

x

2 2 2 1x x

Solve by Factoring Simple Trinomials

Simple Trinomials: The coefficient of the x2 term is 1.Recall: (x + 6)(x + 4) = x(x + 4) + 6(x + 4) = x2 + 4x + 6x + (6)(4) = x2 + (4 + 6)x + (6)(4) = x2 + 10x + 24

The middle term is the sum of the constant terms of the binomial.

The last term is theproduct of the constant terms.

Factor: x2 + 9x + 20

(x + 5)(x + 4)

x2 + 5x + 4x + (4)(5)x(x + 5) + 4(x + 5)

x2 + (a + b)x + ab = (x + a)(x + b)

4.2A.3

Factors of 20

1 20

2 10

4 5

2x bx c

Explain each step going backwards

x2 + (5 + 4)x + (4)(5)

212 x x Factors of 121 122 63 4

212 4 3x x x

4(3 ) (3 )x x x

(3 )(4 )x x

4.2A.4

Recall: (3x + 2)(x + 5) = 3x(x + 5) +2(x + 5) = 3x2 + 15x + 2x + 10 = 3x2 + 17x + 10

To factor 3x2 + 17x + 10, find two numbers

that have a product of 30 and a sum of 17.

Factoring General Trinomials Decomposition

4.2A.5

2ax bx c

Factors of 30

1 30

2 15

3 10

Explain each step going backwards

Solving General Trinomials - the Decomposition Method

3x2 - 10x + 8 The product is 3 x 8 = 24.The sum is -10.

Rewrite the middle term of the polynomial using -6 and -4. (-6x - 4x is just another way ofexpressing -10x.)

3x2 - 6x - 4x + 8

Factor by grouping. 3x(x - 2) (x - 2)(3x - 4)

- 4(x - 2)

4.2A.6

Factors of 24

1 24

2 12

3 8

4 6

2( 3) 5 3 4x x

Factor Factors of 4

1 42 2

2 5 4B B 2 1 4 4B B B ( 1) 4( 1)B B B

( 4)( 1)B B

( 4)( 13 3 )x x

( )(1 2)x x

4.2A.7

Difference of SquaresRecall that, when multiplying conjugate binomials, the product is a difference of squares.

(x - 7)(x + 7) = x2 + 7x – 7x - 49 = x2 - 49

Therefore, when factoring a difference of squares, thefactors will be conjugate binomials.

Factor:

x 2 - 81 = (x - 9)(x + 9) 16x2 - 121 = (4x - 11)(4x + 11)

5x2 - 80y2 = 5(x2 - 16y2)= 5(x - 4y)(x + 4y)

(x)2 - (9)2 (4x)2 - (11)22 1

4x

1 1

2 2x x

4.2A.8

= [(x + y) - 4]

(x + y)2 - 16Factor completely:

= (x + y - 4)(x + y + 4)[(x + y) + 4]

Factoring a Difference of Squares with a Complex Base

= [B - 4] [B + 4]= B2 - 16

25 - (x + 3)2 = [5 - (x + 3)]

= (-x + 2)(8 + x)= (5 - x - 3)(5 + x + 3)

[5 + (x + 3)]

4.2A.9

25(x - 1)2 - 4(3x + 2)2

= [5(x - 1) - 2(3x + 2)] [5(x - 1) + 2(3x + 2)]= (5x - 5 - 6x - 4)(5x - 5 + 6x + 4)= (-x - 9)(11x - 1)

AssignmentSuggested Questions:Page 229# 1 - 5

4.2A.10