 # Factoring Polynomials

## Presentation on theme: "Factoring Polynomials"— Presentation transcript:

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:

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) ∙ 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.

Factoring Using the (Reverse) Distributive Property
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

Factoring Using the (Reverse) Distributive Property
First step is to find the prime factors of each term. 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

Factoring Using the (Reverse) Distributive Property
First step is to find the prime factors of each term. Next step is to find the GCF of the terms in the polynomial. 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)

Factoring Using the (Reverse) Distributive Property
First step is to find the prime factors of each term. Next step is to find the GCF of the terms in the polynomial. 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 · d 12c²d = 2 · 2 · 3 · c · c · d 9cd = 3 · 3 · c · d GCF = 3cd Answer: 3cd(6d + 4c + 3)

FOIL Review x²+ -1x + -6 Using FOIL: First Outer (x + 2)(x – 3) Inner
Last 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)

Group with common factors. Use inverse property to match up the factors. Ex) 35x – 5xy + 3y – 21

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, 2, 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

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 M A -1, -2, -4,

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 M A . -1, -2, -4, 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

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, -1, 2, -2, 3, -3,

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, -1, 2, -2, 3, -3, 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

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 M A 1, -1, 2, -2, 3, -3,

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 M A 1, -1, 2, -2, 3, -3, Answer: (x+2)(X-9)

Now you try… 3x² + 24x + 45

Now you try… 3x² + 24x (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

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 M A 1, 2, 3, 5,

Factoring Trinomials when a>1
2, 15 product = 30, sum = 17 6x²+ 17x + 5 Re write the first and last terms. 6x² Fill in the middle with the two numbers you found, followed by the variable. 6x²+ 2x + 15x + 5 Now factor by grouping.

Factoring Trinomials when a>1
(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 M A -2, , , , ,

Now you try… 10x²- 43x M A -2, , , , ,

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… n²- 25 36x²- 49y² 48a³- 12a 9x²+ 1

Now You Try… n²- 25 (n+5)(n-5) 36x²- 49y² (6x+7y)(6x-7y) 48a³- 12a
12a(4a²-1) 12a(2a+1)(2a-1) 4) 9x² 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 = x²+ 5x – 6 = 0 M A 1, , 6 5 2, , 3 1

Example x²+ 5x - 6 = 0 _M A 1, , 6 5 2,-3 -1 (x-1)(x+6) = 0 -2, 3 1 x - 1 = 0 x + 6 = x = 1 x = -6 Solution 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 = d + 4 = 0 d = d = -4 d = -4/3 Solution 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= 50x 18x³-50x = 0 2x(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) = (d – 5)(3d + 4) = 0 (5 – 5)(3 ∙ 5 + 4) = 0 ( - 5)(3∙ + 4) = 0 (0)(15 + 4) = (-19/3)(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 = 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 – original problem

Multi Step Factoring 5x³+ 15x²- 5x – 15 original problem 5(x³+ 3x²-1x – 3) find GCF

Multi Step Factoring 5x³+ 15x²- 5x – 15 original problem 5(x³+ 3x²-1x – 3) find GCF 5[(x³+ 3x²) + (-1x – 3)] group

Multi Step Factoring 5x³+ 15x²- 5x – 15 original problem 5(x³+ 3x²-1x – 3) find GCF 5[(x³+ 3x²) + (-1x – 3)] group 5[x²(x + 3) + -1(x + 3)] factor GCF

Multi Step Factoring 5x³+ 15x²- 5x – 15 original problem 5(x³+ 3x²-1x – 3) find GCF 5[(x³+ 3x²) + (-1x – 3)] group 5[x²(x + 3) + -1(x + 3)] factor GCF 5(x + 3)(x²-1) perfect square 5(x + 3)(x + 1)(x – 1) final answer

Multi Step Factoring 5x³+ 15x²- 5x – 15 original problem 5(x³+ 3x²-1x – 3) find GCF 5[(x³+ 3x²) + (-1x – 3)] group 5[x²(x + 3) + -1(x + 3)] factor GCF 5(x + 3)(x²-1) perfect square