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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

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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

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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

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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

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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

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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

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Chapter 6 – Factoring Polynomials and Solving Equations 6.1 Introduction To Factoring

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● 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

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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)

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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)

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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)

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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)

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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)

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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)

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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

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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)

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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)

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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

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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

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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

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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)

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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

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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)

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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?

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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.

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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

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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

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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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???? Questions ????

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6.2 Factoring Trinomials – x 2 + bx + c

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

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FOIL F (x + m) (x + n) = x 2

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

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

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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

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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

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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

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

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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)

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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

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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)

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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.

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

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

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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

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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

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

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

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

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

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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

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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

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

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

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

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

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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

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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

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

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

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

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

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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

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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

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

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

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

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

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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

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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

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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

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

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

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

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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

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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

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

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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

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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)

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

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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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???? Questions ????

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6.3 and 6.4 next class

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