Polynomial Review What is a polynomial An algebraic

Polynomial Review What is a polynomial? An algebraic expression consisting of one or more summed terms, each term consisting of a coefficient and one or more variables raised to natural number exponent. Examples: X 2 + 3 x – 3 2 x + 1 5 y 3 + 2 y 2 - 4 y - 8

Terms What is a term? A term is the product of a coefficient and one or more variables raised to a natural number exponent Examples: X 2 xy 3 y 3 x 4

Terms cont. Identify the coefficient and degree for each example X Coefficient – 1 Degree – 1 Coefficient – 2 Degree – 2 Coefficient – 3 Degree – 4 Coefficient – 4 Degree - 0 2 xy 3 y 3 x 4

Types of Polynomials Monomial Has 1 term 3 ab 2 y 3 xy Binomial Has 2 terms combined with addition or subtraction ax 2 + bx 2 x + 3 Trinomial Has 3 terms combined with addition or subtraction ax 2 + bx + c 2 x + 3 x +1

Multiplying Polynomials Binomial x Binomial – FOIL Method First Outer Inner Last F–x*x O–x*2 I– 1*x L– 1*2

FOIL cont. F: O: I: L: Add the terms: x 2 + 2 x + 2 Combine like terms: x 2 + 3 x + 2 x * x = x 2 x * 2 = 2 x 1*x=x 1*2=2

Chapter 6 – Factoring Polynomials and Solving Equations 6. 1 Introduction To Factoring

6. 1 Introduction To Factoring ● Factors – 2 or more numbers that multiply to a new number Factors of 4: 1, 4 and 2, 2 1 * 4 = 4 and 2 * 2 = 4 Factors of 32: 1, 32 2, 16 4, 8 • 1 * 32 = 32 2 * 16 = 32 and 4 * 8 = 32

6. 1 Introduction To Factoring • Factors of a polynomial – 2 or more polynomials that multiply to a higher degree polynomial x 2 – x x 2 and x have a common factor of x – – x 2 x = x*x = 1*x – x(x – 1) We can use the distributive property to pull the common factor out of each term.

6. 1 Introduction To Factoring • Find the common factors of the following terms: 8 x 2 + 6 x • • 8 x 2 = 2 x * 4 x 6 x = 2 x * 3 The common factor is 2 x Pull out the common factor using the distributive property. • 2 x(4 x + 3)

6. 1 Introduction To Factoring 10 x + 6 Step 1: Find the common factors for the terms • Common factor: 2 Step 2: Pull the common factor out from each term • • 10 x = 2 * 5 x 6 =2*3 Step 3: Multiply the common factor by the sum of the other factors • 2(5 x + 3)

6. 1 Introduction To Factoring 6 x 2 + 15 x Step 1: Find the common factors for the terms • Common factor: 3 x Step 2: Pull the common factor out from each term • • 6 x 2 = 3 x * 2 x 15 x = 3 x * 5 Step 3: Multiply the common factor by the sum of the other factors • 3 x(2 x + 5)

6. 1 Introduction To Factoring • Factor the polynomial: 3 z 3 + 9 z 2 – 6 z Find the common factor: 3 z Pull out the common factor: z 2 +3 z – 2 Write the polynomial as the product of the factors • 3 z (z 2 + 3 z – 2)

6. 1 Introduction To Factoring • Factor the polynomial: 2 x 2 y 2 + 4 xy 3 Find the common factor: 2 xy 2 Pull out the common factor: x + 2 y Write the polynomial as the product of the factors: • 2 xy 2(x + 2 y)

6. 1 Introduction To Factoring • Find the GCF and write each polynomial as the product of the factors. 9 x 2 + 6 x 8 a 2 b 3 – 16 a 3 b 2 66 t + 16 t 2

6. 1 Introduction To Factoring • Find the GCF and write each polynomial as the product of the factors. 9 x 2 + 6 x • 3 x (3 x + 2) 8 a 2 b 3 – 16 a 3 b 2 • 8 a 2 b 2(b - 2 a) 66 t + 16 t 2 • 2 t(33 + 8 t)

6. 1 Introduction To Factoring • Factoring by Grouping What is common between the following terms? 5 x(x + 3) + 6(x+ 3) The common factor is the binomial (x + 3) We can rewrite the polynomial as the product of the common factor and the sum of the other factors: (x + 3)(5 x + 6)

6. 1 Introduction To Factoring • Identify the common factors x 2(2 x – 5) – 4 x(2 x – 5) 5 x(3 x – 2) + 2(3 x – 2) (z + 5)z + (z + 5)4

6. 1 Introduction To Factoring • Identify the common factors x 2(2 x – 5) – 4 x(2 x – 5) • 2 x – 5 5 x(3 x – 2) + 2(3 x – 2) • 3 x – 2 (z + 5)z + (z + 5)4 • z+5

