Alg. 2 Honors – Unit 3 Notes – Algebra behind Solving Quadratics

Factoring by Greatest Common Factor and Grouping

Objectives: SWBAT factor out a Greatest Common Factor from polynomials.

SWBAT identify polynomials that are prime.

SWBAT factor polynomials by grouping.

SWBAT use the zero product property to solve a quadratic.

GCF – Greatest Common Factor Prime – the GCF is only 1

A number/quantity that two terms

or more share

Find the greatest common factor.

1. – 6 and – 15 2. 16𝑥, 24𝑥 and 36𝑥 3. 3𝑥2 and 12𝑥

–3 4x 3x

Factor out the greatest common factor.

1 – Find the GCF, 2 – Divide it out, and 3 – Write as GCF (Remainder)

4. 6𝑥 − 14 5. −4𝑥2𝑦 − 6𝑥𝑦2 6. 7𝑥2 − 28𝑥 + 14

6 14

2

2 3 7

x

x

2 24 6

2

2 2 3

x y xy

xy

xy x y

2

2

7 28 14

7

4 2

x x

y x x

7. 8𝑥2 + 9𝑧2 8. 18𝑤3 − 15𝑤4 + 5𝑤5 9. (2𝑥2)(4𝑥2 − 12𝑥)

Prime

3 4 5

3

3 2

18 15 5

18 15 5

w w w

w

w w w w

2

2

3

4 12

4

4 3

2 4 3

8 3

x x

x

x x

x x x

x x

Factor by Grouping

Factor our a ( ) they two terms have in common

OR

Put ( ) around the 1st two terms and last two terms

Factor out a GCF of each parenthesis

They should have a ( ) in common

10. 𝑎(𝑏 + 4) + 𝑐(𝑏 + 4) 11. 𝑎(𝑥 − 3) + 𝑥2(𝑥 − 3) 12. 𝑥2 − 3𝑥 + 4𝑥 − 12

4 4

4

4

a b c b

b

b a c

2

2

3 3

3

3

a x x x

x

x a x

2

2

3 4 12

3 4 12

4

3 4 3

3 4 3

3

4 3

x x x

x x x

x

x x x

x x x

x

x x

13. 3𝑦2 − 8𝑦 + 12𝑦 − 32 14. 𝑎𝑏 + 𝑏𝑐 + 𝑎 + 𝑐 15. 3𝑥2 + 3𝑥𝑦 − 2𝑥𝑦 − 2𝑦2

2

2

3 8 12 32

3 8 12 32

4

3 8 4 3 8

3 8 4 3 8

3 8

3 8 4

y y y

y y y

y

y y y

y y y

y

y y

1

1

1

1

ab bc a c

ab bc a c

b

b a c a c

b a c a c

a c

a c b

2 2

2 2

3 3 2 2

3 3 2 2

3 2

3 2

3 2

3 2

x xy xy y

x xy xy y

x y

x x y y x y

x x y y x y

x y

x y x y

ZERO PRODUCT PROPERTY

0

0 and/or 0

a b

then

a b

Finding the Z.A.R.S.

Zeros~ Answers~ Roots~ Solutions~

When the directions ask you to find the Zeroes, Answers, Roots, or Solutions – you first

factor then set each parenthesis equal to zero and solve each parenthesis.

Solve the following.

