 # 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.

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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 + 3x – 3 2x + 1 5y 3 + 2y 2 - 4y - 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 2xy 3y 3 x 4

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

Types of Polynomials Monomial Has 1 term 3ab 2y 3 xy Binomial Has 2 terms combined with addition or subtraction ax 2 + bx2x + 3 Trinomial Has 3 terms combined with addition or subtraction ax 2 + bx + c2x + 3x +1

Multiplying Polynomials Binomial x Binomial – FOIL Method F irst O uter I nner L ast F – x * xO – x * 2 I – 1 * x L – 1 * 2

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

Chapter 6 – Factoring Polynomials and Solving Equations 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 We can use the distributive property to pull the common factor out of each term. – x(x – 1)

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

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

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

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

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

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

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

6.1 Introduction To Factoring Factoring by Grouping  What is common between the following terms? 5x(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)(5x + 6)

6.1 Introduction To Factoring Identify the common factors  x 2 (2x – 5) – 4x(2x – 5)  5x(3x – 2) + 2(3x – 2)  (z + 5)z + (z + 5)4

6.1 Introduction To Factoring Identify the common factors  x 2 (2x – 5) – 4x(2x – 5) 2x – 5  5x(3x – 2) + 2(3x – 2) 3x – 2  (z + 5)z + (z + 5)4 z + 5

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

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

6.1 Introduction To Factoring Factor by Grouping: 3x + 3y + ax + ay  (3x + 3y) + (ax + ay)  3(x + y) + a(x + y)  (x + y) (3 + a)  Check by using FOIL: (x + y) (3 + a) 3x + ay + 3y + ay

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

3x + 3y + 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  6x 3 – 12x 2 – 3x + 6  2x 5 – 8x 4 + 6x 3 – 24x 2  4t 3 – 12t 2 + 3t – 9 Check by multiplying the factors

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

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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 (x + m) (x + n) = x 2 O (x + m) ( x + n) = nx

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

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

FOIL F (x + m) (x + n) = x 2 O (x + m) ( x + n) = nx I (x + m) (x + n) = mx L (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 + 7x + 12 Step 1: Factor x 2 x 2 + 7x + 12 (x )(x )

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

x 2 + 7x + 12 Step 3: Identify the factors of 12 that add to 7  1 * 12 = 12 1 + 12 = 13  2 * 6 = 122 + 6 = 8  3 * 4 = 123 + 4 = 7

x 2 + 7x + 12 Step 3: Identify the factors of 12 that add to 7  1 * 12 = 12 1 + 12 = 13  2 * 6 = 122 + 6 = 8  3 * 4 = 123 + 4 = 7

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

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

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

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

x 2 + 13x + 30 Step 3: Identify the factors of 30 that add to 12 1 * 30 = 301 + 30 = 31 2 * 15 = 302 + 15 = 17 3 * 10 = 303 + 10 = 13 6 * 5 = 306 + 5 = 11

x 2 + 13x + 30 Step 3: Identify the factors of 30 that add to 12 1 * 30 = 301 + 30 = 31 2 * 15 = 302 + 15 = 17 3 * 10 = 303 + 10 = 13 6 * 5 = 306 + 5 = 11

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

x 2 + 9x + 12 Step 3: Identify the factors of 12 that add to 9 1 * 12 = 121 + 12 = 13 2 * 6 = 122 + 6 = 8 3 * 4 = 123 + 4 = 7

x 2 + 9x + 12 Step 3: Identify the factors of 12 that add to 9 1 * 12 = 121 + 12 = 13 2 * 6 = 122 + 6 = 8 3 * 4 = 123 + 4 = 7

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

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

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

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

7x 2 + 35x + 42 Step 4: Identify the factors of 6 that add to 5 1 * 6 = 61 + 6 = 7 2 * 3 = 62 + 3 = 5

7x 2 + 35x + 42 Step 4: Identify the factors of 6 that add to 5 1 * 6 = 61 + 6 = 7 2 * 3 = 62 + 3 = 5

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

7x 2 + 35x + 42 Step 6: Check by FOIL 7(x + 2)(x + 3) 7(x 2 + 3x + 2x + 6) 7(x 2 + 5x + 6) 7x 2 + 35x + 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 + 9z + 20 t 2 – 2t – 24 x 2 + 5x – 4 2x 4 – 4x 3 – 6x 2

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

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6.3 and 6.4 next class

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