REVIEW OF FACTORING

Chapters 51 ndash 56

Factors

Factors are numbers or variables that are multiplied in a multiplication problem

Factor an expression means to write the expression as a product of its factors

a b = c then a and b are said to be factors of c

Examples

3 5 = 15 then 3 and 5 are factors of the product 15 x3 x4 = x7 then x3 and x4 are factors of x7 x(x + 2) = x2 + 2x then x and x + 2 are factors of x2 + 2x (x ndash 1) (x + 3) = x2 + 2x ndash 3 then x ndash 1 and x + 3 are factors of x2 + 2x ndash 3

Prime Numbers and Composite Numbers

Prime Number is an integer greater than 1 that has exactly two factors itself and one

The first 15 prime numbers are

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47

Composite is a positive integer (other than 1) that is not a prime number

The number 1 is neither prime nor composite it is called a unit

Greatest Common Factor

Greatest Common Factor (GCF) of two or more numbers is the greatest number that divides into all the numbers

GCF (two or more numbers) ndash Write each number as a product of prime factors Determine the prime factors common to all numbers Multiply the common factors

Example Find the GCF of 40 and 140

40 = 2 20 = 2 2 10 = 2 2 2 5 = 23 5

140 = 2 70 = 2 2 35 = 2 2 5 7 = 22 5 7

GCF = 22 5 = 4 5 = 20

Greatest Common Factor

GCF (two or more terms) ndash Take each factor the largest number of times that it appears in all of the terms

Example Find the GCF of 18y2 15 y3 and 27y5

18y2 = 2 9 y y = 2 3 3 y y15 y3 = = 3 5 y y y27y5 = 3 9 y y y y y = 3 3 3 y y y y y

GCF = 3 y y = 3 y2

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factors

Factors are numbers or variables that are multiplied in a multiplication problem

Factor an expression means to write the expression as a product of its factors

a b = c then a and b are said to be factors of c

Examples

3 5 = 15 then 3 and 5 are factors of the product 15 x3 x4 = x7 then x3 and x4 are factors of x7 x(x + 2) = x2 + 2x then x and x + 2 are factors of x2 + 2x (x ndash 1) (x + 3) = x2 + 2x ndash 3 then x ndash 1 and x + 3 are factors of x2 + 2x ndash 3

Prime Numbers and Composite Numbers

Prime Number is an integer greater than 1 that has exactly two factors itself and one

The first 15 prime numbers are

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47

Composite is a positive integer (other than 1) that is not a prime number

The number 1 is neither prime nor composite it is called a unit

Greatest Common Factor

Greatest Common Factor (GCF) of two or more numbers is the greatest number that divides into all the numbers

GCF (two or more numbers) ndash Write each number as a product of prime factors Determine the prime factors common to all numbers Multiply the common factors

Example Find the GCF of 40 and 140

40 = 2 20 = 2 2 10 = 2 2 2 5 = 23 5

140 = 2 70 = 2 2 35 = 2 2 5 7 = 22 5 7

GCF = 22 5 = 4 5 = 20

Greatest Common Factor

GCF (two or more terms) ndash Take each factor the largest number of times that it appears in all of the terms

Example Find the GCF of 18y2 15 y3 and 27y5

18y2 = 2 9 y y = 2 3 3 y y15 y3 = = 3 5 y y y27y5 = 3 9 y y y y y = 3 3 3 y y y y y

GCF = 3 y y = 3 y2

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Prime Numbers and Composite Numbers

Prime Number is an integer greater than 1 that has exactly two factors itself and one

The first 15 prime numbers are

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47

Composite is a positive integer (other than 1) that is not a prime number

The number 1 is neither prime nor composite it is called a unit

Greatest Common Factor

Greatest Common Factor (GCF) of two or more numbers is the greatest number that divides into all the numbers

GCF (two or more numbers) ndash Write each number as a product of prime factors Determine the prime factors common to all numbers Multiply the common factors

Example Find the GCF of 40 and 140

40 = 2 20 = 2 2 10 = 2 2 2 5 = 23 5

140 = 2 70 = 2 2 35 = 2 2 5 7 = 22 5 7

GCF = 22 5 = 4 5 = 20

Greatest Common Factor

GCF (two or more terms) ndash Take each factor the largest number of times that it appears in all of the terms