16. (𝑥 + 3)(𝑥 − 5) = 0 17. 𝑦(4𝑦 − 9) = 0 18. 𝑎2 − 12𝑎 = 0

3 0 5 0

3 5

x x

x x

4 9 0

0 4 9 0

93

4

y y

y y

y y

2 12

12 0

12 0

0 12 0

0 12

Factor

a a

a

a a

Solve

a a

a a

x x

19. 5𝑧2 = 30𝑧 20. 7𝑚2 − 15𝑚 = −25𝑚2 21. 24 12 3 9 0m m m

2

2

5 30 0

5 30

5

5 6 0

5 6 0

5 0 6 0

0 6

z z

Factor

z z

z

z z

Solve

z z

z z

z z

2

2

32 12 0

32 12

4

4 8 3 0

4 8 3 0

4 0 8 3 0

80

3

m m

Factor

m m

m

m m

Solve

m m

m m

m m

Not Factorable

22. 2 5 5 25 0y y y 23. 28 6 12 9 0x x x 24. 27 21 8 24t t t

2

2

5 5 25

5 5 25

5

5 5 5

5 5 5

5

5 5

5 5 0

5 0 5 0

5 5

Factor

y y y

y y y

y

y y y

y y y

y

y y

Solve

y y

y y

y y

2

2

8 6 12 9

8 6 12 9

2 3

2 4 3 3 4 3

2 4 3 3 4 3

4 3

4 3 2 3

4 3 2 3 0

4 3 0 2 3 0

3 3

4 2

Factor

x x x

x x x

x

x x x

x x x

x

x x

Solve

x x

x x

x x

2

2

2

7 21 8 24 0

7 21 8 24

7 21 8 24

7 8

7 3 8 3

7 3 8 3

3

3 7 8

3 7 8 0

3 0 7 8 0

83

7

t t t

Factor

t t t

t t t

t

t t t

t t t

t

t t

Solve

t t

t t

t x

Factoring Trinomials when a > 1

Objectives: SWBAT factor “hard” trinomials, trinomials with a leading coefficient greater than 1.

SWBAT solve a quadratic equation with a leading coefficient greater than 1.

Factoring Trinomial when a > 1

Put the equation in Standard form Find the two numbers that multiply to Factor by grouping / box

the Special number and add to the B

Factor by Grouping

1. 22 9 7x x

Fill in the “x” Rewrite the polynomial Factor first

two terms

Factor last

two terms

22 72 7xx x

2

2

2 2

2 2

2

2 1

x x

x x

x

x x

7 7

7 7

7

7 1

x

x

x

Answer 2 7 1x x

2. 23 2x x

Fill in the “x” Rewrite the polynomial Factor first two terms

Factor last two terms

23 23 2xx x

2

2

3 3

3 3

3

3 1

x x

x x

x

x x

2 2

2 2

2

2 1

x

x

x

Answer 3 2 1x x

2ax bx c /

Grouping

Box Method

3. 23 7 20x x

Fill in the “x” Rewrite the polynomial Factor first

two terms

Factor last

two terms

2 5 20123 xx x

2

2

3 12

3 12

3

3 4

x x

x x

x

x x

5 20

5 20

4

4 4

x

x

x

Answer 3 4 4x x

Factor by X Box Method

6. 22 21y x x

Fill in the “x”

Fill in the box Factor out

Answer 2 7 3x x

7. 2( ) 4 25 6f x x x

Fill in the “x”

Fill in the box Factor out

Answer 4 1 6x x

A

Term

C

Term

1

Factor

2

Factor

A

Term

C

Term

1

Factor

2

Factor

22x

21

6x

7x

3x

2x

7

24x

6

24x

1x

6x

4x

1

ZERO PRODUCT PROPERTY

0

0 and/or 0

a b

then

a b

Finding the Z.A.R.S.

Zeros~ Answers~ Roots~ Solutions~

When the directions ask you to find the Zeroes, Answers, Roots, or Solutions – you first

factor then set each parenthesis equal to zero and solve each parenthesis.

Find the roots of ax2 + bx + c = 0 (By factoring and using the zero product rule).

9. 26 7 2x x 10. 2 8 12w w 11. 2 23 18 2 3 4x x x x

Put all equation in standard form first / Factor any way you would like

26 7 2 0x x 2 8 12 0w w 27 15 2x x

2

2

6 2

6 2

2 3 2 1 3 2

2 1 3 2

2 1 0 3 2 0

1 2

2 3

4 3

4 3

x

x

x x x

x x

x x

x

x x

x x

x

2

2

12

12

2 6 2

2 6

2 6

2 0 6 0

2

6

6

2

w

w

w w w

w w

w w

x x

w w

w w

27x

2

14x

1x

2x

7x

1

7 1 2 0

7 1 0 2 0

1 2

7

x x

x x

x x

Factoring “Easy” Trinomials (a = 1) and Factoring by Substitution

Objectives: SWBAT factor “easy” trinomials, trinomials with a leading coefficient of 1.