6. 1 Introduction To Factoring • Factor by grouping: 2 x 3 – 4 x 2 + 3 x – 6 Step 1: Group the terms that have common factors (2 x 3 – 4 x 2) + (3 x – 6) Step 2: Identify the common factors for each group • • Common factor of 2 x 3 – 4 x 2 : 2 x 2 Common factor of 3 x – 6: 3

2 x 3 – 4 x 2 + 3 x – 6 cont. Step 3: write each grouping as the product of the factors • 2 x 2(x – 2) + 3(x – 2) Note: the parenthesis are the same Step 4: Distribute the common factor from each grouping • 2 x 2(x – 2) + 3(x – 2) • (x – 2)(2 x 2 + 3)

6. 1 Introduction To Factoring • Factor by Grouping: 3 x + 3 y + ax + ay (3 x + 3 y) + (ax + ay) 3(x + y) + a(x + y) (3 + a) Check by using FOIL: (x + y) (3 + a) • 3 x + ay + 3 y + ay

3 x + 3 y + ax + ay cont. • Let's Factor this polynomial again by grouping the x terms and y terms 3 x + 3 y + ax + ay = 3 x + ax + 3 y + ay (3 x + ax) + (3 y + ay) x(3 + a) + y(3 + a) (x + y)

3 x + 3 y + ax + ay cont. • Compare Grouping 1 and Grouping 2 Grouping 1: (x + y) (3 + a) Grouping 2: (3 + a) (x + y) • Are these the same?

Reflect • Does it matter how you group the terms? No, you will get the same answer However, you need to group terms that have a common factor other than 1.

Reflect • Do the parenthesis have to be the same for each term after you have factored the GCF? Yes • How many terms do you need to factor by grouping? At least 4

Practice • Factor the following Polynomials 6 x 3 – 12 x 2 – 3 x + 6 2 x 5 – 8 x 4 + 6 x 3 – 24 x 2 4 t 3 – 12 t 2 + 3 t – 9 • Check by multiplying the factors

Practice cont. • Answers: 3(2 x 2 – 1)(x – 2) 2 x 2 (x 2 + 3)(x – 4) (4 t 2 + 3)(t – 3)

? ? Questions ? ?

6. 2 Factoring Trinomials – 2 x + bx + c

6. 2 Factoring Trinomials – x 2 + bx + c • When factoring a polynomial in the form of x 2 + bx + c, we will be reversing the FOIL process • So, let's review FOIL

FOIL • F (x + m) (x + n) = x 2

FOIL • • F O I L (x + m) (x + n) = x 2 (x + m) ( x + n) = nx (x + m) (x + n) = mx (x + m) (x + n) = nm x 2 + nx + mx + nm

FOIL • Can we combine like terms? Yes, O and I in FOIL give us like terms • x 2 + nx + mx + nm What is the common factor? • X Pull out the x: • x 2 + (n + m)x + nm

Factoring Let's • Standard Form: x 2 + bx + c Compare • (x +n)(x + m): x 2 + (n + m)x + nm

Factoring • Based on the Standard Form: x 2 + bx + c, what is the b in the polynomial we found: x 2 + (n + m)x +nm b = (n + m)

Factoring • Based on the Standard Form: x 2 + bx + c, what is the c in the polynomial we found: x 2 + (n + m)x +nm c = nm

6. 2 Factoring Trinomials – x 2 + bx + c • To factor the trinomial x 2 + bx + c, find two numbers m and n that satisfy the following conditions: m*n=c m+n=b Such that x 2 + bx + c = (x + m)(x + n)

6. 2 Factoring Trinomials – x 2 + bx + c • Remember that (m + n)x was the sum of the O and I in FOIL and nm was the L in FOIL.

x 2 + 7 x + 12 • Step 1: Factor x 2 • x 2 + 7 x + 12 • (x )

x 2 + 7 x + 12 • Step 2: Identify all of the factors of 12 1 * 12 = 12 2 * 6 = 12 3 * 4 = 12

x 2 + 7 x + 12 • Step 3: Identify the factors of 12 that add to 7 1 * 12 = 12 2 * 6 = 12 3 * 4 = 12 1 + 12 = 13 2+6=8 3+4=7

x 2 + 7 x + 12 • Step 4: Complete the factors (x + 3)(x + 4)

x 2 + 7 x + 12 • Step 5: Check by FOIL (x + 3)(x + 4) • x 2 + 4 x + 3 x + 12 • x 2 + 7 x +12

x 2 + 13 x + 30 • Step 1: Factor x 2 • x 2 + 13 x + 30 • (x )