Example Find the GCF of 18y2 15 y3 and 27y5

18y2 = 2 9 y y = 2 3 3 y y15 y3 = = 3 5 y y y27y5 = 3 9 y y y y y = 3 3 3 y y y y y

GCF = 3 y y = 3 y2

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Greatest Common Factor

Greatest Common Factor (GCF) of two or more numbers is the greatest number that divides into all the numbers

GCF (two or more numbers) ndash Write each number as a product of prime factors Determine the prime factors common to all numbers Multiply the common factors

Example Find the GCF of 40 and 140

40 = 2 20 = 2 2 10 = 2 2 2 5 = 23 5

140 = 2 70 = 2 2 35 = 2 2 5 7 = 22 5 7

GCF = 22 5 = 4 5 = 20

Greatest Common Factor

GCF (two or more terms) ndash Take each factor the largest number of times that it appears in all of the terms

Example Find the GCF of 18y2 15 y3 and 27y5

18y2 = 2 9 y y = 2 3 3 y y15 y3 = = 3 5 y y y27y5 = 3 9 y y y y y = 3 3 3 y y y y y

GCF = 3 y y = 3 y2

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Greatest Common Factor

GCF (two or more terms) ndash Take each factor the largest number of times that it appears in all of the terms

Example Find the GCF of 18y2 15 y3 and 27y5

18y2 = 2 9 y y = 2 3 3 y y15 y3 = = 3 5 y y y27y5 = 3 9 y y y y y = 3 3 3 y y y y y

GCF = 3 y y = 3 y2

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor3 Use the distributive property to factor out the GCF

a (b + c) = ab + ac

Example 9 pg 348 7x ndash 35 GCF = 77(x ndash 5)

FOIL = First Outer Inner Last (a + b) (c + d)

First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factor a Monomial from a Polynomial

1 Determine the greatest common factor (GCF) of all the terms in the polynomial

2 Write each term as the product of the GCF and its other factor

3 Use the distributive property to factor out the GCFDistributive property says the if a b and c are all real numbers then a (b + c) = ab + ac

Monomial has one term Exp 5 4x -6x2

Binomial has two terms Exp x + 4 x2 ndash 6 2x2 ndash 5x Trinomial has three terms Exp x2 ndash 2x + 3 5x2 ndash 6x +7 Polynomial has infinite number of terms Exp 2x4 ndash 4x2 ndash 6x + 3

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factor a Monomial from a Polynomial

Example

6a4 + 27a3 - 18a2 GCF = 3a2

3a2(2a2 + 9a ndash 6)

Example

x(5x-2) + 7(5x-2) (5x-2) is a factor

(x+7)(5x-2)

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factor a Four-Term Polynomial by Grouping

1 Determine if there is a common factor if yes then factor out2 Arrange the four terms so that the first two terms and the last

two terms have a common factor3 Use distributive property to factor each group of terms

(first two last two)4 Factor GCF from the results

Example 27 pg 348 Example 22 pg 348x2 + 3x ndash 2xy ndash 6y x2 ndash 5x + 4x ndash 20x(x + 3) ndash 2y(x + 3) x(x ndash 5) + 4(x ndash 5)(x + 3) (x ndash 2y) (x ndash 5) (x + 4)

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Signs

If the 3rd term is positive the factor of last two terms will be positive

If the 3rd term is negative the factor of the last two terms will be negative

2 positives ndash Factor positive x2 + b + c ( _ + _ ) ( _+ _ )

b = Negative c = Positive ndash Factor Negative x2 ndash b + c ( _ - _ ) ( _- _ )

b = Positive c = Negative ndash Factor Positive Negative x2 + b ndash c ( _ + _ ) ( _- _ )

b = Negative c = Negative ndash Factor Positive Negative x2 ndash b ndash c ( _ + _ ) ( _- _ )

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factoring Trinomials a = 1

In the form of ax2 + bx + c where a = 11 Find two numbers whose product equals the constant c and whose

sum equals the coefficient of the x-term b

2 Use the two numbers found including their signs to write the trinomial in factored form (x + one number)(x + second number)

3 Check using the FOIL method

Example 37 pg 348 Factors of c that add to b

x2 + 11x + 18 8 = (2)(9)

(x + 9) (x+ 2) 2 + 9 = 11

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factoring Trinomials

Example 40 pg 348 Factors of c (56)that add to b (-15) x2 ndash 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15

A prime polynomial is a polynomial that cannot be factored using only integer coefficients