SWBAT solve a quadratic equation with a leading coefficient of 1.

SWBAT factor quadratics using substitution.

The “Short cut” when a=1

When a=1, you can go from the x straight to the ( ) when factoring

Factor trinomials in the form x2 + bx + c (using the x-factor).

1. x2 + 7x – 8 2. n2 + 7n +12 3. x2 – 4x – 12

1 8x x 3 4n n 2 6x x

Find the roots of ax2 + bx + c = 0 (By factoring and using the zero product rule).

4. x2 – 2x – 15 = 0 5. x2 + 2 = 3x 6. Find the zeros of:

𝑓(𝑥) = 𝑥2 + 3𝑥 − 40

2 3 2 0x x Zeroes just means your answer are

points

5 3 0

5 0 3 0

5 3

x x

x x

x x

2 1 0

2 0 1 0

2 1

x x

x x

x x

8 5 0

8 0 5 0

8 5

8,0 & 5,0

x x

x x

x x

7. 2x2 + 8x – 24 = 0 8. 4k2 + 14k = 30 9. 0 = 3x2 + 33x + 36

2

2

2

2 8 24 0

2 2

4 12 0

GCF

x x

x x

2

2

2

4 14 30 0

2

2 8 24 0

2 2

2 7 15 0

k k

GCF

k k

k k

2

2

3

3 33 36 0

3 3

11 12 0

GCF

x x

x x

Can’t go any further at this

Point (need Quadratic Formula)

U substitution

You can substitute a “u” for a ( ), and then factor the new statement using the u. Then at

the end of the factoring, back substitute the ( ) for the u.

Factor quadratics using substitution.

10. (x + 1)2 + 4(x + 1) + 4 11. (x – 2)2 + 4(x – 2) – 21

2

1

4 4

u x

u u

2

2

4 21

u x

u u

2 2

1

1 2 1 2

3 3

u u

u x

x x

x x

7 3

2

2 7 2 3

5 5

u u

u x

x x

x x

2

2

2 10 3 15

2 10 3 15

2 5 3 5

5 2 3

5 0 2 3 0

35

2

k k k

k k k

k k k

k k

k k

k k

Special Factoring Patterns

Objectives: SWBAT identify different types of factoring problems.

SWBAT identify and use special factoring patterns.

SWBAT factor the “un-factorable”.

Perfect Squares Addition Rule Subtraction Rule

2 2a b a b a b 2 22a b a b a ab b 2 22a b a b a ab b

The key important concept is the perfect square rule and being about to see that if a

binomial (two terms) can be factored.

Factor with special patterns.

1. x2 – 25 2. m2 – 22m + 121 3. d2 – 64 4. x2 + 12x + 36

Perfect Square Subtraction Rule Perfect Square Addition Rule

2

11 11

11

m m

or

m

2

6 6

6

x x

or

x

2

25

5 5

5 5

two numbers that make x

x and x

two numbers that make

and

x x

2

64

8 8

8 8

two numbers that make d

d and d

two numbers that make

and

d d

5. 9x2 – 25 6. 4x2 + 36 7. d2 – 12

2 9

3 3

25

5 5

3 5 3 5

two numbers that make x

x and x

two numbers that make

and

x x

2

9

3 3

3 3

two numbers that make x

x and x

two numbers that make

i and i

x i x i

2

2

4

4 36

4

4 9

GCF

x

x

2

12

12 12

12 2 3

2 3 2 3

two numbers that make d

d and d

two numbers that make

and

d d

Find the roots of the following equations.

