Factoring Polynomials Algebra I. Vocabulary Factors – The numbers used to find a product. Prime...

Factoring Polynomials

Algebra I

Vocabulary

• Factors – The numbers used to find a product.

• Prime Number – A whole number greater than one and its only factors are 1 and itself.

• Composite Number – A whole number greater than one that has more than 2 factors.

Vocabulary

• Factored Form – A polynomial expressed as the product of prime numbers and variables.

• Prime Factoring – Finding the prime factors of a term.

• Greatest Common Factor (GCF) – The product of common prime factors.

Prime or Composite?

Ex) 36

Ex) 23

Prime or Composite?

Ex) 36 Composite. Factors: 1,2,3,4,6,9,12,18,36

Ex) 23 Prime. Factors: 1,23

Prime Factorization

Ex) 90 = 2 ∙ 45 = 2 ∙ 3 ∙ 15 = 2 ∙ 3 ∙ 3 ∙ 5 OR use a factor tree: 90 9 10 3 3 2 5

Prime Factorization of Negative Integers

Ex) -140 = -1 ∙ 140 = -1 ∙ 2 ∙ 70 = -1 ∙ 2 ∙ 7 ∙ 10 = -1 ∙ 2 ∙ 7 ∙ 2 ∙ 5

Now you try…

Ex) 96

Ex) -24

Now you try…

Ex) 96 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3

Ex) -24 -1 ∙ 2 ∙ 2 ∙ 2 ∙ 3

Prime Factorization of a Monomial

12a²b³= 2 · 2 · 3 · a · a · b · b · b

-66pq²= -1 · 2 · 3 · 11 · p · q · q

Finding GCF

Ex) 48 = 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 60 = 2 ∙ 2 ∙ 3 ∙ 5 GCF = 2 · 2 · 3 = 12

Ex) 15 = 3 · 5 16 = 2 · 2 · 2 · 2 GCF – none = 1

Now you try…

Ex) 36x²y 54xy²z

Now you try…

Ex) 36x²y = 2 · 2 · 3 · 3 · x · x · y 54xy²z = 2 · 3 · 3 · 3 · x · y · y · z

GCF = 18xy

Factoring Using the (Reverse) Distributive Property

• Factoring a polynomial means to find its completely factored form.


• First step is to find the prime factors of each term.

Ex) 12a²+ 16a 12a²= 2 · 2 · 3 · a · a 16a = 2 · 2 · 2 · 2 · a



• Next step is to find the GCF of the terms in the polynomial.

Ex) 12a²+ 16a 12a²= 2 · 2 · 3 · a · a 16a = 2 · 2 · 2 · 2 · a GCF = 4a




• Now write what is left of each term and leave in parenthesis.

Ex) 12a²+ 16a 12a²= 2 · 2 · 3 · a · a 16a = 2 · 2 · 2 · 2 · a 4a(3a + 4)




• Now write what is left of each term and leave in parenthesis.

Ex) 12a²+ 16a 12a²= 2 · 2 · 3 · a · a 16a = 2 · 2 · 2 · 2 · a 4a(3a + 4) Final Answer 4a(3a + 4)

Another Example:

18cd²+ 12c²d + 9cd

Another Example:

18cd²+ 12c²d + 9cd

18cd² = 2 · 3 · 3 · c · d · d12c²d = 2 · 2 · 3 · c · c · d 9cd = 3 · 3 · c · dGCF = 3cdAnswer: 3cd(6d + 4c + 3)

FOIL Review

Using FOIL:FirstOuter (x + 2)(x – 3)InnerLast

x²+ -3x + 2x + -6 x²+ -1x + -6

Factoring by Grouping

• Factor some polynomials having 4 or more terms. Pairs of terms are grouped together and factored using GCF.

Ex) 4ab + 8b + 3a + 6

Factoring by Grouping

• Factor some polynomials having 4 or more terms. Pairs of terms are grouped together and factored using GCF.

Ex) 4ab + 8b + 3a + 6 (4ab + 8b) + (3a + 6) 4b(a + 2) + 3(a + 2) These must be the same!

(a + 2)(4b + 3) *** check by using FOIL

Grouping more than one way

• Group more than one way to get the same answer.

• Use the commutative property to move terms to group.

Ex) 4ab + 8b + 3a + 6

Grouping more than one way

• Group more than one way to get the same answer.

• Use the commutative property to move terms to group.