x 2 + 13 x + 30 • Step 2: Find all factors of 30 • • 1 * 30 = 30 2 * 15 = 30 3 * 10 = 30 6 * 5 = 30

x 2 + 13 x + 30 • Step 3: Identify the factors of 30 that add to 12 • • 1 * 30 = 30 2 * 15 = 30 3 * 10 = 30 6 * 5 = 30 1 + 30 = 31 2 + 15 = 17 3 + 10 = 13 6 + 5 = 11

x 2 + 13 x + 30 • Step 4: Complete the factors (x + 3)(x + 10)

x 2 + 13 x + 30 • Step 5: Check by FOIL (x + 3)(x + 10) • x 2 + 10 x + 30 • x 2 + 13 x + 30

z 2 – 8 z + 15 • Step 1: Factor z 2 • z 2 – 8 z + 20 • (z )

z 2 – 8 z + 15 • Step 2: Find all factors of 15 • • 1 * 15 = 15 -1 * -15 = 15 3 * 5 = 15 -3 * -5 = 15

z 2 – 8 z + 15 • Step 3: Identify the factors of 15 that add to -8 • • 1 * 15 = 15 -1 * -15 = 15 3 * 5 = 15 -3 * -5 = 15 1 + 15 = 16 -1 + -15 = -16 3+ 5=8 -3 + -5 = -8

z 2 – 8 z + 15 • Step 4: Complete the factors (z – 3)(z – 5)

z 2 – 8 z + 15 • Step 5: Check by FOIL (z – 3)(z – 5) • z 2 – 3 z – 5 z + 15 • z 2 – 8 z + 15

y 2 – 3 y – 4 • Step 1: Factor y 2 • y 2 – 3 y – 4 • (y )

y 2 – 3 y – 4 • Step 2: Find all factors of -4 • 1 * -4 = -4 • -1 * 4 = -4 • -2 * 2 = -4

y 2 – 3 y – 4 • Step 3: Identify the factors of -4 that add to -3 • 1 * -4 = -4 • -1 * 4 = -4 • -2 * 2 = -4 1 + -4 = -3 -1 + 4 = 3 -2 + 2 = 0

y 2 – 3 y – 4 • Step 4: Complete the factors (y + 1)(y – 4)

y 2 – 3 y – 4 • Step 5: Check by FOIL (y + 1)(y – 4) • y 2 – 4 y + 1 y – 4 • y 2 – 3 y – 4

x 2 + 9 x + 12 • Step 1: Factor x 2 • x 2 + 9 x + 12 • (x )

x 2 + 9 x + 12 • Step 2: Find all factors of 12 • 1 * 12 = 12 • 2 * 6 = 12 • 3 * 4 = 12

x 2 + 9 x + 12 • Step 3: Identify the factors of 12 that add to 9 • 1 * 12 = 12 • 2 * 6 = 12 • 3 * 4 = 12 1 + 12 = 13 2+6 =8 3+4 =7

x 2 + 9 x + 12 • No Factors of 12 will add to 9 • What do you think this means? The Trinomial, x 2 + 9 x + 12, is prime • The only factors are itself and 1

7 x 2 + 35 x + 42 • Step 1: Factor 7 from all 3 terms • 7*x 2 + 7*5 x + 7 *6 • 7(x 2 + 5 x + 6)

7 x 2 + 35 x + 42 • Step 2: Factor x 2 • 7(x 2 + 5 x + 6) • 7(x )

7 x 2 + 35 x + 42 • Step 3: Find all factors of 6 • 1*6=6 • 2*3=6

7 x 2 + 35 x + 42 • Step 4: Identify the factors of 6 that add to 5 • 1*6=6 • 2*3=6 1+6=7 2+3=5

7 x 2 + 35 x + 42 • Step 5: Complete the factors 7(x + 2)(x + 3)

7 x 2 + 35 x + 42 • Step 6: Check by FOIL 7(x + 2)(x + 3) • 7(x 2 + 3 x + 2 x + 6) • 7(x 2 + 5 x + 6) • 7 x 2 + 35 x + 42

Reflect • Is it possible to factor every trinomial? No, if a trinomial is prime it cannot be factored • How do you know if a trinomial is prime? If you cannot find 2 numbers that multiply to nm and add to (m + n)

Practice • z 2 + 9 z + 20 • t 2 – 2 t – 24 • x 2 + 5 x – 4 • 2 x 4 – 4 x 3 – 6 x 2

Practice Answers • z 2 + 9 z + 20 (z + 4)(z + 5) • t 2 – 2 t – 24 (t – 6)(t + 4) • x 2 + 5 x – 4 No factors – Trinomial is prime • 2 x 4 – 4 x 3 – 6 x 2 2 x 2(x – 3)(x + 1)

? ? Questions ? ?

6. 3 and 6. 4 next class