8. x2 + 18 = 9x 9. 2x2 – 5x + 3 = 0 10. 4x2 + 36 = 0 11. x2 – 121 = 0

2 9 18 0x x

6 3 0

6 0 3 0

6 3

x x

x x

x x

12. 4x3 – 4x2 = 224x 13. 2x2 = 50 14. 3y3 – 12y2 – 2y2 + 8 = 0

2

2

2 2 3 3

2 2 3 3

2 1 3 1

2 3 1

2 3 0 1 0

3 1

2

x x x

x x x

x x x

x x

x x

x x

2

9

3 3

3 3 0

3 0 3 0

3 3

two numbers that make x

x and x

two numbers that make

i and i

x i x i

x i x i

x i x i

2

2

4

4 36 0

4 4

9 0

GCF

x

x

2

121

11 11

11 11 0

11 0 11 0

11 11

two numbers that make x

x and x

two numbers that make

and

x x

x x

x x

3 2

2

4 4 224 0

4 56

x x x

x x x

factor just inside

4 8 7

4 0 8 0 7 0

0 8 7

x x x

x x x

x x x

22 50 0

2 25

'

x

x x

can t factor just inside

go straight to solving

2 0 25 0

0 25

x x

x x

3 2 2

3 2 2

2

2

2

4

3 12 2 8

3 12 2 8

3 2

3 4 2 4

3 2 4

'

T erms Factor by Grouping

y y y

y y y

y

y y y

y y

can t factor just inside

go straight to solving

2

2

2

2

3 2 4

3 2 0 4 0

3 2

2

3

2 4

3

6 4

3

y y

y y

y

y

y y

y y

Solve Quadratic Equations by Finding Square Roots

Objective: SWBAT solve quadratic equations by using square roots.

Square root - Principal square root -

Radical - Radicand -

Simplify the radical expression.

1. 2. 3.

10 15

150

5 6

4. 5 6 2 18 5. 17

16

/ \

4 6

/ \

2 2

24

2 6

/ \

49 2

/ \

7 7

98

7 2

10 108

10 6 3

60 3

17

16

17

4

Rationalizing the denominator - Conjugates -

Can’t have a square root in the denominator Multiple terms in the denominator

Multiply the top and bottom to make a perfect square Multiply the top and bottom by the

on the bottom a perfect square of the denominator

Simplify the radical expression.

6. 6

5 7.

8

12 8.

2

4 5

Solving Quadratics by Square Roots

Isolate the variable or parenthesis squared

Take a square root of both sides (don’t forget the plus or minus if it is an even root) Solve for the variable

Use square roots to solve the following quadratic equations.

10. 3x2 = 75 11. z2 – 7 = 25

Reduce First

2 3

3 3

6

9

6

3

6 5

5 5

30

25

30

5

2 4 5

4 5 4 5

4 2 10

16 4 5 4 5 25

4 2 10

16 5

4 2 10

11

2

2

2

3 75

25

25

5

x

x

x

x

2

2

2

7 25

32

32

4 2

z

z

z

z

12. 2x

6 1025

13. 24(x 1) 8

2

2

2

2

6 1025

425

100

100

10

x

x

x

x

x i

2

2

2

4( 1) 8

( 1) 2

( 1) 2

1 2

1 2

x

x

x

x

x

Complete the Square

Objective: SWBAT complete the square.

SWBAT factor by completing the square.

SWBAT solve quadratics by completing the square.

SWBAT derive the quadratic formula.

Perfect square trinomials - 2

x a x a x a

𝒙𝟐 – 𝟏𝟎𝒙 + 𝟏𝟐 = 𝟎

Find the c value that would make the following a perfect trinomial.

1. 2x 10x c 2. 2x 8x c 3. 2x 13x c

2

2

2

2

2

10___ ___

2

+5

5

10 ___ 25 ___

10 25

5 2

xx c

x x

x x

x c

2

2

2

2

2

8___ ___

2

4

4

8 ___16 ___

8 16

4 16

xx c

x x

x x

x c

2

2

2

2

2

13___ ___

2

13

2

13

2

1698 ___ ___

4

16913

4

13 169

2 4

xx c

x x

x

x c

2

2

2

2

2

2

2

2

10 12

10 ______ 12 _______

10______ 12 _______

2

5

5

10 ___ 25 ___ 12 ___ 25 ____

10 25 13

5 13

5 13

5 13

5 13

x x

x x

xx

x x

x x

x

x

x

x

Solve the following by completing the square.