Ex) 4ab + 8b + 3a + 6 4ab + 3a + 8b + 6 (4ab + 3a) + (8b + 6) a(4b + 3) + 2(4b + 3)

(4b + 3)(a + 2)

Additive Inverse Property

• Group with common factors.• Use inverse property to match up the

factors.Ex) 35x – 5xy + 3y – 21

Additive Inverse Property

• Group with common factors.• Use inverse property to match up the

factors.Ex) 35x – 5xy + 3y – 21 (35x – 5xy) + (3y – 21) 5x(7 – y) + 3(y – 7) inverse property

5x(-1)(y – 7) + 3(y – 7) (7 – y) = -1(y – 7)

-5x(y – 7) + 3(y – 7) (-5x + 3)(y – 7)

Factoring Trinomials When a=1

• ALWAYS check for GCF first!• Factor trinomials in the standard

form ax²+ bx + c• Solve equations in the standard form ax²+ bx + c = 0

Factoring when b and c are positive

x²+ 6x + 8• factors(Multiply) sum(Add) 1, 8 9 2, 4 6

• 2 and 4 multiply to give you 8 and add together to give you 6.

• Answer: (x+2)(x+4)• Check using FOIL

Factoring when b is negative and c is positive

• Both factors need to be negative to have a positive product and a negative sum.

x²- 10x + 16



x²- 10x + 16 M A -1,-16 -17 -2,-8 -10 -4,-4 -8



x²- 10x + 16 M A . -1,-16 -17 -2,-8 -10 -4,-4 -8

Answer: (x-2)(x-8)

Factoring when b is positive and c is negative

• One factor has to be positive and one has to be negative to get a negative product. x²+ x – 12



M A 1,-12 -11 -1, 12 11 2, -6 -4 -2, 6 4 3,-4 -1 -3, 4 1



M A 1,-12 -11 -1, 12 11 2, -6 -4 -2, 6 4 3,-4 -1 -3, 4 1 Answer: (x-3)(x+4)

Factoring when b is negative and c is negative

• One factor has to be positive and one has to be negative to get a negative product.

x²-7x – 18



x²-7x – 18 M A 1,-18 -17 -1, 18 17 2,-9 -7 -2, 9 7 3,-6 -3 -3, 6 3



x²-7x – 18 M A 1,-18 -17 -1, 18 17 2,-9 -7 -2, 9 7 3,-6 -3 -3, 6 3 Answer: (x+2)(X-9)

Now you try…

3x² + 24x + 45

Now you try…

3x² + 24x + 45 3(x²+ 8x + 15) GCF 3(x + 3)(x + 5) final answer

Factoring Trinomials when a>1

• Multiply a and c.• Need to find two numbers where the

product is equal to a∙c (30) and the sum is equal to b (17).

6x²+ 17x + 5


• Multiply a and c.• Need to find two numbers where the product

is equal to a∙c (30) and the sum is equal to b (17).

6x²+ 17x + 5M A1, 30 312, 15 173, 10 135, 6 11


2, 15 product = 30, sum = 17 6x²+ 17x + 5• Re write the first and last terms. 6x² + 5• Fill in the middle with the two numbers you found,

followed by the variable.

6x²+ 2x + 15x + 5• Now factor by grouping.


(6x²+ 2x) + (15x + 5) group 2x(3x + 1) + 5(3x + 1) GCF (3x + 1)(2x + 5) final answer

***check by using FOIL

Now you try…

10x²- 43x + 28

Now you try…

10x²- 43x + 28 280 -43 M A -2,-140 -142 -4,-70 -74 -8,-35 -43 -10,-28 -38 -14,-20 -34

Now you try…

10x²- 43x + 28

(10x²-8x) + (-35x + 28)

Now you try…

10x²- 43x + 28

(10x²-8x) + (-35x + 28) 2x(5x – 4) + 7(-5x + 4)

Now you try…

10x²- 43x + 28

(10x²-8x) + (-35x + 28) 2x(5x – 4) + 7(-5x + 4) 2x(5x – 4) + (-1)(7(5x – 4))

Now you try…

10x²- 43x + 28

(10x²-8x) + (-35x + 28) 2x(5x – 4) + 7(-5x + 4) 2x(5x – 4) + 7(-1)(5x – 4) 2x(5x – 4) + (-7)(5x – 4) (5x – 4)(2x – 7) final answer

Prime Polynomials

• Some polynomials cannot be factored in any way and they are considered prime.