Example 36 pg 348 Factors of c (-15) that add to b (+4) x2 + 4x ndash 15 -15 = (-3)(5)PRIME -15 = (3)(-5)

-3 + 5 ne 4 5 + -3 ne 4

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factoring Trinomials a ne 1

In the form of ax2 + bx + c where a ne1 by Trial and Error1 Factor out any common factors to all three terms2 Write all pairs of factors of the coefficient of the squared term a3 Write all pairs of factors of the constant term c4 Try various combinations of these factors until the correct middle term

bx is found

Example Factors Possible Sum of 12 Factors InnerOuter

3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x

(3)(4) (3x+3)(x+4) 12x + 3x = 15x

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factoring Trinomials a ne 1

Example 1 Common Factor = 22x2 + 2x ndash 12 2 Factors of -6 that add to 12(x2 + x ndash 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x ndash 2) (-2)(3) add -2 + 3 = 1

Remember you can check with FOIL2(x + 3) (x ndash 2)2(x2 ndash 2x + 3x ndash 6) 2(x2 + x ndash 6)2x2 + 2x ndash 12

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Factoring Trinomials

In the form of ax2 + bx + c where a ne1 by Grouping1 Factor out any common factors to all three terms2 Fine two numbers whose product is equal to the product of a times c and whose sum

is equal to b3 Rewrite the middle term bx as the sum or difference of two terms using the numbers

found 4 Factor by grouping5 HINT If no factors of (a)(c) add up to (b) then cannot factor

Example No common factors3x2 + 20x + 12 a = 3 b = 20 c = 123x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 363x(x + 6) + 2(x + 6) factors of 36 that add to 20(3x + 2) (x + 6) (1)(36) 1 + 36 = 37

(2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Difference of Two Squares

a2 ndash b2 = (a + b) (a ndash b)

Example 64 pg 349 Example 70 page 349

x2 ndash 36 64x6 ndash 49y6

x2 ndash 62 (8x 3) 2 ndash (7y3) 2

(x + 6) (x ndash 6) (8x 3 + 7y3 )(8x 3 ndash 7y3 )

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Sum of Two Cubes

a3 + b3 = (a + b) (a2 ndash ab + b2)

Example 74 pg 349 Example 77 page 349

x3 + 8 125a3 + b3

x3 + 23 a = x b = 2 5a3 + b3 a = 5a b = b

(x + 2) (x 2 ndash 2x + 22) (5a + b) ((5a) 2 ndash 5ab + b2)

(x + 2) (x 2 ndash 2x + 4) (5a + b) (25a 2 ndash 5ab + b2)

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Difference of Two Cubes

a3 ndash b3 = (a - b) (a2 + ab + b2)

Example 76 pg 349 Example 73 page 349

b3 ndash 64 x3 ndash 1

b3 ndash 43 a = b b = 4 x3 ndash 13 a = x b = 1

(b ndash 4) (b2 + 4b + 42) (x ndash 1)(x2 + 1x + 12)

(b ndash 4)(b2 + 4b + 16) (x ndash 1)(x2 + x + 1)

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

General Procedure for Factoring a Polynomial

1 Factor any GCF of all terms

2 If a two term polynomial determine if it is a special factor If so factor using the formula

3 If three term polynomial factor according to methods discussed for a = 1 or a ne 1

4 If more than three terms try factoring by grouping

5 Determine if there are any common factors and factor them out

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Quadratic Equation

Standard Form a + bx + c = 0 a b and c are real numbers

Zero factor Property ndash if ab = 0 then a = 0 or b = 0

Solve the quadratic Equation by factoring 1 Write the equation in standard form

2 Factor the side of the equation that is not 0

3 Set each factor equal to 0 and solve

4 Check each solution

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Quadratic Equation

Example Check (x = 1)

x2 ndash 3x = -2 x2 ndash 3x = -2

x2 ndash 3x + 2 = 0 12 ndash 3(1) = -2

Factors of 2 that sup to -3 1 ndash 3 = -2

(-1)(-2) = 2 -2 = -2 True

(-1) + (-2) = -3 Check (x = 2)

(x ndash 1)(x ndash 2) x2 ndash 3x = -2

x ndash 1 = 0 and x ndash 2 = 0 22 ndash 3(2) = -2

x = 1 x = 2 4 ndash 6 = -2

-2 = -2 True

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85

Homework ndash Review Factoring

Page 348 ndash 349

11 13 15 35 39 41 55 63 69 85