4. 2x 12x 18 0 5. 2x 10x 21 0

2

2

2

2

2

2

2

2

12 18

12 ______ 18 _______

12______ 18 _______

2

6

6

12 ___ 36 ___ 18 ___ 36 ____

12 36 18

6 18

6 18

6 3 2

6 3 2

x x

x

xx

x x

x x

x

x

x

x

2

2

2

2

2

2

2

2

10 21

10 ______ 21_______

10______ 21_______

2

5

5

10 ___ 36 ___ 21___ 25 ____

10 36 4

5 4

5 4

5 2

5 2

3

7

x x

x

xx

x x

x x

x

x

x

x

x

x

6. 2x 8x 1 7. 23x 42x 24

8. 2x 5x 12 0

2

2

2

2

2

2

2

5 12

5______ 12 ______

2

5

2

5

2

25 255 ___ ___ 12 __ ____

4 4

255

4

5 23

2 4

5 23

2 4

5 23

2 2

5 23

2 2

x x

xx

x x

x x

x

x

ix

ix

2

2

2

2

2

2

2

2

8 1

8 ______ 1_______

8______ 1_______

2

4

4

8 ___16 ___ 1___16 ____

8 16 15

4 15

4 15

4 15

4 15

x x

x

xx

x x

x x

x

x

x

x

2

2

2

2

2

2

2

2

2

3 42 24

3

14 8

14 ______ 8 _______

14______ 8 _______

2

7

7

14 ___ 49 ___ 8 ___ 49 ____

14 16 41

7 41

7 41

7 41

7 41

x x

x x

x x

xx

x x

x x

x

x

x

x

Use the Quadratic Formula and the Discriminant

Objective: SWBAT find and use the discriminant.

SWBAT use the quadratic formula.

Quadratic Formula - Discriminant -

2 4

2

b b acx

a

2 4b ac

Value of the

discriminant Number and type of solutions

Graph

Zero

1 real solution

Positive discriminant

2 real solutions

Negative discriminant

2 imaginary solutions

Find the discriminant and describe the solutions.

1. 23x 11x 4 0 2. 22a 13a 21

2

2

3

11

4

4

11 4 3 4

121 48

73

2

a

b

c

b ac

real solutions

2

2

2

13

21

4

13 4 3 21

169 252

83

2 imaginary

a

b

c

b ac

solutions

Use the quadratic formula to solve the following equations. Give exact answers, no decimals.

1. 2x 12x 32 0 2. 2x 4x 1 0

2

2

1

12

32

4

2

12 12 4 1 32

2 1

12 16

2

12 4

2

6 2

4 8

a

b

c

b b acx

a

x

x

x

x

x x

2

2

1

4

1

4

2

4 4 4 1 1

2 1

4 12

2

4 2 3

2

2 3

a

b

c

b b acx

a

x

x

x

x

3. 23x 11x 4 0 4. 22a 13a 7

2

2

3

11

4

4

2

11 11 4 3 4

2 3

11 73

6

11 73

6

a

b

c

b b acx

a

x

x

x

2

2

2

13

7

4

2

13 13 4 2 7

2 2

13 225

4

13 15

4

17

2

a

b

c

b b acx

a

x

x

x

x x

5. 2x x 1 0 6. 2x 12x 1 4x 3

2

2

1

1

1

4

2

1 1 4 1 1

2 1

1 3

2

1 3

2

a

b

c

b b acx

a

x

x

ix

2

2

2

16 4 0

1

16

4

4

2

16 16 4 1 4

2 1

16 240

2

16 4 15

2

8 2 15

x x

a

b

c

b b acx

a

x

x

x

x