Ex) 2x²+ 5x – 2• No GCF• No grouping• No factors that equal -4, will also equal

the sum of 5• Prime

Perfect Square

• In factored form, if both factors are the same, write it as one factor squared.

Ex) x²+ 10x + 25 (x + 5)(x + 5) (x + 5)²

Factoring Difference of Squares

• This only works with binomials that are being subtracted (difference of squares).

• Find the square root of the first term and the second term.

• One factor is the sum of the square roots and the other factor is the difference of the square roots.

a²- b² (a + b)(a – b)

Now You Try…

1) n²- 25

2) 36x²- 49y²

3) 48a³- 12a

4) 9x²+ 1

Now You Try…

1) n²- 25 (n+5)(n-5)

2) 36x²- 49y² (6x+7y)(6x-7y)

3) 48a³- 12a 12a(4a²-1) 12a(2a+1)(2a-1)

4) 9x²+ 1 prime – can’t be factored

Solve Equations by Factoring

• Set the equation equal to zero• Factor• Set each separate factor equal to

zero• Solve for the variable

Example

x²+ 5x = 6 -6 -6 x²+ 5x – 6 = 0M A 1,-6 -5-1, 6 5 2,-3 -1-2, 3 1

Example

x²+ 5x - 6 = 0_M A 1,-6 -5-1, 6 5 2,-3 -1 (x-1)(x+6) = 0 -2, 3 1 x - 1 = 0 x + 6 = 0 +1 +1 -6 -6 x = 1 x = -6Solution Set {1,-6}***Check by substituting in original equation

Solve an Equation in Factored Form

• Equation will be set equal to zero.• Once factored, set each factor to zero and

solve for the variable.Ex) (d – 5)(3d + 4) = 0 d – 5 = 0 3d + 4 = 0 + 5 + 5 -4 -4 d = 5 3d = -4 3 3 d = -4/3Solution set {5,-4/3}

Practice Problems:

Ex) 8a²- 9a – 5 = 4 – 3a

Practice Problems:

Ex) 8a²- 9a – 5 = 4 – 3a

(4a + 3)(2a – 3) = 0

Solution set {-¾, 1½}

More practice…

18x³= 50x

More practice…

18x³= 50x18x³-50x = 02x(9x²-25) 2x(3x + 5)(3x – 5)2x = 0 3x + 5 = 0 3x – 5 = 0 x = 0 x = -1⅔ x = 1⅔

Solution set {-1⅔, 0, 1⅔}

Check Equations

• Substitute solution set back into the factored equation.

• Solution set {5, } (d – 5)(3d + 4) = 0 (d – 5)(3d + 4) = 0 (5 – 5)(3 ∙ 5 + 4) = 0 ( - 5)(3∙ + 4) = 0 (0)(15 + 4) = 0 (-19/3)(0) = 0 0 = 0 0 = 0

Special Cases

• Some equations have to be set equal to zero.

Ex) x² = 7x (subtract 7x from both sides)

x²- 7x = 0 (factor using GCF)

x(x – 7) = 0 x = 0 x – 7 = 0 x = 7**Solution set {0, 7}

Multi Step Factoring

• Some problems will require more than one method to completely factor the polynomial.

Ex)

5x³+ 15x²- 5x – 15 original problem


5x³+ 15x²- 5x – 15 original problem5(x³+ 3x²-1x – 3) find GCF


5x³+ 15x²- 5x – 15 original problem5(x³+ 3x²-1x – 3) find GCF5[(x³+ 3x²) + (-1x – 3)] group


5x³+ 15x²- 5x – 15 original problem5(x³+ 3x²-1x – 3) find GCF5[(x³+ 3x²) + (-1x – 3)] group5[x²(x + 3) + -1(x + 3)] factor GCF


5x³+ 15x²- 5x – 15 original problem5(x³+ 3x²-1x – 3) find GCF5[(x³+ 3x²) + (-1x – 3)] group5[x²(x + 3) + -1(x + 3)] factor GCF5(x + 3)(x²-1) perfect square5(x + 3)(x + 1)(x – 1) final answer


5x³+ 15x²- 5x – 15 original problem5(x³+ 3x²-1x – 3) find GCF5[(x³+ 3x²) + (-1x – 3)] group5[x²(x + 3) + -1(x + 3)] factor GCF5(x + 3)(x²-1) perfect square

Factoring Polynomials Algebra I. Vocabulary Factors – The numbers used to find a product. Prime...

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